Joseph UVA 1452 Jump

题目传送门

 /*
     数学：约瑟夫环问题的变形，首先定义f[i]表示剩下i个人时，最后一个选出的人，有个公式：f[i] = (f[i-1] + m) % i
         f[1] = 0(编号从0开始)，那么类似最后一个数的求法，先找到剩2个人和剩3个人时，最后的编号，然后跟着最后一个的一起递推
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-8 14:26:38
 * File Name     :UVA_1459.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 5e5 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;

 void Joseph(int n, int m)    {
     int ans1 = , ans2 = , ans3 = ;
     for (int i=; i<=n; ++i)    {
         ans1 = (ans1 + m) % i;
         if (i == ) {       //2个人就是0或1
             ans2 = !ans1;
         }
         else if (i == )    {
             ans2 = (ans2 + m) % i;
             bool vis[];    memset (vis, false, sizeof (vis));
             vis[ans1] = vis[ans2] = true;
             for (int i=; i<; ++i) {
                 if (!vis[i])    {
                     ans3 = i;   break;
                 }
             }
         }
         else    {
             ans2 = (ans2 + m) % i;
             ans3 = (ans3 + m) % i;
         }
     }

     printf ("%d %d %d\n", ans3 + , ans2 + , ans1 + );
 }

 int main(void)    {     //UVA 1452 Jump
     int T;  scanf ("%d", &T);
     while (T--) {
         int n, m;   scanf ("%d%d", &n, &m);
         Joseph (n, m);
     }

     return ;
 }