自动AC机直接过

//就想知道凭什么难度是7
#include <iostream>         //[2003普及组-A]乒乓球
#include <algorithm>
#include <string>
#include <cmath>
using namespace std;

void Print(string str, int k)
{
    int a = 0, b = 0;
    for (int i = 0; i < str.length(); i++)
        if(str[i] != '\n')
        {
            if(str[i] == 'W')
                a++;
            else
                b++;
            if((a >= k || b >= k) && abs(a - b) >= 2)
            {
                cout << a << ":" << b << endl;
                a = 0, b = 0;
            }
        }

    cout << a << ":" << b << endl;
}

int main()
{
    string str;
    getline(cin, str, 'E');

    Print(str, 11);
    cout << endl;
    Print(str, 21);

    return 0;
}

#include<iostream>
#include<string>
using namespace std;

int main() {
    char str[1000005];
    char ch;
    int ww = 0,  ll = 0, len = 0;
    while((ch = getchar()) != 'E') {
        if(ch == '\n' || ch == ' ')
            continue;           
        str[len++] = ch;            
    }
    
    
    if(len == 0) {
        cout << "0:0" <<endl;
        cout << endl;
        cout << "0:0" <<endl;
        return 0;   
    } 
    
    //处理11分制 
    for(int i = 0 ; i < len; ++i) {
        bool flag = true;
        if(str[i] == 'W')
            ww++;
        else
            ll++;
        
        if(ww >= 11 && ww-ll > 1){
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0;
        }
        
        if(ll >= 11 && ll-ww > 1) {
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0;
        }
        
        if(i == len - 1 && flag) {
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0;

        }   
            
    }   
    cout << endl;   
    for(int i = 0 ; i < len; ++i) {
        bool flag = true;
        if(str[i] == 'W')
            ww++;
        else
            ll++;
        
        if(ww >= 21 && ww-ll > 1){
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0;                     
        }
        
        if(ll >= 21 && ll-ww > 1) {
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0;
        }
        
        if(i == len - 1) {
            cout <<ww<<':'<<ll<<endl;
            ww = 0, ll = 0; 
        }
    }
    
    return 0;
}

思路：循环逐个读字符，同时判断比赛情况，结束就储存，读到E跳出循环，循环外输出对局情况，记得输出没结束的对局

注意：题目没给好数据范围，一般考试不会这样，一般CE又WA就是数组开的不够大之类

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    int w11_tmp = 0, l11_tmp = 0, l21_tmp = 0, w21_tmp = 0,
        w11[10000], l11[10000], w21[10000], l21[10000], j11 = 0, j21 = 0;
    char c;
    for (;(c=getchar())!='E';)
    {
        if (c == 'W')++w11_tmp, ++w21_tmp;
        if (c == 'L')++l11_tmp, ++l21_tmp;
        if ((w11_tmp > 10 || l11_tmp > 10) && abs(w11_tmp - l11_tmp) > 1)
            w11[j11] = w11_tmp, l11[j11] = l11_tmp, ++j11, w11_tmp = l11_tmp = 0;
        if ((w21_tmp > 20 || l21_tmp > 20) && abs(w21_tmp - l21_tmp) > 1)
            w21[j21] = w21_tmp, l21[j21] = l21_tmp, ++j21, w21_tmp = l21_tmp = 0;
    }
    for (int i = 0; i < j11; ++i)
        printf("%d:%d\n", w11[i], l11[i]);
    printf("%d:%d\n\n", w11_tmp, l11_tmp);
    for (int i = 0; i < j21; ++i)
        printf("%d:%d\n", w21[i], l21[i]);
    printf("%d:%d", w21_tmp, l21_tmp);
}

