20分求助0.o

LegendPei @ 2024-10-07 22:30:52

#include <stdio.h>  
#include<math.h>
int main()
{
    int str, numbers[100000] = { 0 }, i = 0;
    while (1)
    {
        str = getchar();
        if (str == '\n')continue;
        if (str == 'W')numbers[i] = 1;
        if (str == 'L')numbers[i] = 2;
        i++;
        if (str == 'E')break;
    }
    int w = 0, l = 0;
    for (int j = 0; 1; j++)
    {
        if (numbers[j] == 1)w++;
        if (numbers[j] == 2)l++;
        if ((abs(w - l) >= 2 ) && (w >= 11 || l >= 11))
        {
            printf("%d:%d\n", w, l);
            w = 0, l = 0;
        }
        if (numbers[j] == 0)
        {
            printf("%d:%d\n\n", w, l);
            break;
        }
    }
    w = 0, l = 0;
    for (int j = 0; 1; j++)
    {
        if (numbers[j] == 1)w++;
        if (numbers[j] == 2)l++;
        if ((abs(w - l) >= 2 ) && (w >= 21 || l >= 21))
        {
            printf("%d:%d\n", w, l);
            w = 0, l = 0;
        }
        if (numbers[j] == 0)
        {
            printf("%d:%d", w, l);
            break;
        }
    }
    return 0;
}

by Martin_L @ 2024-10-08 20:38:49

这题的问题在于，万一它的换行不是\n，你的代码就行不通了；因此，建议用 if else-if else 的结构排除。

完整代码：

#include <stdio.h>  
#include<math.h>
int main()
{
    int str, numbers[100000] = { 0 }, i = 0;
    while (1)
    {
        str = getchar();
        if (str == 'E')break;
        if (str == 'W')numbers[i] = 1;
        else if (str == 'L')numbers[i] = 2;
        else continue;
        i++;
    }
    int w = 0, l = 0;
    for (int j = 0; 1; j++)
    {
        if (numbers[j] == 1)w++;
        if (numbers[j] == 2)l++;
        if ((abs(w - l) >= 2 ) && (w >= 11 || l >= 11))
        {
            printf("%d:%d\n", w, l);
            w = 0, l = 0;
        }
        if (numbers[j] == 0)
        {
            printf("%d:%d\n\n", w, l);
            break;
        }
    }
    w = 0, l = 0;
    for (int j = 0; 1; j++)
    {
        if (numbers[j] == 1)w++;
        if (numbers[j] == 2)l++;
        if ((abs(w - l) >= 2 ) && (w >= 21 || l >= 21))
        {
            printf("%d:%d\n", w, l);
            w = 0, l = 0;
        }
        if (numbers[j] == 0)
        {
            printf("%d:%d", w, l);
            break;
        }
    }
    return 0;
}

by Bennyboy2024 @ 2024-10-08 22:39:27

看了下大佬的代码。我刚上过一个夏令营，天赋不行。学的很差。不敢用getchar(), 等。

看了下，这里吧str 改 char 类型，然后 #include <iostream> 最后把 getchar() 改 std::cin 就可以了。

#include <stdio.h>  
#include<math.h>
#include <iostream> //用下大佬不用的cin
int main()
{
    int numbers[100000] = { 0 }, i = 0;
    char str; // int 改成 char
    while (1)
    {
        std::cin >> str; // getchar() 变 cin 就 AC 了
        if (str == '\n')continue;
        if (str == 'W')numbers[i] = 1;
        if (str == 'L')numbers[i] = 2;
        i++;
        if (str == 'E')break;
    }