这题目有3点需要注意：1.当分数为3/1时要输出3；2.当连分数为[3;]时要输出[3]；3.在分数转成连分数的时候不能使用浮点数，精度会出现问题，必须要用整数处理

#include <bits/stdc++.h>
using namespace std;

int gcd(int x,int y)//求最大公约数
{
    if(x<y) swap(x,y);
    int r=x%y;
    while(r){
        x=y;
        y=r;
        r=x%y;
    }
    return y;
}
int main()
{
    //freopen("input.txt","r",stdin);
    string s;
    char c;
    while(getline(cin,s)){
        if(!isdigit(s[0])){//第一个不是数字即连分数，否则是分数
            int d;
            stack<int>ans;
            stringstream ss(s);
            ss>>c;
            while(ss>>d>>c) ans.push(d);//将连分数中的数字进栈
            int fz=1,fm=0;
            while(!ans.empty()){//将数字依次处理转换成分数
                int x=ans.top();
                ans.pop();
                swap(fz,fm);
                fz+=x*fm;
            }
            int gg=gcd(fz,fm);
            fz/=gg;fm/=gg;//约分为最简
            if(fm==1) printf("%d\n",fz);//考虑分母为1的情况，否则第4个点会报错
            else printf("%d/%d\n",fz,fm);
        }
        else{
            int fz,fm;
            queue<int>ans;
            stringstream ss(s);
            ss>>fz>>c>>fm;//提取分子和分母
            while(1){
                /*
                int tt=(int)t;
                ans.push(tt);
                t=t-tt;
                if(fabs(t)<=1e-6) break;
                t=1/t;
                */
                //在转换成连分数的时候不能用浮点数而要用整数，否则精度出现问题！！
                int t=fz/fm;
                ans.push(t);
                fz=fz-t*fm;
                if(fz==0) break;
                swap(fz,fm);
            }
            if(ans.size()==1) {printf("[%d]",ans.front());continue;}//注意格式问题
            printf("[%d;",ans.front());
            ans.pop();
            while(ans.size()!=1){
                printf("%d,",ans.front());
                ans.pop();
            }
            printf("%d]\n",ans.front());
        }
    }
    return 0;
}

基于连分数的结构,可以采用链表的结构进行储存和处理

连分数

上图连分数即是[292;1,15,7,3]
分解为[192;[1;[15;[7;[3;NULL]]]]]

C
若只是理解链表,看test()函数之上即可

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Ctnfrct {
    int base;
    struct Ctnfrct *next;
} ctnfrct;
typedef struct Frct {
    int up;
    int down;
} frct;

void swap(int *a, int *b) {
    int t = *a;
    *a = *b;
    *b = t;
}

void c2f(ctnfrct *ctn, frct *frc) {
    if (ctn->next) {
        c2f(ctn->next, frc);
        swap(&frc->down, &frc->up);
        frc->up = frc->down * ctn->base + frc->up;
    } else {
        frc->up = ctn->base;
        frc->down = 1;
    }
}

ctnfrct *f2c(frct frc) {
    ctnfrct *curr = (ctnfrct *) malloc(sizeof(ctnfrct));
    if (!frc.down) return (ctnfrct *) 0;
    if (frc.up != 1) {
        curr->base = frc.up / frc.down;
        frc.up %= frc.down;
        swap(&frc.up, &frc.down);
        curr->next = f2c(frc);
    } else {
        curr->base = frc.down;
        curr->next = (ctnfrct *) 0;
    }
    return curr;
}

//此函数与题无关,仅方便理解
void test() {
    ctnfrct a = {3, (ctnfrct *) 0};
    ctnfrct b = {7, &a}, c = {15, &b}, d = {1, &c}, e = {292, &d};
    frct x;
    c2f(&e, &x);

    printf("%d/%d\n", x.up, x.down);
    x.up *= 2;
    x.down *= 2;
    ctnfrct *n = f2c(x);
    while (n) {
        printf("%d ", n->base);
        n = n->next;
    }
}

int main() {
    char str[1000], num_s[10];
    int num;
    while (scanf("%s", str) != EOF) {
        if (str[0] == '[') {
            ctnfrct head, *p = &head;
            int cnt = 0;
            for (int i = 1; i < strlen(str); i++) {
                if (str[i] < '0' || str[i] > '9')