/*
这个题的测试数据太大了，数组一定要足够大才能全部通过
不然就会Runtime Error
*/
#include<iostream>
#include<cmath>

using namespace std;

int main()
{
char contest[100000];
int total, win = 0, lose = 0;
for (total = 0;;) {
cin >> contest[total];
if (contest[total] == 'E' || contest[total] == EOF)break;
else if (contest[total] != ' '&& contest[total] != '\n')total++;
}
//11回合制
for (int i = 0; i < total; i++) {
if (contest[i] == 'W')win++;
else lose++;
if ((win >= 11 || lose >= 11) && abs(win - lose) >= 2) {
cout << win << ":" << lose << endl;
win = 0; lose = 0;
}
}
cout << win << ":" << lose << endl << endl;
//21回合制
win = 0; lose = 0;
for (int i = 0; i < total; i++) {
if (contest[i] == 'W')win++;
else lose++;
if ((win >= 21 || lose >= 21) && abs(win - lose) >= 2) {
cout << win << ":" << lose << endl;
win = 0; lose = 0;
}
}
cout << win << ":" << lose;
return 0;
}

#include <bits/stdc++.h>
using namespace std;
char ch[300000000];
int grade_w=0,grade_l=0;
void work(int n)
{
grade_w=0;grade_l=0;
int i=1;
while(ch[i]!='E')
{
if(ch[i]==' ')
continue;
if(ch[i]=='W') ++grade_w;
else if(ch[i]=='L') ++grade_l;

if((grade_w>=n || grade_l>=n) && abs(grade_w-grade_l)>=2)
{
printf("%d:%d\n",grade_w,grade_l);
grade_w=0; grade_l=0;
}
++i;
}
printf("%d:%d\n",grade_w,grade_l);
}
int main()
{
int i=0;
do
scanf("%c",&ch[++i]);
while(ch[i]!='E');

work(11);
printf("\n");
work(21);
return 0;
}

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
string C[100000];//我也不知道它一共打了多少局，反正开始开100和1000都RE了 

int main()
{
    int Win1 = 0, Los1 = 0;
    int Win2 = 0, Los2 = 0;//21分制 
    string A;
    int cursor = 0;
    bool flag = true;
    while(cin>>A){//很多行字符串 
        for(int i = 0 ; i < A.length() ; i++){//每行 
            if(A[i] == 'W'){ 
                Win1++;
                Win2++;
            }
            if(A[i] == 'L'){
                Los1++;
                Los2++;
            }
            if(abs(Win1 - Los1) >= 2 && max(Win1, Los1) >= 11){
                cout<<Win1<<":"<<Los1<<endl;
                Win1 = 0;
                Los1 = 0;
            }
            if(abs(Win2 - Los2) >= 2 && max(Win2, Los2) >= 21){
                C[cursor++] = to_string(Win2) + ':' + to_string(Los2);
                Win2 = 0;
                Los2 = 0;
            }
            if(A[i] == 'E'){
                flag = false;
                break;
            } 
        }
        if(flag == false)
            break;
    }
    cout<<Win1<<":"<<Los1<<endl;//不完整的一段 
    cout<<endl;
    
    for(int i = 0 ; i < cursor ; i++)
        cout<<C[i]<<endl;
    cout<<Win2<<":"<<Los2<<endl;//不完整的一段 
    return 0;
 } 

修复一下 2016wudi 的答案

#include<iostream>
#include<cstdlib>
#include<string>
using namespace std;
int main()
{
string s1, s2;//方便输入
//cout << "请输入"<<endl;
while (true) {
getline(cin, s2);
s1 += s2;//输入字符串
if (s2.find('E') != string::npos)break;
}
int i = 0, a = 0, b = 0;//i是指针，a统计了赢的次数，b是输的次数
while (s1[i] != 'E')//只要不是结束符,现在是判断11分制
{
switch (s1[i])//switch判断s1[i]
{
case 'W':a++; break;//如果s1[i]是‘W’,累计到a里
case 'L':b++; break;//如果s1[i]是‘E’,累计到b里
}
if ((a >= 11 || b >= 11) && (abs(a - b) >= 2))//是否到达条件，一盘有没有结束
{
cout << a << ":" << b << endl;//输出
a = 0; b = 0;//清零，下一盘
}
i++;//下一个字符
}
cout << a << ":" << b << endl;//剩下的也要输完
cout << endl;//换行隔开
a = b = 0; i = 0;//清零,换了一个分制
while (s1[i] != 'E')//同理
{
switch (s1[i])
{
case 'W':a++; break;
case 'L':b++; break;
}
if ((a >= 21 || b >= 21) && (abs(a - b) >= 2))
{
cout << a << ":" << b << endl;
a = 0; b = 0;
}
i++;
}
cout << a << ":" << b << endl;
return 0;
}

emmmm......

