题解 - Vijos

5

2016wudi LV 7 @ 2017-10-10 22:34:33

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
struct P{
    string name;
    string mark;
}a[1005];
bool cmp(P p,P q){
    if(p.mark.size()==q.mark.size()){
        if(p.mark==q.mark){
            return p.name<q.name;
        }
        return p.mark>q.mark;
    }
    return p.mark.size()>q.mark.size();
}
int main(){
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i].name>>a[i].mark;
    }
    sort(a,a+n,cmp);
    for(int i=0;i<n;i++){
        cout<<a[i].name<<endl;
    }
    return 0;
}

string类

3

zsh2333 LV 8 @ 2017-09-18 11:21:21

在sort中的cmp中直接定义水过。。。。。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;

struct node{
    char a[25];
    char b[10005];
    int len;
}shui[1005];

int n;

int work(char *x,char *y,int lan){//比较数字大小
    for(int i=0;i<lan;i++){
        if(x[i]<y[i]) return 2;
        if(x[i]>y[i]) return 1;
    }
    return 0;
}

bool work2(char *x,char*y){//比较名字的大小
    int lenx=strlen(x);
    int leny=strlen(y);
    for(int i=0;i<min(lenx,leny);i++){
        if(x[i]<y[i]) return true;
        if(x[i]>y[i]) return false;
    }
    if(lenx<leny) return true;
    return false;
}

bool cmp(node x,node y){//先比较数字长度
    if(x.len>y.len) return true;
    else if(x.len<y.len) return false;
    else{
        int aa;
        if((aa=work(x.b,y.b,x.len))==1) return true;
        else if(aa==2) return false;
        else return work2(x.a,y.a);
    }
}

int main(){
    scanf("%d\n",&n);
    for(int i=1;i<=n;i++){
        scanf("%s\n%s\n",shui[i].a,shui[i].b);
        shui[i].len=strlen(shui[i].b);
    }
    sort(shui+1,shui+n+1,cmp);
    for(int i=1;i<=n;i++){
        for(int j=0;j<strlen(shui[i].a);j++)
        printf("%c",shui[i].a[j]);
        printf("\n");
    }
    return 0;
}

2

乔彦博 LV 8 @ 2018-08-18 14:47:41

只是个结构体排序嘛。。。
sort+cmp+string水过
AC


#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
struct hehe{
    string a;
    string b;
}a[1001];
bool cmp(hehe a,hehe b)
{
    if(a.b==b.b) return a.a<b.a;
    if(a.b.length()!=b.b.length()) return a.b.length()>b.b.length();
    return a.b>b.b;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i].a>>a[i].b;
    }
    sort(a+1,a+n+1,cmp);
    for(int i=1;i<=n;i++)
    {
        cout<<a[i].a<<endl;
    }
    return 0;
}

2

弑神丿海皇 LV 7 @ 2017-11-03 22:14:07

最顶上的没有这个格式，不方便，希望顶上去
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;

struct node{
char a[25];
char b[10005];
int len;
}shui[1005];

int n;

int work(char *x,char *y,int lan){//比较数字大小
for(int i=0;i<lan;i++){
if(x[i]<y[i]) return 2;
if(x[i]>y[i]) return 1;
}
return 0;
}

bool work2(char *x,char*y){//比较名字的大小
int lenx=strlen(x);
int leny=strlen(y);
for(int i=0;i<min(lenx,leny);i++){
if(x[i]<y[i]) return true;
if(x[i]>y[i]) return false;
}
if(lenx<leny) return true;
return false;
}

bool cmp(node x,node y){//先比较数字长度
if(x.len>y.len) return true;
else if(x.len<y.len) return false;
else{
int aa;
if((aa=work(x.b,y.b,x.len))==1) return true;
else if(aa==2) return false;
else return work2(x.a,y.a);
}
}

int main(){
scanf("%d\n",&n);
for(int i=1;i<=n;i++){
scanf("%s\n%s\n",shui[i].a,shui[i].b);
shui[i].len=strlen(shui[i].b);
}
sort(shui+1,shui+n+1,cmp);
for(int i=1;i<=n;i++){
for(int j=0;j<strlen(shui[i].a);j++)
printf("%c",shui[i].a[j]);
printf("\n");
}
return 0;
}