                {
                    sscanf(num_s,"%d",&num);
                    ctnfrct *curr = (ctnfrct *) malloc(sizeof(ctnfrct));

                    curr->base = num;
                    curr->next = (ctnfrct *) 0;
                    p->next = curr;
                    p = p->next;
                    memset(num_s,0,sizeof(num_s));
                    cnt = 0;
                } else {
                    num_s[cnt++] = str[i];
                }
            }
            frct frc;
            c2f(head.next, &frc);
            if (frc.down != 1) printf("%d/%d\n", frc.up, frc.down);
            else printf("%d\n", frc.up);
        } else {
            frct frc;
            frc.down = 1;
            sscanf(str,"%d/%d",&frc.up,&frc.down);
            ctnfrct * ctn = f2c(frc);
            printf("[%d",ctn->base);
            if(ctn->next){
                printf(";");
                ctn = ctn->next;
                while(ctn->next){
                    printf("%d,",ctn->base);
                    ctn = ctn->next;
                }
                printf("%d]\n",ctn->base);
            } else {
                printf("]\n");
            }

        }
    }
}

主要注意的是将整数位和分数位分开的时候要判断一下是否存在，比如4测试点[2]应该输出2;2/1应该输出[2]而不是[2；]

```#!/usr/bin/env python
# coding:utf-8
"""
Name : main.py
Author  : Tang
"""
def fen2lian(fen: str) -> str:
    lian = '['
    a = int(fen.split('/')[0])
    b = int(fen.split('/')[1])
    lian += str(a // b)
    lian += ';'
    r = a % b
    a, b = b, r
    while r:
        lian += str(a // b)
        lian += ','
        r = a % b
        a, b = b, r
    lian = lian[:-1]
    lian += ']'
    return lian


def lian2fen(lian: str) -> str:
    if ';' in lian:
        zheng = lian.split(';')[0][1:]
    else:
        return str(lian[1:-1])
    try:
        xiao = lian.split(';')[1].split(',')
        xiao[-1] = xiao[-1][:-1]
        b = int(xiao[-1])
        old_r = 1
        a = int(xiao[-2])
        new_r = b
        b = a * b + old_r
        old_r = new_r
        for i in range(len(xiao) - 2):
            try:
                a = int(xiao[-3 - i])
            except IndexError:
                break
            new_r = b
            b = a * b + old_r
            old_r = new_r
    except:
        b = int(lian.split(';')[1][:-1])
        new_r = 1

    A = int(zheng) * b + new_r
    return str(A) + "/" + str(b)


if __name__ == '__main__':
    while True:
        try:
            s = input()
            if s[0] == '[':
                print(lian2fen(s))
            else:
                print(fen2lian(s))
        except EOFError:
            break

python可以考虑用try-except来判断EOF

import re


def to_fraction(values):
    numerator = 1
    denominator = 0
    for x in reversed(values):
        numerator, denominator = denominator, numerator
        numerator += denominator * x
    return "%d/%d" % (numerator, denominator) if denominator > 1 else numerator


def to_continued(numerator, denominator):
    res_str = "[%d;" % (numerator / denominator)
    numerator %= denominator
    if numerator == 0:
        return res_str.replace(";", "]")
    while numerator > 0:
        numerator, denominator = denominator, numerator
        res_str += "%d," % (numerator / denominator)
        numerator %= denominator
    return res_str.rstrip(",") + "]"


while True:
    try:
        line = raw_input()
    except EOFError:
        break
    if line[0] == "[":
        print to_fraction(map(int, re.findall(r'\d+', line)))
    else:
        print to_continued(*map(int, re.findall(r'\d+', line)))

字符串处理+模拟

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 105;
int a[maxn], len, x, y;
char ch[1000];

void swap(int& a,int& b){
    a=a^b, b=b^a, a=a^b;
}
int getin(){
    len=x=y=0;
    memset(a,0,sizeof(a));
    int lent=strlen(ch);
    if( ch[0]=='[' ){
        for ( int i=0; i<lent; i++ ){
            if( isdigit(ch[i]) ){
                a[len]=a[len]*10+ch[i]-'0';
            }else len++;
        }
        len--;
        return 0;
    } else{
        for ( int i=0; i<lent; i++ ){
            if( isdigit(ch[i]) ){