#include<iostream>
#include<math.h>
using namespace std;

char s[1024*1024];

void result(int length,int full){
int W=0,L=0;
for(int i=0;i<length;i++){
if(s[i]=='W')
W++;
else if(s[i]=='L')
L++;
if(max(W,L)>=full && abs(W-L)>=2){
cout<<W<<':'<<L<<endl;
W=L=0;
}
}
cout<<W<<':'<<L<<endl;
}

int main(){
int i=0;
while((s[i] = getchar()) != EOF){
if(s[i]=='E')
break;
i++;
}
result(i,11);
cout<<endl;
result(i,21);
return 0;
}

用上max()和abs()函数，会比较简洁

#include <iostream>
#define f(x) ((x) < 0 ? (-(x)) : (x))
using namespace std;
void match(string &s, int limit){
    int k = 0, w = 0, l = 0;
    while(k < s.length()){
        if(s[k] == ' '){
            k++;
            continue;
        }
        if(s[k] == 'W') w++;
        if(s[k] == 'L') l++;
        k++;
        if((w >= limit || l >= limit) && f(w - l) > 1){
            cout << w << ':' << l << endl;
            w = l = 0;
        }
    }
    cout << w << ':' << l << endl;
    putchar('\n');
    return ;
}
int main(){
    string s;
    string ans;
    while(getline(cin, s)){
        if(s.find('E') == string::npos)
            ans.append(s);
        else{
            ans.append(s.substr(0, s.find('E')));
            break;
        }
    }
    match(ans, 11);
    match(ans, 21);
    return 0;
}

被坑了两次😂😂😂第一次没看清输出，第二次数组太小😂😂😂
菜了菜了，还是来水一下题解吧

#include <cstdio>
#include <cstring>
char s[100000];

int abs(int a,int b)
{
int c;
if (a>b)
{
c=a-b;
}
if (b>a)
{
c=b-a;
}
return c;
}

void output(int total,int type)
{
int w=0,l=0;
for (int i=0;i<=total;i++)
{

if (s[i]=='W')
{
w++;
}
if (s[i]=='L')
{
l++;
}
if (s[i]=='E')
{
printf("%d:%d\n",w,l);
}
if ((w>=type||l>=type)&&(abs(w,l)>1))
{
printf("%d:%d\n",w,l);
w=0;l=0;
}
}
}

int main()
{
int i=0;
while (scanf("%c",&s[i])!=EOF)
{
if (s[i]=='E') break;
i++;
}
output(i,11);
printf("\n");
output(i,21);
return 0;
}

小胖，你就骗下去吧，我也不揭穿你了。

这是我能想到的最简单的代码了，MAX=2000的时候有一个RE所以设置到了4000。

#include <iostream>
using namespace std;

const int MAX=4000;
int a[MAX*2], b[MAX*2], c[MAX], d[MAX];

int main(){
char ch;
int m=0, n=0;
while (cin>>ch && ch!='E'){
if (ch=='W'){
a[m]++;
c[n]++;
if (a[m]>=11 && a[m]-b[m]>1) m++;
if (c[n]>=21 && c[n]-d[n]>1) n++;
}
else{
b[m]++;
d[n]++;
if (b[m]>=11 && b[m]-a[m]>1) m++;
if (d[n]>=21 && d[n]-c[n]>1) n++;
}
}

for (int i=0; i<=m; i++)
cout<<a[i]<<":"<<b[i]<<endl;
cout<<endl;
for (int i=0; i<=n; i++)
cout<<c[i]<<":"<<d[i]<<endl;

return 0;
}

fn process(contest_info: &str, max_score: i32) {
    let mut p1_score = 0;
    let mut p2_score = 0;
    for c in contest_info.chars() {
        if let 'W' = c {
            p1_score += 1;
        } else {
            p2_score += 1;
        }
        if (p1_score >= max_score || p2_score >= max_score) && (p1_score - p2_score).abs() >= 2 {
            println!("{}:{}", p1_score, p2_score);
            p1_score = 0;
            p2_score = 0;
        }
    }
    println!("{}:{}", p1_score, p2_score);
}

