题目链接:uva 1394 - And Then There Was One
题目大意:给出n,k和m,表示有n个人围成一个圈,从第m个人开始(m也要去掉),每次走k步删除掉,问最后剩下人的序号。
解题思路:约瑟夫环的小变形,套公式dp[i] = (dp[i-1] + k)%i。
#include <stdio.h> int main () { int n, k, m; while (scanf("%d%d%d", &n, &k, &m) == 3 && n + k + m) { int ans = 0; for (int i = 2; i < n; i++) ans = (ans + k) % i; printf("%d\n", (ans + m) % n + 1); } return 0; }
uva 1394 - And Then There Was One(约瑟夫环)
原文:http://blog.csdn.net/keshuai19940722/article/details/19342605