您的位置：首页 > 产品设计 > UI/UE

湘潭市赛 Josephus Problem 线段树

2012-07-22 23:43 471 查看

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1149

Josephus Problem
Accepted : 44		Submit : 334
Time Limit : 5000 MS		Memory Limit : 65536 KB

Josephus Problem

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?
Input
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.
Output
For each test case, output a line containing m integers, the people executed in the q[i]-th round.
Sample Input
2 1 0 1 2 3

1

41 5 1 1 0 2

1 2 3 4 40

Sample Output
1

2 4 6 8 35

#include<iostream>

#include<stdio.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

int ans[100001];

int xw[100001];

const int maxn=200000;

int w,sum[maxn<<2];

void build(int l,int r,int rt){///建立线段树

    sum[rt] = r - l + 1;

    if(l == r) return ;

    int m = (l+r) >> 1;

    build(lson);

    build(rson);

}

int update(int p,int l,int r,int rt){///更新单个节点



    sum[rt]--;

    if(l == r) return l ;

    int m = (l + r) >> 1;

    if(p <= sum[rt<<1])

        return    update(p,lson);

    else

        return   update(p-sum[rt<<1],rson);



}

int main()

{

    int n,m,i;

    int x,A,B,M;

    while(scanf("%d%d%d%d%d%d",&n,&m,&x,&A,&B,&M)!=EOF)

    {

        build(1,n,1);

        int z = 1;

        for(i = 1; i <= n; i++)

        {

            z = ((int)x+z)%sum[1];

            if(z == 0) z = sum[1];

            int s = update(z,1,n,1);

            ans[i] = s;

            x = (int)(((__int64)x * A + B) % M);

        }

        for(i = 0; i < m; i++)

            scanf("%d",&xw[i]);





        for(i=0;i<m-1;i++)

            printf("%d ",ans[xw[i]]);

        if(m!=0) printf("%d",ans[xw[m-1]]);

        printf("\n");

    }

    return 0;

}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签： each build generator output input

相关文章推荐

新的分享

章节导航