fn main() {
    let mut buf = String::new();
    let mut contest_info = String::new();
    let mut flag = true;
    while flag {
        std::io::stdin().read_line(&mut buf).unwrap();
        contest_info.push_str(match buf.find('E') {
            Some(i) => {
                flag = false;
                &buf[..i]
            }
            None => buf.trim(),
        });
        buf.clear();
    }
    process(contest_info.as_str(), 11);
    println!();
    process(contest_info.as_str(), 21);
}

string=''
def do_work(src):
    w = l = 0
    for ch in string:
        if ch == 'E':
            print("{}:{}".format(w,l))
            return
        elif ch == 'W':
            w += 1
        elif ch == 'L':
            l += 1
        if abs(w-l)>1 and (w>=src or l>=src):
            #print(w,':',l)
            print("{}:{}".format(w,l))
            w = l = 0


while 1:
    s = input()
    string += s
    if s.find('E') != -1:
        break

do_work(11)
print()
do_work(21)

这才叫简洁 kakaka

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

int main()
{
    int i = 0,game1 = 0,game2 = 0,flag = 0;
    int score1[10000][2],score2[10000][2];
    char a[100000];
    while((a[i] = getchar()) != 'E')
    {
        i++;
    }
    i = 0;
    do
    {
        score1[game1][0] = 0;
        score1[game1][1] = 0;
        while((score1[game1][0] < 11 && score1[game1][1] < 11)||abs(score1[game1][0] - score1[game1][1]) <= 1)
        {
            if(a[i] == 'W')
            {
                score1[game1][0]++;
                i++;
            }
            else if(a[i] == 'L')
            {
                score1[game1][1]++;
                i++;
            }
            else if(a[i] == 'E')
            {
                flag = 1;
                break;
            }
            else
            {
                i++;
                continue;
            }
        }
        game1++;
        if(!flag && a[i] == 'E')
        {
            game1++;
            score1[game1][0] = 0;
            score1[game1][1] = 0;
        }
    }while(a[i] != 'E');

    i = 0;
    do
    {
        score2[game2][0] = 0;
        score2[game2][1] = 0;
        while((score2[game2][0] < 21 && score2[game2][1] < 21)||abs(score2[game2][0] - score2[game2][1]) <= 1)
        {
            if(a[i] == 'W')
            {
                score2[game2][0]++;
                i++;
            }
            else if(a[i] == 'L')
            {
                score2[game2][1]++;
                i++;
            }
            else if(a[i] == 'E')
            {
                flag = 1;
                break;
            }
            else
            {
                i++;
                continue;
            }
        }
            game2++;
        if(!flag && a[i] == 'E')
        {
            game2++;
            score2[game2][0] = 0;
            score2[game2][1] = 0;
        }
    }while(a[i] != 'E');
    for(i = 0; i < game1; i++)
        printf("%d:%d\n",score1[i][0],score1[i][1]);
    printf("\n");
    for(i = 0; i < game2; i++)
        printf("%d:%d\n",score2[i][0],score2[i][1]);
    return 0;
}

//数组开10000会越界，要开100000
#include <iostream>
#include <cmath>
using namespace std;
int shu[100000];
int main(int argc, char** argv)
{
int s=0;//统计输入字母个数
while((shu[s]=getchar())!=EOF)
{
if(shu[s]=='E')
{
break;
}
s++;
}
int w=0;
int l=0;
for(int i=0;i<s;i++)
{
if(shu[i]=='W') w++;
if(shu[i]=='L') l++;
if(((w==11||l==11)&&abs(w-l)>=2)||((w>11||l>11)&&abs(w-l)==2))
{
cout<<w<<':'<<l<<endl;
w=0;
l=0;
}
}
cout<<w<<':'<<l<<endl<<endl;
w=0;
l=0;
for(int i=0;i<s;i++)
{
if(shu[i]=='W') w++;
if(shu[i]=='L') l++;
if((w==21||l==21)&&abs(w-l)>=2||(w>21||l>21)&&abs(w-l)==2)
{
cout<<w<<':'<<l<<endl;
w=0;
l=0;
}
}
cout<<w<<':'<<l<<endl;
return 0;
}

