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  • UVa 725 Division --- 简单枚举

    题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=666 

    /* UVa 725 Division --- 简单枚举 */
    #include <cstdio>
    #include <cstring>
    
    bool used[10];
    
    /* 判断传进来的两个数是否满足条件 */
    bool judge(int a, int b)
    {
        if (a > 98765)
            return 0;
        memset(used, 0, sizeof used);
        if (b < 10000)
            used[0] = 1;
        while (b){
            used[b % 10] = 1;
            b /= 10;
        }
        while (a){
            used[a % 10] = 1;
            a /= 10;
        }
        int sum = 0;
        for (int i = 0; i < 10; ++i)
            sum += used[i];
        return sum == 10;
    }
    
    int main()
    {
        int n, kase = 0;
    
        while (scanf("%d", &n)== 1 && n){
            if (kase++){
                printf("
    ");
            }
            int cnt = 0;
            for (int i = 1234; i <= 54321; ++i){
                if (judge(i*n, i)){
                    printf("%05d / %05d = %d
    ", i*n, i, n);
                    ++cnt;
                }
            }//for(i)
            if (cnt == 0){
                printf("There are no solutions for %d.
    ", n);
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/tommychok/p/5055021.html
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