2

吴哥——传奇 LV 8 @ 2017-08-17 11:03:19

冒泡轻松过，用字符串，strcmp+strcpy+strlen！！！！！！

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
struct bla
{
    char ID[30];
    char number[10010];
}a[1010];
int main(){
    int n,i,j;
    char L[10010];
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%s\n%s",a[i].ID,a[i].number);
    }
    for(i=1;i<=n-1;i++)
    {
        for(j=1;j<=n-i;j++)
        {
            if(strlen(a[j].number)==strlen(a[j+1].number))
            {
                if(strcmp(a[j].number,a[j+1].number)<0)
                {
                    strcpy(L,a[j+1].number);
                    strcpy(a[j+1].number,a[j].number);
                    strcpy(a[j].number,L);
                    strcpy(L,a[j+1].ID);
                    strcpy(a[j+1].ID,a[j].ID);
                    strcpy(a[j].ID,L);
                }
                else if(strcmp(a[j].number,a[j+1].number)==0)
                {
                    if(strcmp(a[j].ID,a[j+1].ID)>0)
                    {
                        strcpy(L,a[j+1].ID);
                        strcpy(a[j+1].ID,a[j].ID);
                        strcpy(a[j].ID,L);
                    }
                }
            }
            else if(strlen(a[j].number)<strlen(a[j+1].number))
            {
                strcpy(L,a[j+1].number);
                strcpy(a[j+1].number,a[j].number);
                strcpy(a[j].number,L);
                strcpy(L,a[j+1].ID);
                strcpy(a[j+1].ID,a[j].ID);
                strcpy(a[j].ID,L);
            }
        }
    }
    for(i=1;i<=n;i++)
        puts(a[i].ID);
    return 0;
}

1

Sky1231 (sky1231) LV 10 @ 2021-03-18 13:16:55

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <set>
#include <limits>
#include <string>
#include <sstream>
using namespace std;

const int oo_min=0xcfcfcfcf,oo_max=0x3f3f3f3f;