Java表示，数据范围不需要的，这辈子都不可能需要的，Pascal，C,CPP会踩的坑，我们都可以跳过，数据我们可以边读边做，我们的数组还能自己变长，就是时间长了一点，但是能不用考虑坑就一次过那就都还好。

import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;

public class P1217 {
    static FileReader fr;
    static int n;
    
    static int score11a, score11b;
    static int score21a, score21b;
    
    static class Match{
        int sc1, sc2;
        Match(int a, int b){
            this.sc1 = a;
            this.sc2 = b;
        }
        
        public String toString() {
            return String.format("%d:%d", sc1, sc2);
        }
    }
    
    static ArrayList<Match> result11 = new ArrayList<Match>();
    static ArrayList<Match> result21 = new ArrayList<Match>();
    
    public static Scanner getInput() {
        return new Scanner(System.in);
    }

//  public static Scanner getInput() throws IOException {
//      File f = new File("input.txt");
//      fr = new FileReader(f);
//      Scanner sc = new Scanner(fr);
//
//      return sc;
//  }
    
    public static void input() throws IOException {
        Scanner sc = getInput();
        
        boolean finished = false;
        while (!finished) {
            String s = sc.nextLine();
            char[] c = s.toCharArray();
            for (int i = 0; i < c.length; i++) {
                if (c[i] == 'E') {
                    finished = true;
                    result11.add(new Match(score11a, score11b));
                    result21.add(new Match(score21a, score21b));
                    break;
                }
                if (c[i] == 'W') {
                    score11a++;
                    score21a++;
                    if (score11a >= 11 && score11b <= score11a - 2) {
                        result11.add(new Match(score11a, score11b));
                        score11a = score11b = 0;
                    }
                    if (score21a >= 21 && score21b <= score21a - 2) {
                        result21.add(new Match(score21a, score21b));
                        score21a = score21b = 0;
                    }
                }else if (c[i] == 'L') {
                    score11b++;
                    score21b++;
                    if (score11b >= 11 && score11a <= score11b - 2) {
                        result11.add(new Match(score11a, score11b));
                        score11a = score11b = 0;
                    }
                    if (score21b >= 21 && score21a <= score21b - 2) {
                        result21.add(new Match(score21a, score21b));
                        score21a = score21b = 0;
                    }
                }
            }
            
        }
        if (fr != null) {
            fr.close();
        }
    }


    public static void output() {
        for (int i = 0; i<result11.size(); i++) {
            System.out.println(result11.get(i));
        }
        System.out.println();
        for (int i = 0; i<result21.size(); i++) {
            System.out.println(result21.get(i));
        }
    }

    public static void main(String args[]) throws IOException {
        input();
    //  algorithm();
        output();
    }

}

做的麻烦了点 乒乓球分制不太懂 看了一下题解第一个的明白了！

#include <iostream>
#include <cmath>
using namespace std;
int main(){
char a[100000];
int hua=0,dui=0,i=0;
while ((a[i] = getchar()) != EOF) {
if (a[i] == 'E')
break;
i++;
}
for(int i=0;;i++){
if(a[i]=='W') hua++;
else if(a[i]=='L') dui++;
if(((hua==11||dui==11)&&(abs(hua-dui)>=2))||((hua>11||dui>11)&&(abs(hua-dui)==2))||a[i]=='E'){
cout<<hua<<":"<<dui<<endl;
hua=0;
dui=0;
}

if(a[i]=='E') break;
}
cout<<endl;
for(int i=0;;i++){
if(a[i]=='W') hua++;
else if(a[i]=='L') dui++;
if(((hua==21||dui==21)&&(abs(hua-dui)>=2))||((hua>21||dui>21)&&(abs(hua-dui)==2))||a[i]=='E'){
cout<<hua<<":"<<dui<<endl;
hua=0;
dui=0;
}
if(a[i]=='E') break;
}

}