为什么我TEL了（洛谷，关于乒乓球）

Question

问题遇到的现象和发生背景

为什么我TEL了

用代码块功能插入代码，请勿粘贴截图

#include<stdio.h>
int win[100000];
int w=0, l=0;
int main()
{
    char s[100];
    int count=0;
    for (int i = 1;; i++)
    {
        scanf("%c", &s[i]);
        count++;
        {
            if (s[i] == 'W')win[i] = 1;
        else if (s[i] == 'L') win[i] = 2;
        else if (s[i] == 'E') win[i] = 0;/* break;*/
        else   win[i] = 0; 
        }
    }
    for (int i = 1;i<=count; i++)
    {
        if (win[i] == 1)w++; 
        if (win[i] == 2)l++; 
        if (win[i] == 0)
        {
            /*printf("%d:%d", w, l);*/
            break;
        }
        if (w - l >= 2 || l - w >= 2)
            if (w >= 11 || l >= 11)
            {
                printf("%d:%d\n", w, l);
                w = 0;
                l = 0;
            }
    }
    w = 0;
    l = 0;
    printf("\n");
    for (int i = 1;i<=count; i++)
    {
        if (win[i] == 1)w++;
        if (win[i] == 2)l++;
        if (win[i] == 0)
        {
            break;
        }
        if (w - l >= 2 || l - w >= 2)
            if (w >= 21 || l >= 21)
            {
                printf("%d:%d\n", w, l);
                w = 0;
                l = 0;
            }
    }
    return 0;
}

运行结果及报错内容

代码编译错误

Answer 1

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>    
#include<math.h>
#include<list>
#include<stack>

using namespace std;

struct R
{
    int ll, rr;
}R1[10005],R2[10005];

int main(int argc, char* argv[])
{
    string s;
    int l1 = 0, l2 = 0, r1 = 0, r2 = 0;
    int c1 = 0, c2 = 0;
    while (cin >> s)
    {
        int len = s.length();
        int i;
        for (i = 0; i < len; ++i)
        {
            if (s[i] == 'W')
            {
                l1++; l2++;
            }
            else if (s[i] == 'L')
            {
                r1++; r2++;
            }
            else break;
            //分差大于或者等于 2
            if ((l1 >= 11 || r1 >= 11) && abs(l1 - r1) >= 2)
            {
                R1[c1].ll = l1;
                R1[c1].rr = r1;
                c1++;
                l1 = r1 = 0;
            }
            if ((l2 >= 21 || r2 >= 21) && abs(l2 - r2) >= 2)
            {
                R2[c2].ll = l2;
                R2[c2].rr = r2;
                c2++;
                l2 = r2 = 0;
            }
        }
        if (s[i] == 'E')
        {
            R1[c1].ll = l1;
            R1[c1].rr = r1;
            c1++;
            

            R2[c2].ll = l2;
            R2[c2].rr = r2;
            c2++;
            break;
        }
    }
    for (int i = 0; i < c1; ++i)
    {
        printf("%d:%d\n", R1[i].ll, R1[i].rr);
    }
    printf("\n");
    for (int i = 0; i < c2; ++i)
    {
        printf("%d:%d\n", R2[i].ll, R2[i].rr);
    }
    return 0;
}

Answer 2

#include<stdio.h>
#define N 25*2501
 
 
char str[N];
 
int main()
{
    int length,cnt1=0,cnt2=0,cnt3=0,cnt4=0,i;
    for(i=0;;i++)
    {
        scanf("%c",str+i);
        if(str[i] == 'E')
            break;
    }
    length = i;
    for(i=0;i<=length;i++)//11分制
    {
        //printf("str[%d]=%c\n",i,str[i]);
        if(str[i]=='W')
            cnt1++;
        if(str[i]=='L')
            cnt2++;
        if(str[i]=='E')
        {
            printf("%d:%d\n",cnt1,cnt2);
            break;
        }
        if(cnt1>=11&&(cnt1-cnt2)>=2||cnt2>=11&&(cnt2-cnt1)>=2)
        {
            printf("%d:%d\n",cnt1,cnt2);
            cnt1=0;
            cnt2=0;
        }    
                
     } 
     printf("\n");
     for(i=0;i<=length;i++)//同21分制一样
    {
        if(str[i]=='W')
            cnt3++;
        if(str[i]=='L')
            cnt4++;
        if(str[i]=='E')
        {
            printf("%d:%d\n",cnt3,cnt4);
            break;
        }
        if(cnt3>=21&&(cnt3-cnt4)>=2||cnt4>=21&&(cnt4-cnt3)>=2)
        {
            printf("%d:%d\n",cnt3,cnt4);
            cnt3=0;
            cnt4=0;
            }            
     } 
    return 0;
}

Answer 3

记得采纳哦