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  • 20140222

    #include<stdio.h>
    void main()
    {
        int n, m, i, p, k, w[100];
        scanf("%d%d", &n, &m);
        for (i = 0; i < n; i++)
            w[i] = i + 1;
        i = 0;
        k = 1;
        p = 0;
        while (p < n - 1)
        {
            if (w[i] == -1)
            {
                i = (i + 1) % n;
                continue;
            }
            if (k == m)
            {
                w[i] = -1;
                p++;
                k = 0;
            }
            k++;
            i = (i + 1) % n;
        }
        for (i = 0; i < n;i++)
        if (w[i] != -1)
        {
            printf("%d
    ", w[i]);
            break;
        }
    }
    
    while (!(a == R || b == R || c == R))
    {
        if (!b)
        {
            a -= bv;
            b = bv;
        }
        else if (c == cv)
        {
            a += cv;
            c = 0;
        }
        else if (b > cv - c)
        {
            b -= cv - c;
            c = cv;
        }
        else
        {
            c += b;
            b = 0;
        }
        printf(a, b, c);
    }
    
    #include<stdio.h>
    void main()
    {
        int a, b, c, bv, cv, R, n;
        scanf("%d%d%d", &a, &bv, &cv);
        R = a / 2;//平分
        b = c = 0;
        n = 0;
        while (!(a == R || b == R || c == R))
        {
            if (!b)
            if (a < bv)
            {
                n = -1;
                break;
            }
            else//无法倒满b,则不能分
            {
                a -= bv;
                b = bv;
            }
            else if (c == cv)//c倒满a
            {
                a += cv;
                c = 0;
            }
            else if (b > cv - c)//b能倒满c
            {
                b -= cv - c;
                c = cv;
            }
            else//b不能倒满c
            {
                c += b;
                b = 0;
            }
            n++;
            printf("%d  %d  %d
    ", a, b, c);
        }
        printf("%d
    ", n);
    }
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  • 原文地址:https://www.cnblogs.com/yuanqi/p/3561201.html
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