                x=x*10+ch[i]-'0';
            }else swap(x,y);
        }
        swap(x,y);
        return 1;
    }
}
int gcd(int x, int y){
    return y==0?x:gcd(y,x%y);
}
int main(){
    while( scanf("%s", ch)!=EOF ){
        if( getin() ){
            printf("[%d", x/y );
            x=x%y;
            if( x!=0 && y!=0 ) printf(";");
            swap(x,y);
            while( x!=1 && x!=0 && y!=0){
                printf("%d",x/y);
                x=x%y;
                if( x==0 ) break;
                swap(x,y);
                if( x!=1 ) printf(",");
            }
            printf("]\n");
        }else {
            if( len>1 ){
                x=1, y=a[len];
                for ( int i=len-1; i>=1; i-- ){
                    x=y*a[i]+x;
                    if( i!=1 ) swap(x,y);
                }
                int d=gcd(x,y);
                x/=d, y/=d;
                if( y==0 ) printf("%d\n", x);
                else printf("%d/%d\n",x/d,y/d);
            }else printf("%d\n", a[len] ); 
        }
    }
    return 0;
}

果然被恶心到了

。。。。。。真的好恶心的题目

很不讲道理的输入输出

岛儿小过分了点

给出几个猥琐的范例

1/1 输出 【1】

2/1 输出 【2】

1 输出【1】

2 输出【2】

【1】 输出 1

【2】 输出 2

其他的应该没什么问题吧

4/2 输出【2】

就是这样了

惨痛经历啊- -

难点：
1.未知行数
2.约分
3.特殊情况：
n/1: [n]
n: [n]
连分数中数字不一定小于10
然后就是算法：
递归函数处理
分数->连分数：基例-分母为1的时候直接输出分子;
连分数->分数：基例-数组最后只剩分子和分母;
递归链通过简单地推导很容易得出
下面给出代码：

#include<iostream>
#include<string>
#include<string.h>
using namespace std;
void reduction(int& a,int& b) {//约分
    int c = a < b ? a : b;
    int i;
    for (i = 2; i <= c; i++)if (a % i == 0 && b % i == 0) {
        a /= i;
        b /= i;
        reduction(a, b);
        break;
        }
}
void fraction(int num[], int n) {//连分数[数组]->分数
    if (n == 2) {
        reduction(num[0], num[1]);
        if (num[1] == 1)cout << num[0] << endl;
        else cout << num[0] << '/' << num[1] << endl;
    }
    else {
        num[n - 3] = num[n - 3] * num[n - 2] + num[n - 1];
        fraction(num, n - 1);
    }
}
void confrac(int a, int b) {//分数->连分数
    if (b == 1)cout << a << "]" << endl;
    else {
        cout << a / b << ",";
        int t;
        t = a;
        a = b;
        b = t % b;
        confrac(a, b);
    }
}
int main()
{
    char x[100][256] = {'0'};
    for (int i = 0;cin>>x[i]; i++)
    {
        int num[100];
        if (x[i][0] == '[') {//连分数->分数
            int j = 0, k;
            num[0] = 0;
            for (k = 0;; k++)
            {
                num[k] = 0;
                for (; x[i][j] != ';' && x[i][j] != ','; j++) {
                    if (x[i][j] == ']')goto LOOP;
                    if (x[i][j] > 48 && x[i][j] <= 57)\
                        num[k] = 10 * num[k] + x[i][j] - 48;
                }
                j++;
            }
        LOOP:k++;num[k] = 1;k++;
            fraction(num, k);
        }
        else {//分数->连分数
            if (strstr(x[i], "/") == NULL)cout << "[" << atoi(x[i]) << "]" << endl;
            else {
                int a, b, j;
                a = atoi(x[i]);
                string y;
                y = string(x[i]);
                for (j = 0; x[i][j] != '/'; j++);
                y.replace(0, j + 1, "");
                b = stoi(y, 0);
                reduction(a, b);
                if (b == 1)cout << "[" << a << "]" << endl;
                else {
                    cout << "[" << a / b << ";";
                    int t;
                    t = a;
                    a = b;
                    b = t % b;
                    confrac(a, b);
                }
            }
        }
    }
    return 0;
}

#include <iostream>
#include <string>
#include <memory.h>

using namespace std;

int main()
{
    std::ios::sync_with_stdio(false);
    string s;
    while (cin >> s)
    {
        int a[100], t = 0;
        bool flag = false;
        memset(a, 0, sizeof(a));
        if (s[0] == '[')
        {
            for (int i = 1; i < s.size(); i++)
            {
                if (s[i] >= '0'&&s[i] <= '9')
                {
                    flag = true;
                    a[t] = a[t] * 10 + (int)(s[i] - '0');
                }
                else
