CF792B Counting-out Rhyme(约瑟夫环）

B. Counting-out Rhyme

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

n children are standing in a circle and playing the counting-out game. Children are numbered clockwise from 1 to n. In the beginning, the first child is considered the leader. The game is played in k steps. In the i-th step the leader counts out ai people in clockwise order, starting from the next person. The last one to be pointed at by the leader is eliminated, and the next player after him becomes the new leader.

For example, if there are children with numbers [8, 10, 13, 14, 16] currently in the circle, the leader is child 13 and ai = 12, then counting-out rhyme ends on child 16, who is eliminated. Child 8 becomes the leader.

You have to write a program which prints the number of the child to be eliminated on every step.

Input

The first line contains two integer numbers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ n - 1).

The next line contains k integer numbers a1, a2, ..., ak (1 ≤ ai ≤ 109).

Output

Print k numbers, the i-th one corresponds to the number of child to be eliminated at the i-th step.

Examples

input

Copy

7 5
10 4 11 4 1

output

4 2 5 6 1 

input

Copy

3 2
2 5

output

3 2 

Note

Let's consider first example:

• In the first step child 4 is eliminated, child 5 becomes the leader.
• In the second step child 2 is eliminated, child 3 becomes the leader.
• In the third step child 5 is eliminated, child 6 becomes the leader.
• In the fourth step child 6 is eliminated, child 7 becomes the leader.
• In the final step child 1 is eliminated, child 3 becomes the leader.

就是一个简单的约瑟夫环问题，

要求输出每次被淘汰的人。

估计很多人交错的原因就是时间超限（我也T了一发）

由于给的Ai。数据会很大1 ≤ ai ≤ 109，

可以每次把Ai对对列长度取一下余;

就可以解决时间超限的问题。

AC代码：

#include <iostream>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<deque>
using namespace std;
int main()
{
     int n,i,j,k,a[210000],b[210000],x,y,z,m;
    deque<int>p;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=1;i<=n;i++)
        p.push_back(i);
        for(i=0;i<m;i++)
        {
            a[i]%=p.size();
            for(j=1;j<=a[i];j++)
            {
                p.push_back(p.front());
                p.pop_front();
            }
            if(i==0)
            printf("%d",p.front());
            else
            printf(" %d",p.front());
            p.pop_front();

        }
    }
    return 0;
}点击打开链接