NOIP2003乒乓球问题

Question

这是题目和题目所给的测试样本，看不懂题目的输入输出之间的关系，也不知道这个字母E表示跳过循环，还是跳出程序

Answer 1

输入字符串后，循环遍历，统计每行L和W的个数，11分制就是11个字符一组，21分制就是21个字符一组

Answer 2

代码如下：

#define _CRT_SECURE_NO_WARNINGS 1

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

int main()
{
    char s[2502][26];
    int i, j, nmbW,nmbL;
    int n = 0;
    int flag = 0;
    while (scanf("%s", s[n]) != EOF)
        n++;

    //11分制
    nmbW = 0; nmbL = 0;
    for (i = 0; i < n; i++)
    {
        for (j = 0; s[i][j] != '\0'; j++)
        {
            if (s[i][j] == 'W')
            {
                flag = 0;
                nmbW++;
            }
            else if (s[i][j] == 'L')
            {
                flag = 0;
                nmbL++;
            }
            else if (s[i][j] == 'E')
                break;
            //11分制度
            if (nmbW + nmbL == 11)
            {
                printf("%d:%d\n", nmbW, nmbL);
                nmbW = 0; nmbL = 0;
                flag = 1;
            }

        }
    }
    if(flag == 0) //避免刚好结束比赛的时候重复输出
        printf("%d:%d\n", nmbW, nmbL);
    printf("\n"); //空行
    nmbW = 0; nmbL = 0;
    flag = 0;
    for (i = 0; i < n; i++)
    {
        for (j = 0; s[i][j] != '\0'; j++)
        {
            if (s[i][j] == 'W')
            {
                flag = 0;
                nmbW++;
            }
            else if (s[i][j] == 'L')
            {
                flag = 0;
                nmbL++;
            }
            else if (s[i][j] == 'E')
                break;
            //11分制度
            if (nmbW + nmbL == 21)
            {
                printf("%d:%d\n", nmbW, nmbL);
                nmbW = 0; nmbL = 0;
                flag = 1;
            }
        }
    }
    if (flag == 0) //避免刚好结束比赛的时候重复输出
        printf("%d:%d\n", nmbW, nmbL);
    
    return 0;
}

Answer 3

运行结果

Answer 4

代码更新如下：

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
typedef struct _score
{
    int sw;
    int sl;
}Score;
int main()
{
    char ch;
    Score sa[7000], sb[7000]; //11分和21分比赛记录
    int na = 0, nb = 0; //组数
    int cw1 = 0, cl1 = 0; //11分制
    int cw2 = 0, cl2 = 0; //21分制
    int i;

    while (scanf("%c",&ch)!=EOF)
    {
        if (ch == 'E')
            break;
        else if (ch == 'W')
        {
            cw1++; cw2++;
        }
        else if (ch == 'L')
        {
            cl1++; cl2++;
        }

        if ( (cw1 >= 11 && cw1-cl1>=2) || (cl1>=11 && cl1-cw1>=2) )
        {
            sa[na].sw = cw1;
            sa[na].sl = cl1;
            na++;
            cw1 = 0; cl1 = 0;
        }
        if ( (cw2 >= 21 && cw2-cl2>=2) || (cl2>=21 && cl2-cw2 >=2) )
        {
            sb[nb].sw = cw2;
            sb[nb].sl = cl2;

            nb++;
            cw2 = 0; cl2 = 0;
        }
    }

    //没有输入，输出0:0
    if (na == 0 && nb==0 && cw1==0 && cl1==0)
    {
        printf("0:0\n\n0:0\n");
        return 0;
    }

    //输出11分制结果
    for (i = 0; i < na; i++)
        printf("%d:%d\n", sa[i].sw, sa[i].sl);
    if (cw1 != 0 || cl1 != 0)
        printf("%d:%d\n", cw1, cl1);

    printf("\n");
    //输出21分制结果
    for (i = 0; i < nb; i++)
        printf("%d:%d\n", sb[i].sw, sb[i].sl);
    if (cw2 != 0 || cl2 != 0)
        printf("%d:%d\n", cw2, cl2);

    return 0;
}