                {
                    t++;
                    flag = false;
                }
            }
            int x = a[t], y = 1;
            for (int i = t; i >= 0; i--)
            {
                int temp = x;
                x = y + a[i] * x;
                y = temp;
            }
            int max = x, min = y;
            while (max%min != 0)
            {
                int temp = min;
                min = max%min;
                max = temp;
            }
            x = x / min, y = y / min;
            if (y == 1)
                cout << x << endl;
            else
                cout << x << "/" << y << endl;
        }
        else
        {
            for (int i = 0; i < s.size(); i++)
            {
                if (s[i] >= '0'&&s[i] <= '9')
                {
                    flag = true;
                    a[t] = a[t] * 10 + (int)(s[i] - '0');
                }
                else
                    t++;
            }
            if (t == 0)
                cout << '[' << a[0] << ']' << endl;
            else
            {
                int max = a[0], min = a[1];
                while (max%min != 0)
                {
                    int temp = min;
                    min = max%min;
                    max = temp;
                }
                int x = a[0] / min, y = a[1] / min;
                t = 0;
                cout << "[";
                while (x != 1)
                {
                    int div = x / y, rem = x%y;
                    if (t == 0)
                        cout << div;
                    else
                        if (t == 1)
                            cout << ";" << div;
                        else
                            cout << "," << div;
                    x = rem;
                    if (x != 1)
                    {
                        int temp = y;
                        y = x;
                        x = temp;
                    }
                    t++;
                }
                if (y != 0)
                    cout << ',' << y;
                cout << ']' << endl;
            }
        }
    }
    return 0;
}

如果第四个测试数据过不了的话就是因为【4】-4/1；实际上应该是【4】-4；
如果后三个没过的话可，就是你没有约分

注意能整除的情况

#include<iostream>
#include<string.h>
#include<string>
#include<queue>
using namespace std;
queue<int>q;
int gcd(int a,int b){
    if(a<b) swap(a,b);
    return b==0?a:gcd(b,a%b);
}
struct node{
    int x;
    int y;
    node(int a,int b){
        x=a;
        y=b;
    }
};
node operator+(node a,node b){
    int tx=a.x*b.y+b.x*a.y;
    int ty=a.y*b.y;
    int k=gcd(tx,ty);
    node c(tx/k,ty/k);
    return c;
}
node operator+(node a,int b){
    int tx=a.x+b*a.y;
    node c(tx,a.y);
    return c;
}
void cal1(char *a){
    int s=strlen(a),t=0,i=0;
    node ans(0,1);
    for(i=1;a[i]!=';'&&a[i]!=']';i++){
        t*=10;
        t+=a[i]-'0';
    }
    if(a[i]==0){
        cout<<t<<endl;
        return;
    }
    ans.x=t;
    i++;
    int u=i;
    node tn(0,1);
    for(i=s-2;i>=u;i--){
        tn=tn+(a[i]-'0');
        swap(tn.x,tn.y);
        if(a[i-1]==',') i--;
    }
    ans=ans+tn;
    if(ans.y==1) cout<<ans.x<<endl;
    else cout<<ans.x<<'/'<<ans.y<<endl;
}
void cal2(char *a){
    int tx=0,ty=0,i=0,s=strlen(a);
    for(i=0;a[i]!='/'&&a[i]!=0;i++){
        tx*=10;
        tx+=a[i]-'0';
    }
    if(s==i+1){
        cout<<'['<<tx<<']'<<endl;
        return;
    }
    i++;
    for(i;a[i]!=0;i++){
        ty*=10;
        ty+=a[i]-'0';
    }
    int tt=gcd(tx,ty);
    tx=tx/tt;
    ty=ty/tt;
    if(ty==1){
        cout<<'['<<tx<<']'<<endl;
        return;
    }
    cout<<'['<<tx/ty<<';';
    tx=tx%ty;
    swap(tx,ty);
    while(!q.empty()) q.pop();
    while(1){
        tt=gcd(tx,ty);
        tx=tx/tt;
        ty=ty/tt;
        q.push(tx/ty);
        tx=tx%ty;
        swap(tx,ty);
        if(tx%ty==0){
            q.push(tx/ty);
            break;
        }
    }
    while(!q.empty()){
        cout<<q.front();
        q.pop();
        if(!q.empty()) cout<<',';
    }
    cout<<']'<<endl;
}
int main(){
    char a[1000];
