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  • P1328 生活大爆炸版石头剪刀布

    一道模拟题,20个if搞定

    #include<cstdio>
    using namespace std;
    int a[202], b[202];
    int main()
    {
        int n, na, nb, sa = 0, sb = 0;
        scanf("%d %d %d", &n, &na, &nb);
        for(int i = 0; i < na; i++)
            scanf("%d", &a[i]);
        for(int i = 0; i < nb; i++)
            scanf("%d", &b[i]);
        for(int i = 0; i < n; i++)
        {
            if(a[i % na] == 0 && b[i % nb] == 1) sb++;
            if(a[i % na] == 0 && b[i % nb] == 2) sa++;
            if(a[i % na] == 0 && b[i % nb] == 3) sa++;
            if(a[i % na] == 0 && b[i % nb] == 4) sb++;
            if(a[i % na] == 1 && b[i % nb] == 0) sa++;
            if(a[i % na] == 1 && b[i % nb] == 2) sb++;
            if(a[i % na] == 1 && b[i % nb] == 3) sa++;
            if(a[i % na] == 1 && b[i % nb] == 4) sb++;
            if(a[i % na] == 2 && b[i % nb] == 0) sb++;
            if(a[i % na] == 2 && b[i % nb] == 1) sa++;
            if(a[i % na] == 2 && b[i % nb] == 3) sb++;
            if(a[i % na] == 2 && b[i % nb] == 4) sa++;
            if(a[i % na] == 3 && b[i % nb] == 0) sb++;
            if(a[i % na] == 3 && b[i % nb] == 1) sb++;
            if(a[i % na] == 3 && b[i % nb] == 2) sa++;
            if(a[i % na] == 3 && b[i % nb] == 4) sa++;
            if(a[i % na] == 4 && b[i % nb] == 0) sa++;
            if(a[i % na] == 4 && b[i % nb] == 1) sa++;
            if(a[i % na] == 4 && b[i % nb] == 2) sb++;
            if(a[i % na] == 4 && b[i % nb] == 3) sb++;
        }
        printf("%d %d", sa, sb);
    }
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  • 原文地址:https://www.cnblogs.com/njbw7782/p/10356227.html
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