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  • poj2244

    题意:约瑟夫问题变形,给定人数和最后存活的编号,问杀人间隔最少是多少。

    分析:从小到大枚举杀人间隔,然后用解约瑟夫问题的方法,求最后存活的人,如果恰好是要求的那个,则输出结果。

    View Code
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    using namespace std;
    
    int n;
    
    int last(int n, int m)
    {
        int ret = 0;
        for (int i = 2; i <= n; i++)
            ret = (ret + m) % i;
        return ret;
    }
    
    int main()
    {
        //freopen("t.txt", "r", stdin);
        while (scanf("%d", &n), n)
        {
            int i = 1;
            while (last(n - 1, i) != 0)
                i++;
            printf("%d\n", i);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/rainydays/p/2575663.html
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