    while(cin>>a){
    if(a[0]=='['){
        cal1(a);
    }
    else{
        cal2(a);
    }
    }
}

#include<iostream>//不会压行的蒟蒻
#include<cstring>
#include<algorithm>
using namespace std;
int F[120];
char A[10000];
int main()
{
while(scanf("%s",A) != EOF){
if(A[0] == '['){
int I = 0;
int cursor = 0;
for(int i = 1 ; i < strlen(A) ; i++){
if(A[i] >= '0' && A[i] <= '9'){
I = 10*I + A[i] - '0';
}
if((A[i] == ';' || A[i] == ',' || A[i] == ']') && A[i - 1] >= '0' && A[i - 1] <= '9'){//;是连分数的第一段整数。连分数可能没有后面的分数段
F[cursor++] = I;//F[0]是整数段
I = 0;
}
}
I = F[0];//整数部分
if(cursor == 1)//只有第一段的整数
cout<<F[0]<<endl;
else{
int a = 1,b = F[--cursor];//a/b是分数部分
if(cursor > 1){
while(cursor> 1){
int tmp = b;
b = F[cursor - 1] * b + a;
a = tmp;
cursor--;
}
}
I *= b;

int j = __gcd(I + a, b);//约分
cout<<I+a/j<<"/"<<b/j<<endl;
}
}

else{
int I = 0;
int a = 0, b = 0;
bool flag = false;
for(int i = 0 ; i < strlen(A) ; i++){//a/b字符串对应的分子分母
if(A[i] == '/'){
flag = true;// ‘/’前的分子。分母未必有
if(A[0] == '-')
a = -a;
continue;
}
if(flag == false && A[i] >= '0' && A[i] <= '9')
a = 10*a + A[i] - '0';
else if(flag == true && A[i] >= '0' && A[i] <= '9')
b = 10*b + A[i] - '0';
}
if(flag == true){//分数形式，但可能分子分母整除了
cout<<"["<<a/b;
if(a%b == 0){//整除关系
cout<<"]"<<endl;
continue;
}
else{
cout<<";";//整数部分完成
while(1){
a %= b;
cout<<b/a;
swap(a, b);
if(a%b != 0)
cout<<",";
else{//整除即最后一段
cout<<"]"<<endl;
break;
}
}
}
}
else{//整数形式
cout<<"["<<a<<"]"<<endl;
}
}
}
return 0;
}

借鉴了之前的同学的代码;有一点点的改动([2;] -> [2]),我借鉴的那个代码输出了[2;]实际应该输出[2];
#include<vector>
#include<cstdio>
using namespace std;

int gcd(int x, int y){
int tmp;
tmp = x%y;
while(tmp != 0){
x = y;
y = tmp;
tmp = x%y;
}
return y;
}
struct Frac{
int x, y;
Frac(int a){
x = a;
y = 1;
}
Frac(int a, int b){
if(a != 0){
int m = gcd(a, b);
a /= m;
b /= m;
x = a;
y = b;
}else{
x = 0;
y = 1;
}
}
void operator += (const Frac &r){
int x0 = x*r.y + y*r.x;
int y0 = y*r.y;
int m = gcd(x0, y0);
x0 /= m;
y0 /= m;
x = x0;
y = y0;
}
void filp(){
int m = x;
x = y;
y = m;
}
void print(){
if(y == 1){
printf("%d\n", x);
}else{
printf("%d/%d\n", x, y);
}
}
};
char s[100000];
int main(){
while(scanf("%s", s) == 1){
if(s[0] == '['){
vector<int> a;
int i;
for(i = 1; s[i] != 0; ++i){
if(s[i] >= '0' && s[i] <= '9'){
int x = 0;
while(s[i] >= '0' && s[i] <= '9'){
x = x*10 + s[i] - '0';
++i;
}
--i;
a.push_back(x);
}
}
Frac f(0);
for(int i = a.size() - 1; i >= 0; --i){
f += Frac(a[i]);
if(i){
f.filp();
}
}
f.print();
}else{
int x, y;
vector<int> a;
sscanf(s, "%d/%d", &x, &y);
Frac f(x, y);
while(f.y != 1){
a.push_back(f.x/f.y);
f.x %= f.y;
f.filp();
}
a.push_back(f.x);
if(a.size() == 1){
printf("[%d]\n", a[0]);

}else{
printf("[");
for(int i = 0; i < a.size(); ++i){
printf("%d", a[i]);
if(i == 0){
printf(";");
}else{
if(i != a.size() - 1){
printf(",");
}
}
}
printf("]\n");
}

}
}
return 0;
}

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
long a[1000];
void aha1(int x,int y,int t)
{
    for(int i=t;i>=1;i--)
    {
        long c=x;
        x=y;
        y=c;
        x=x+y*a[i];
    }
    printf("%ld",x);
    if(y!=1)
        printf("/%ld",y);
    printf("\n");
    return;
}
void aha2(int x,int y)
{
    long e=1,k;
    k=x/y;
    printf("[%ld",k);
    x=x-k*y;
    while(x!=0)
    {
        int c=x;
        x=y;
        y=c;
        k=x/y;
        x=x-k*y;
        if(e==1)
            printf(";%ld",k),e=0;
        else
            printf(",%ld",k);
    }
    printf("]\n");
}
int main(){
    char x[1000];
    while(gets(x))
    {
        memset(a, 0, sizeof(a));
        if(x[0]=='[')
        {
            int t=1,q=1;
            while(x[q]!=']')
            {
                if(x[q]>='0'&&x[q]<='9')
                {
                    a[t]=a[t]*10+x[q]-'0';
                }
                else
                    t++;
                q++;
            }
            aha1(a[t],1,t-1);
        }
        else
        {
            int t=1,q=0;
            while(x[q]!='\n'&&x[q]!=EOF&&x[q]!='\0')
            {
                if(x[q]>='0'&&x[q]<='9')
                {
                    a[t]=a[t]*10+x[q]-'0';
                }
                else
