Topic link: https://nanti.jisuanke.com/t/41352

The meaning of the title is easy to understand. Not many people have seen it. They feel frightened by the amount of passage. In fact, it's a water problem. Just reverse simulation.

Using queue simulation, reverse simulation, it needs to put m cards in the back, then I put m cards in the front. At first, there was only one card in the queue, and slowly added to n cards.

First increase the card, then all the way to 1 card. By the way, there may be only five cards, but m is 6, 7, 8 or 9, 10. So in order to reduce unnecessary simulation,

Use mod to reduce, because some simulation will make the card compare with before, it will remain intact.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <stack>
 7 #include <string>
 8 #include <map>
 9 #include <cmath>
10 #include <iomanip>
11 using namespace std;
12  
13 typedef long long LL;
14 #define inf 1e9
15 #define rep(i,j,k) for(int i = (j); i <= (k); ++i)
16 #define rep__(i,j,k) for(int i = (j); i < (k); ++i)
17 #define per(i,j,k) for(int i = (j); i >= (k); --i)
18 #define per__(i,j,k) for(int i = (j); i > (k); --i) 
19 
20 const int N = 40000010;
21 int arr[N];
22 int mod,n,q;
23 queue<int> que;
24 
25 void work(){
26 
27     int tmp;
28     que.push(n);//Press in the largest card
29     int num = 1;
30     int end = n - 1;//The last 1 card is taken away directly, so there is no need to simulate that round.
31     while(num <= end){
32         tmp = mod % num;
33 
34         rep(o,1,tmp){
35             que.push(que.front());
36             que.pop();
37         }
38         que.push(n-num);//Press in the previous card
39         ++num;
40     }
41 //    que.push(1);
42 }
43 
44 int main(){
45 
46     int T;
47     scanf("%d",&T);
48 
49     while(T--){
50         scanf("%d%d",&n,&mod);
51 
52         rep(i,1,n){
53             arr[i] = i;//1-n Card
54         }
55         scanf("%d",&q);
56 
57         work();
58 
59         per(i,n,1){//Remove card
60             arr[i] = que.front();
61             que.pop();
62         }
63 
64         int o;
65         rep(i,1,q){
66             scanf("%d",&o);
67             printf("%d\n",arr[o]);
68         }
69 
70 //       rep(i,1,n) printf("%d ",arr[i]);
71     }
72 
73     getchar(); getchar();
74     return 0;
75 }