class ubigint{
public:
    vector<int> num;
    int operator == (ubigint b)
    {
        if (num.size()!=b.num.size())
            return 0;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
                if (num[i]!=b.num[i])
                    return 0;
            return 1;
        }
    }
    int operator != (ubigint b)
    {
        return ((!(*this==b))?1:0);
    }
    int operator < (ubigint b)
    {
        if (num.size()<b.num.size())
            return 1;
        else if (num.size()>b.num.size())
            return 0;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
            {
                if (num[size-1-i]<b.num[size-1-i])
                    return 1;
                else if (num[size-1-i]>b.num[size-1-i])
                    return 0;
            }
            return 0;
        }
    }
    int operator > (ubigint b)
    {
        if (num.size()<b.num.size())
            return 0;
        else if (num.size()>b.num.size())
            return 1;
        else
        {
            for (unsigned long long i=0,size=num.size();i<size;i++)
            {
                if (num[size-1-i]<b.num[size-1-i])
                    return 0;
                else if (num[size-1-i]>b.num[size-1-i])
                    return 1;
            }
            return 0;
        }
    }
    int operator <= (ubigint b)
    {
        return ((!(*this>b))?1:0);
    }
    int operator >= (ubigint b)
    {
        return ((!(*this<b))?1:0);
    }
    ubigint mov(long long x)
    {
        ubigint ans;
        ans.num.clear();
        if (x+num.size()>0)
        {
            ans.num.resize(num.size()+x,0);
            for (long long i=0,size=num.size();i<size;i++)
                if (i+x>=0)
                    ans.num[i+x]=num[i];
        }
        else
            ans=0;
        return ans;
    }
    ubigint operator = (long long b)
    {
        char s[20+1];
        sprintf(s,"%lld",b);
        num.clear();
        num.resize(strlen(s));
        for (unsigned long long i=0,size=num.size();i<size;i++)
            num[i]=(s[size-1-i]-'0');
        return (*this);
    }
    ubigint operator = (string b)
    {
        num.clear();
        num.resize(b.size());
        for (unsigned long long i=0,size=num.size();i<size;i++)
            num[i]=(b[size-1-i]-'0');
        return (*this);
    }
    ubigint operator + (ubigint b)
    {
        ubigint ans;
        ans.num.clear();
        ans.num.resize(max(num.size(),b.num.size()));
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            ans.num[i]=((i<num.size())?num[i]:0)+((i<b.num.size())?b.num[i]:0);
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
        {
            if (ans.num[i]>=10&&i==size-1)
                ans.num.resize(++size);
            ans.num[i+1]+=(ans.num[i]/10);
            ans.num[i]%=10;
        }
        return ans;
    }
    ubigint operator += (ubigint b)
    {
        return (*this=(*this+b));
    }
    ubigint operator - (ubigint b)
    {
        ubigint ans;
        if ((*this)==b)
        {
            ans=0;
            return ans;
        }
        ans.num.clear();
        ans.num.resize(max(num.size(),b.num.size()));
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            ans.num[i]=((i<num.size())?num[i]:0)-((i<b.num.size())?b.num[i]:0);
        for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            while (ans.num[i]<0)
                ans.num[i]+=10,ans.num[i+1]--;
        while (ans.num[ans.num.size()-1]==0)
            ans.num.resize(ans.num.size()-1);
        return ans;
    }
    ubigint operator -= (ubigint b)
    {
        return (*this=(*this-b));
    }
    ubigint tm (ubigint b,unsigned long long pos)
    {
        ubigint ans,num_0;
        num_0=0;
        ans.num.clear();
        if (*this!=num_0&&b!=num_0)
        {
            ans.num.resize(num.size()+pos);
            for (unsigned long long i=0,size=num.size();i<size;i++)
                ans.num[i+pos]=num[i]*b.num[pos];
            for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            {
                if (ans.num[i]>=10&&i==size-1)
                    ans.num.resize(++size);
                ans.num[i+1]+=(ans.num[i]/10);
                ans.num[i]%=10;
            }
        }
        else
            ans=0;
        return ans;
    }
    ubigint operator * (ubigint b)
    {
        ubigint ans,num_0;
        ans=0,num_0=0;
        ans.num.clear();
        if (*this!=num_0&&b!=num_0)
        {
            for (unsigned long long i=0,size=b.num.size();i<size;i++)
                ans+=(*this).tm(b,i);
        }
        return ans;
    }
    ubigint operator *= (ubigint b)
    {
        return (*this=(*this*b));
    }
    ubigint operator / (ubigint b)
    {
        ubigint ans,num_0;
        num_0=0;
        if (*this<b)
            return num_0;
        if (*this!=num_0&&*this>num_0)
        {
            ubigint x,y;
            x=*this,y=b.mov(num.size()-b.num.size());
            ans.num.resize(num.size()-b.num.size()+1,0);
            for (long long i=ans.num.size()-1;i>=0;i--,y=y.mov(-1))
                while (x>=y)
                    x-=y,ans.num[i]++;
            for (unsigned long long i=0,size=ans.num.size();i<size;i++)
            {
                if (ans.num[i]>=10&&i==size-1)
                    ans.num.resize(++size);
                ans.num[i+1]+=(ans.num[i]/10);
                ans.num[i]%=10;
            }
            while (ans.num.size()>1&&ans.num[ans.num.size()-1]==0)
                ans.num.resize(ans.num.size()-1);
        }
        else
            ans=0;
        return ans;
    }
    ubigint operator /= (ubigint b)
    {
        return (*this=(*this/b));
    }
    void print()
    {
        for (unsigned long long i=0,size=num.size();i<size;i++)
            printf("%d",num[size-1-i]);
    }
};
class waterking{
public:
    char name[1<<5];
    ubigint water;
};

int n,num[1<<10];
waterking wk[1<<10];