                    t++;
                q++;
            }
            int m=a[1],n=a[2];
            aha2(m,n);
        }
    }
    return 0;
}

//可能比较方便理解?
#include <iostream>
#include <string>
#include <cmath>
#include <vector>

using namespace std;

vector<int> f2c(long p, long q) {
    vector<int> ret;
    ret.push_back((int)(p / q));
    if (p%q) {
        vector<int> tmp = f2c(q, p%q);
        ret.insert(ret.end(), tmp.begin(), tmp.end());
    }
    return ret;
}

void c2f(vector<int> c, long *p, long *q) { // 1/p/q
    if (c.size() == 1) {
        *p = 1; *q = c[0];
    } else {
        long np = 0, nq = 0;
        vector<int> tmp;
        tmp.insert(tmp.end(), c.begin() + 1, c.end());
        c2f(tmp, &np, &nq);
        *p = nq;
        *q = nq*c[0] + np;
    }
}

int main(void) {
    string input;
    while (cin >> input) {
        if (input[0] == '[') {
            string pure = input.substr(1, input.length() - 2) + ',';
            for (int i = 0; i < input.length(); i++)
                if (pure[i] == ';')
                    if(pure[i+1]==',')
                        pure.at(i) = ' ';
                    else
                        pure.at(i) = ',';

            vector<int> t;
            size_t pos = pure.find(',');
            size_t size = pure.size();

            while (pos != string::npos) {
                string x = pure.substr(0, pos);
                t.push_back(atoi(x.c_str()));
                pure = pure.substr(pos + 1, size);
                pos = pure.find(',');
            }
            
            long p, q;
            c2f(t, &p, &q);
            //cout << q << '/' << p << endl;
            if (p != 1)
                cout << q << '/' << p << endl;
            else
                cout << q << endl;

            
        } else {
            string s1, s2;
            auto it = input.find('/');
            if (it > input.size()) {
                s1 = input; s2 = "0";
            } else {
                s1 = input.substr(0, it);
                s2 = input.substr(it + 1, input.length() - it);
            }
            long p, q;
            p = atoi(s1.c_str());
            q = atoi(s2.c_str());
            vector<int> C = f2c(p, q);
            cout << "[" << C[0] << ";";
            for (int i = 1; i < C.size(); i++) {
                cout << C[i];
                if (i != C.size() - 1)
                    cout << ",";
            }
            cout << "]" << endl;
        }
    }
    return 0;
}

评测结果
编译成功

Free Pascal Compiler version 3.0.0 [2015/11/16] for i386
Copyright (c) 1993-2015 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
Linking foo.exe
62 lines compiled, 0.0 sec, 28352 bytes code, 1268 bytes data
测试数据 #0: Accepted, time = 0 ms, mem = 1272 KiB, score = 20
测试数据 #1: Accepted, time = 0 ms, mem = 1272 KiB, score = 20
测试数据 #2: Accepted, time = 0 ms, mem = 1308 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 1272 KiB, score = 20
测试数据 #4: Accepted, time = 15 ms, mem = 1236 KiB, score = 20
Accepted, time = 15 ms, mem = 1308 KiB, score = 100
代码
program p1696;
var
a:ansistring;
b:array[1..100000]of longint;
i,n,j,k,num1,num2,x,t:longint;
function pd(n:longint):boolean;
begin
if n in [48..57] then pd:=false else pd:=true;
end;
begin
while not eof do begin
readln(a);
n:=length(a);
if pd(ord(a[1])) then begin
i:=2;x:=0;j:=1;num1:=1;
while i<n do
begin
repeat
x:=x*10+(ord(a[i])-ord('0'));
inc(i);
until (pd(ord(a[i]))) or (i>n);
inc(i);
b[j]:=x;
x:=0;
inc(j);
end;
dec(j);
if j=1 then writeln(b[1]) else begin
num2:=b[j];
for k:=j downto 1 do
begin
num1:=b[k-1]*num2+num1;
t:=num1; num1:=num2; num2:=t;
end;
writeln(num1,'/',num2);
end;
end else begin
i:=1;x:=0;j:=0;num1:=0;num2:=0;
while i<=n do
begin
repeat
x:=x*10+(ord(a[i])-ord('0'));
inc(i);
until (pd(ord(a[i]))) or (i>n);
inc(i);
if j=0 then begin num1:=num1+x; inc(j); x:=0;
end else num2:=num2+x;
end;
if num2=0 then writeln('[',num1,']')else
if num1 mod num2=0 then writeln('[',num1 div num2,']')
else begin
write('[');
for k:=1 to 1000 do begin
b[k]:=num1 div num2;
num1:=num1 mod num2;
t:=num1; num1:=num2; num2:=t;
