题目传送门
1 /*
2 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
3 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
4 */
5 /************************************************
6 * Author :Running_Time
7 * Created Time :2015-8-8 14:26:38
8 * File Name :UVA_1459.cpp
9 ************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 5e5 + 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36
37 void Joseph(int n, int m) {
38 int ans1 = 0, ans2 = 0, ans3 = 0;
39 for (int i=2; i<=n; ++i) {
40 ans1 = (ans1 + m) % i;
41 if (i == 2) { //2个人就是0或1
42 ans2 = !ans1;
43 }
44 else if (i == 3) {
45 ans2 = (ans2 + m) % i;
46 bool vis[3]; memset (vis, false, sizeof (vis));
47 vis[ans1] = vis[ans2] = true;
48 for (int i=0; i<3; ++i) {
49 if (!vis[i]) {
50 ans3 = i; break;
51 }
52 }
53 }
54 else {
55 ans2 = (ans2 + m) % i;
56 ans3 = (ans3 + m) % i;
57 }
58 }
59
60 printf ("%d %d %d
", ans3 + 1, ans2 + 1, ans1 + 1);
61 }
62
63 int main(void) { //UVA 1452 Jump
64 int T; scanf ("%d", &T);
65 while (T--) {
66 int n, m; scanf ("%d%d", &n, &m);
67 Joseph (n, m);
68 }
69
70 return 0;
71 }