int cmpwk(waterking a,waterking b)
{
    return (a.water>b.water)||(a.water==b.water&&strcmp(a.name,b.name)<0);
}
void qs(int l,int r)
{
    int i,j;
    waterking mid=wk[num[(l+r)>>1]];
    for (i=l,j=r;i<j;)
    {
        while (cmpwk(wk[num[i]],mid))
            i++;
        while (cmpwk(mid,wk[num[j]]))
            j--;
        if (i<=j)
        {
            swap(num[i],num[j]);
            i++,j--;
        }
    }
    if (l<j)
        qs(l,j);
    if (i<r)
        qs(i,r);
}
int main()
{
    scanf("%d\n",&n);
    for (int i=0;i<n;i++)
    {
        num[i]=i;
        char s[1<<14];
        memset(s,0,sizeof(s));
        string ss;
        memset(wk[i].name,0,sizeof(wk[i].name));
        scanf("%s\n%s\n",wk[i].name,s);
        ss=s,wk[i].water=ss;
    }
    qs(0,n-1);
    for (int i=0;i<n;i++)
        printf("%s\n",wk[num[i]].name);
}

1

feidaoluoye LV 9 @ 2017-10-02 19:21:32

#include<iostream>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#define maxa 1000+30
#define inf 10000000
using namespace std;
struct node
{
string id,num;
}e[maxa];
bool comp(node a,node b)
{
int i;
if(a.num==b.num)
{
for(i=0;i<a.id.length()&&i<b.id.length();++i)
if(a.id[i]!=b.id[i])
return a.id[i]<b.id[i];
if(a.id.length()<b.id.length())
return true;
}
else
{
if(a.num.length()!=b.num.length())
return a.num.length()>b.num.length();
else for(i=0;i<a.num.length()&&i<b.num.length();++i)
if(a.num[i]!=b.num[i])
return a.num[i]>b.num[i];
}
return false;
}
int main()
{
int n,i = 0;
cin>>n;
for(i=0;i<n;++i){
cin>>e[i].id>>e[i].num;
}
sort(e,e+n,comp);
for(i=0;i<n;++i)
cout<<e[i].id<<endl;
return 0;
}

0

scarletocean LV 6 @ 2020-06-06 02:19:27

用了自定义排序规则的multimap，第3个测试WA，其它都AC，请教为什么

#include<set>
#include<algorithm>
#include<cstring>
#include <iostream>

using namespace std;

struct Up {
char name[20];
unsigned long long n;
};

struct Rule {
bool operator()(const Up & u1, const Up & u2) const {
if (u1.n != u2.n) return u1.n > u2.n;
else return (strcmp(u1.name, u2.name) < 0);
}
};

int main(void) {
multiset<Up, Rule> ms;
Up u;
int n = 0;

scanf("%d", &n);
for (int i = 0; i < n; i++) {
//scanf("%s", u.name);
//getchar();
//scanf("%ll", &u.n);
cin >> u.name >> u.n;
ms.insert(u);
}

multiset<Up, Rule>::iterator p;
for (p = ms.begin(); p != ms.end(); p++) {
printf("%s\n", p->name);
}
return 0;
}

0

Johnwegner LV 6 @ 2018-08-12 18:51:59

#include <iostream>
#include<string>

using std::string;
using std::cin;
using std::cout;
using std::endl;

int PARTITION(int p, int r);
void QUICKSORT(int p, int r);

string name[10000]; //储存水王ID
unsigned long long totcomments[10000]; //储存水王发帖数

int main()
{
int totnum; //水王总数
cin>>totnum;
int nail=0;// 标志，记录读取水王数据
while(nail<totnum)
{
cin>>name[nail];
cin>>totcomments[nail];
nail++;
}
QUICKSORT(0, totnum-1);
for(int i=totnum-1;i>=0;i--)
{
cout<<name[i]<<endl;
}
return 0;
}

int PARTITION(int p, int r) //快速排序核心是找到分界点，因此对名字同样进行分解即可
{
unsigned long long tip = totcomments[r];
int i=p-1;
for(int j=p;j<r;j++)
{
if(totcomments[j]<tip)
{
i=i+1;
unsigned long long temp=totcomments[i];
totcomments[i]=totcomments[j];
totcomments[j]=temp;