if k=1 then write(b[k],';')
else if num2<1 then begin writeln(b[k],']');break;end else write(b[k],',');
end;
end;
end;
end;
end.

测试数据 #0: Accepted, time = 0 ms, mem = 796 KiB, score = 20
测试数据 #1: Accepted, time = 0 ms, mem = 800 KiB, score = 20
测试数据 #2: Accepted, time = 0 ms, mem = 800 KiB, score = 20
测试数据 #3: Accepted, time = 0 ms, mem = 800 KiB, score = 20
测试数据 #4: Accepted, time = 15 ms, mem = 800 KiB, score = 20
第四个数据原本老是错，暴力模拟就可以AC了，感觉程序好长

测试数据04 应该是

2/1

[2]

之类的数据吧

输出是

[2]

2

#include<iostream>
#include<cstdio>
#include<cmath>
#include<sstream>
#include<string>
#include<algorithm>
#include<numeric>
#include<vector>
using namespace std;

int p[200] = { 0 };
int q[200] = { 0 };
int fra[200] = { 0 };
int gcd(int a, int b)
{
    if (a < b)
    {
        int temp;
        temp = a;
        a = b;
        b = temp;
    }
    while (a%b)
    {
        int re = 0;
        re = a%b;
        a = b;
        b = re;
        re = a%b;
    }
    return b;
}
void fraction_to_num(string fraction)
{
    int nume = 0;
    int deno = 0;
    int re = 0;
    int quo = 0;
    int i = 0;

    string s1, s2;
    string s3;
    size_t it;
    stringstream str1,str2;
    stringstream str3;

    it = fraction.find('/');
    if (it > fraction.size()){
        s1 = fraction; s2 = "0";
    }
    else
    {
        s1 = fraction.substr(0, it);
        s2 = fraction.substr(it + 1, fraction.length() - it);
    }
    

    str1 << s1;
    str2 << s2;

    str1 >> nume;
    str2 >> deno;


    if (nume >= deno)
    {
        if (deno != 0)
        {
            while ((nume / deno) != 0)
            {
                quo = nume / deno;
                re = nume%deno;
                fra[i++] = quo;
                if (re == 0)break;
                nume = deno;
                deno = re;
            }
        }
        else
        {
            fra[i++] = nume;
        }
        
        str3 << "[";
        for (int k = 0; k < i; k++)
        {
            str3 << fra[k];
            if (k <= (i - 2))
            {
                if (k == 0)str3 << ";";
                else str3 << ",";
            }
        }
        str3 << "]";
        str3 >> s3;
    }
    else
    {
        int temp;
        temp = nume;
        nume = deno;
        deno = temp;
        while ((nume / deno) != 0)
        {
            quo = nume / deno;
            re = nume%deno;
            fra[i++] = quo;
            if (re == 0)break;
            nume = deno;
            deno = re;
        }
        str3 << "[0;";
        for (int k = 0; k < i; k++)
        {
            str3 << fra[k];
            if (k <= (i - 2))
            {
                 str3 << ",";
            }
        }
        str3 << "]";
        str3 >> s3;
    }
    cout << s3<<endl;       
}
void num_to_fraction(string s)
{
    string num;
    int i = 0;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] == '[' || s[i] == ']' || s[i] == ';' || s[i] == ',')
            continue;
        num.push_back(s[i]);
    }
    if (num.size() >= 2)
    {
        
        for (i = 0; i < num.length(); i++)
        {
            if (i == 0)
            {
                p[0] = num[0] - '0';
                q[0] = 1;
            }
            else
            if (i == 1)
            {
                p[1] = (num[0] - '0')*(num[1] - '0') + 1;
                q[1] = (num[1] - '0');
            }
            else
            {
                p[i] = (num[i] - '0')*p[i - 1] + p[i - 2];
                q[i] = (num[i] - '0')*q[i - 1] + q[i - 2];
            }
        }
        cout << p[i - 1] / gcd(p[i - 1], q[i - 1]) << "/" << q[i - 1] / gcd(p[i - 1], q[i - 1]) << endl;
    }
    else
    {
        cout << num << endl;
    }
    
}


int main()
{
    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    string t;

    while (cin >> t)
    {
        if (t[0] == '[')
        {
            num_to_fraction(t);
        }
        else
        {
            fraction_to_num(t);
        }
    }
    return 0;
}

#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<math.h>
#define MAX 100
//声明全局变量
