Josephus Problem

题目描述

Do you know the famous Josephus Problem? There are n people standing in a circle waiting to be executed. The counting out begins at the first people in the circle and proceeds around the circle in the counterclockwise direction. In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom.

In traditional Josephus Problem, the number of people skipped in each round is fixed, so it's easy to find the people executed in the i-th round. However, in this problem, the number of people skipped in each round is generated by a pseudorandom number generator:

x[i+1] = (x[i] * A + B) % M.

Can you still find the people executed in the i-th round?

输入

There are multiple test cases.

The first line of each test cases contains six integers 2 ≤ n ≤ 100000, 0 ≤ m ≤ 100000, 0 ≤ x[1], A, B < M ≤ 100000. The second line contains m integers 1 ≤ q[i] < n.

输出

For each test case, output a line containing m integers, the people executed in the q[i]-th round.

样例输入

2 1 0 1 2 3
1
41 5 1 1 0 2
1 2 3 4 40

样例输出

1
2 4 6 8 35

#include
#include 
#include
using namespace std;
typedef long long LL;
#define maxn 200005
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
LL sum[4*maxn],a[maxn],c[maxn];///一般线段树sum[]开4*n为保守空间大小，这里开小了就可能会报错
void build(LL l,LL r,LL rt)   ///建线段树
{
    sum[rt]=r-l+1;
    LL m=(l+r)>>1;
    if(l==r)
        return;
    build(lson); ///建左树
    build(rson); ///建右树
} 
int update(LL l,LL r,LL rt,LL num) ///单个结点的更新
{ 
    sum[rt]--;///一人出队后包含该结点的区间都减一
    LL m=(l+r)/2; 
    if(l==r) return l;
    LL t; 
    if(num<=sum[rt<<1]) ///左区间的对应结点的更新
        t=update(lson,num); 
    else          ///右区间的对应结点的更新
        t=update(rson,num-sum[rt<<1]); 
    return t; 
} 
int main() 
{ 
    LL n,m,A,B,M,x,i;
    while(~scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&x,&A,&B,&M)) 
    { 
        for(i=0; i