string str=name[i];
name[i]=name[j];
name[j]=str; //同时交换ID
}
else if(totcomments[j] == tip)
{
if(name[j]>name[r])
{
i=i+1;
unsigned long long temp=totcomments[i];
totcomments[i]=totcomments[j];
totcomments[j]=temp;

string str=name[i];
name[i]=name[j];
name[j]=str; //发帖数相同，同时交换ID，按字典顺序,注意输出倒序，因此大的ID在前
}
}
}
unsigned long long temp=totcomments[r];
totcomments[r]=totcomments[i+1];
totcomments[i+1]=temp;

string str=name[r];
name[r]=name[i+1];
name[i+1]=str; //对姓名进行分界点交换，对发帖数排序同时保证对应ID跟着变化
return i+1;
}

void QUICKSORT(int p, int r)
{
if(p<r)
{
int q=PARTITION(p, r);
QUICKSORT(p, q-1);
QUICKSORT(q+1,r);
}
}

0

Johnwegner LV 6 @ 2018-08-12 18:47:32

#include <iostream>
#include<string>

using std::string;
using std::cin;
using std::cout;
using std::endl;

int PARTITION(int p, int r);
void QUICKSORT(int p, int r);

string name[5000]; //储存水王ID
unsigned long long totcomments[5000]; //储存水王发帖数

int main()
{
int totnum; //水王总数
cin>>totnum;
int nail=0;// 标志，记录读取水王数据
while(nail<totnum)
{
cin>>name[nail];
cin>>totcomments[nail];
nail++;
}
QUICKSORT(0, totnum-1);
for(int i=totnum-1;i>=0;i--)
{
cout<<name[i]<<endl;
}
return 0;
}

int PARTITION(int p, int r) //快速排序核心是找到分界点，因此对名字同样进行分解即可
{
unsigned long long tip = totcomments[r];
int i=p-1;
for(int j=p;j<r;j++)
{
if(totcomments[j]<tip)
{
i=i+1;
unsigned long long temp=totcomments[i];
totcomments[i]=totcomments[j];
totcomments[j]=temp;

string str=name[i];
name[i]=name[j];
name[j]=str; //同时交换ID
}
else if(totcomments[j] == tip)
{
if(name[j]>name[r])
{
i=i+1;
unsigned long long temp=totcomments[i];
totcomments[i]=totcomments[j];
totcomments[j]=temp;

string str=name[i];
name[i]=name[j];
name[j]=str; //发帖数相同，同时交换ID，按字典顺序,注意输出倒序，因此大的ID在前
}
}
}
unsigned long long temp=totcomments[r];
totcomments[r]=totcomments[i+1];
totcomments[i+1]=temp;

string str=name[r];
name[r]=name[i+1];
name[i+1]=str; //对姓名进行分界点交换，对发帖数排序同时保证对应ID跟着变化
return i+1;
}

void QUICKSORT(int p, int r)
{
if(p<r)
{
int q=PARTITION(p, r);
QUICKSORT(p, q-1);
QUICKSORT(q+1,r);
}
}

0

noheart LV 5 @ 2018-05-13 16:41:07

用一个数组优化空间，记录起始位置和长度，冒泡过

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
#include<stdlib.h>
using namespace std;
struct node{
char name[24];
int st,len;
}list[1008];
char s[10008];
int n,a[1008],t[10008];

void init(){
cin >> n;
int p=0,k;
for (int i=1;i<=n;i++){
cin >> list[i].name;
cin >> s;
list[i].st=p+1;
k=strlen(s);list[i].len=k;
for (int j=1;j<=k;j++){
t[++p]=s[j-1]-48;
}
}
}

bool small(node x,node y){
if(x.len==y.len){
int x1=x.st,y1=y.st;
for (int i=1;i<=x.len;i++,x1++,y1++){
if (t[x1]>t[y1])return false;
else if(t[y1]>t[x1])return true;
}
if (strcmp(x.name,y.name)<0)return false;
else return true;
}
else return (y.len>x.len);
}