char s[MAX];
int store[MAX];
int m, n;
int *mu = &m, *zi = &n;
//函数swap，用于交换分子分母
void swap(int *a, int *b)
{
    int t;
    t = *a;
    *a = *b;
    *b = t;
}
//函数add，取倒数后相加并约分
void add(int *a, int *b, int x)//*a是分子，*b是分母，分子小于分母
{
    *a += (*b)*x;            //相加
    for (int i = *a; i >= 1; i--)   //找到最大公因数，约分
        if (*b%i == 0 && *a / i == 0)
        {
            *b /= i;
            *a /= i;
            break;
        }
    swap(a, b);              //调换分子分母的位置
}
//函数lianTofen,用于将连分数形式[n;a,b,…]转换为分数形式x/y
void lianTofen(char *s)
{
    int len = strlen(s);
    int i, num = 0;
    int j = 0;
    for (i = 1; i < len; i++)        //连分数表示法中第一个元素是[，故从下标1开始
    {                                //注意到s[len-1]是]，s[len-2]一定是一个数字
        if (isdigit(s[i]))
        {
            num = num * 10 + (s[i] - '0');
            store[j] = num;          //store[]用来存储s[]中的数值的
        }
        else {
            num = 0;
            j++;
        }
    }
    store[j] = 1;                    //运算
    *zi = store[j], *mu = store[j - 1];
    for (i = j - 1; i >= 1; i--)
    {
        add(zi, mu, store[i - 1]);

    }
    swap(zi, mu);
}
//函数fenTolian，把分数形式转换为连分数形式并输出
void fenTolian(char *s)
{
    int len = strlen(s);
    int a = 0, b = 0;
    int i, k = 0, j = 0;
    for (i = 0; i < len; i++)
        if (!isdigit(s[i]))
            k = i;               //找到'/'在数组中的下标k
    for (i = 0; i < k; i++)
        a = 10 * a + s[i] - '0';
    for (i = k + 1; i < len; i++)
        b = 10 * b + s[i] - '0';
    do {
        k = a / b;               //取整
        store[j++] = k;
        a = a - k*b;
        if (a == 0)break;
        swap(&a, &b);
    } while (1);
    if (j == 1)
        printf("[%d]\n", store[0]);
    else
    {
        printf("[%d;", store[0]);
        printf("%d", store[1]);
        for (i = 2; i < j; i++)
            printf(",%d", store[i]);
        printf("]\n");
    }
}
int main()
{
    while (scanf("%s",s)>0)
    {
        if (s[0] == '[')
        {
            lianTofen(s);
            if(m!=1)                      //要考虑分母是1的情况
            {
                printf("%d/%d\n", n, m);  //注意这里的分母和分子是调换的
            }
            else printf("%d\n", n);
        }
        else
            fenTolian(s);
    }
    return 0;
}

强烈要求改描述，反正我想不到最后输出的**不是分数**= =

#include <bits/stdc++.h>
using namespace std;

char s[100000];

struct Frac {
    int upper,downer;
    Frac()
    {
    }
    Frac(int x)
    {
        upper=x;downer=1;
    }
    Frac(int x,int y)
    {
        if (!x) {
            upper=0;downer=1;
        } else {
            int g=__gcd(x,y);
            x/=g;y/=g;
            upper=x;downer=y;
        }
    }
    Frac operator +(const Frac &r)
    {
        return Frac(upper*r.downer+r.upper*downer,downer*r.downer);
    }
    void flip()
    {
        swap(upper,downer);
    }
    void print()
    {
        if (downer==1) {
            printf("%d\n",upper);
        } else {
            printf("%d/%d\n",upper,downer);
        }
    }
};

int main()
{
    while (scanf("%s",s)!=EOF) {
        if (s[0]=='[') {
            vector<int> a;
            for (int i=1;s[i];i++) {
                if ('0'<=s[i]&&s[i]<='9') {
                    int x=0;
                    while ('0'<=s[i]&&s[i]<='9') x=x*10+s[i++]-'0';
                    a.push_back(x);
                }
            }
            Frac t(0);
            for (int i=a.size()-1;i>=0;i--) {
                t=t+Frac(a[i]);
                if (i) t.flip();
            }
            t.print();
        } else {
            int x,y;
            sscanf(s,"%d/%d",&x,&y);
            Frac t(x,y);
            vector<int> ans;
            while (t.downer!=1) {
                ans.push_back(t.upper/t.downer);
                t.upper%=t.downer;
                t.flip();
            }
            ans.push_back(t.upper);
            putchar('[');
            for (int i=0;i<ans.size();i++) {
                printf("%d",ans[i]);
                if (i==0) putchar(';');
                else if (i!=ans.size()-1) putchar(',');
            }
            puts("]");
        }
    }
    return 0;
}