CodeForces - 792B

最新推荐文章于 2018-08-23 17:36:46 发布

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#CodeForces - 792B

本文介绍了一种计数淘汰赛游戏的算法实现，n名儿童围成一圈按顺时针编号从1到n，通过指定步数逐步淘汰玩家直至决出胜者。文章详细描述了游戏规则及示例，并提供了一份AC代码实现。

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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
7 5
10 4 11 4 1

output
4 2 5 6 1 

input
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.

ac代码：

#include <bits/stdc++.h>
#include <queue>
#define ll long long

using namespace std;
int n,k;
int a[105];

int main(){

	while(~scanf("%d%d",&n,&k)){
		queue <int>qu;
		for(int i=1;i<=k;i++){
			scanf("%d",&a[i]);
		}
		for(int i=1;i<=n;i++){
			qu.push(i);
		}
		int aun=n;
		int cnt=0;
		int kai[110];
		for(int i=1;i<=k;i++){
			int tmp=(a[i]+1)%aun;
			if(tmp==0)
				tmp=aun;
			while(tmp--){
				int ss=qu.front();
				qu.pop();
				if(tmp==0){
					kai[cnt]=ss;
					cnt++;
					aun--;
				}
				else{
					qu.push(ss);
				}
			}
		}
		printf("%d",kai[0]);
		for(int i=1;i<cnt;i++){
			printf(" %d",kai[i]);
		}
		printf("\n");
	}
	
	return 0;
}