void sort1(){
for (int i=1;i<=n;i++)a[i]=i;
for (int i=1;i<=n;i++)
for (int j=i+1;j<=n;j++)
if(small(list[a[i]],list[a[j]]))
{int tmp=a[i];a[i]=a[j];a[j]=tmp;}
for (int i=1;i<=n;i++){
cout << list[a[i]].name << endl;
}
}

int main(){
init();
sort1();
return 0;
}

0

Gugumelon (MAYDAYTYH) LV 7 @ 2017-12-07 20:15:29

#include<stdio.h>
#include<string.h>
#define N 100000+5
#define ll long long
struct shui
{
char name[20];
char tie[50];
}ren[1005];
int strcmpp(char x[],char y[])
{
int hh;
int lx=strlen(x),ly=strlen(y);
if(lx>ly) hh=1;
else if(ly>lx) hh=-1;
else hh=strcmp(x,y);
return hh;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)
{
scanf("%s",ren[i].name);
scanf("%s",ren[i].tie);
}
for(int i=1;i<n;++i)
{
for(int j=n;j>i;--j)
{
if(strcmpp(ren[j].tie,ren[j-1].tie)>0)
{
struct shui x;
strcpy(x.name,ren[j].name);strcpy(x.tie,ren[j].tie);
strcpy(ren[j].name,ren[j-1].name);strcpy(ren[j].tie,ren[j-1].tie);
strcpy(ren[j-1].name,x.name);strcpy(ren[j-1].tie,x.tie);
}
else if(strcmpp(ren[j].tie,ren[j-1].tie)==0)
{
if(strcmp(ren[j].name,ren[j-1].name)<0)
{
struct shui x;
strcpy(x.name,ren[j].name);strcpy(x.tie,ren[j].tie);
strcpy(ren[j].name,ren[j-1].name);strcpy(ren[j].tie,ren[j-1].tie);
strcpy(ren[j-1].name,x.name);strcpy(ren[j-1].tie,x.tie);
}
else continue;
}
else continue;
}
}
for(int i=1;i<=n;++i) puts(ren[i].name);
return 0;
}

0

fjlyzpcr LV 8 @ 2009-09-13 20:25:54

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：运行时错误...|错误号: 216

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：75 有效耗时：0ms

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

同样的程序

前N次交 216？。。。

重启 AC？。。。

倭寇啊。。。。

经典~~搞笑~~~吐血~~~爷微笑~~~

0

_orz_ LV 10 @ 2009-08-18 22:55:36

喝矿泉水有利健康~

0

Zintch LV 8 @ 2009-08-17 16:10:24

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

冒泡都可以秒杀，晕……

0

linw1 LV 8 @ 2009-08-12 09:44:48

哪个大哥告诉我，哪里错了？

type shui=record

name:string;

liang:ansistring;

end;

var a:array[1..1000]of shui;

n,b,j,i,c:longint;

k:shui;

begin

readln(n);

for i:=1 to n do begin

readln(a[i].name);

readln(a[i].liang);

b:=length(a[i].liang);

if b>c then c:=b;

end;

for i:=1 to n do begin

for j:=1 to c-length(a[i].liang) do insert('0',a[i].liang,1);

end;

for i:=1 to n-1 do

for j:=n downto i+1 do begin

if a[j-1].liang>a[j].liang then begin

k:=a[j-1];

a[j-1]:=a[j];

a[j]:=k;

end;

end;

for i:=1 to n-1 do begin

if (a[i].name

0

how LV 10 @ 2009-08-01 12:16:29

MMD...水题交了5遍才AC...

0

xyj-flash LV 9 @ 2009-07-31 22:37:18

我竟然用了两遍Qsort来对付一道水题。。。

而且是用选择排序就0msAC的题。。。

0

YY_0227 LV 8 @ 2009-07-25 16:34:36

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

此水我竟WA一次。。。⊙﹏⊙b汗

就先把ID冒泡排一遍

然后比较发帖数先比长度，再直接用大于号比

哦了

0

zhh1993 LV 10 @ 2009-07-24 17:22:23

按字典序输出.......郁闷

题解

136 条题解

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136 条题解

水王争霸

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