B. Counting-out Rhyme（约瑟夫环）

For example, if there are children with numbers [8,?10,?13,?14,?16] currently in the circle, the leader is child 13 and a_i?=?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.

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,?...,?a_k (1?≤?a_i?≤?10⁹).

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

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.

题目意思：

n个人围成一圈坐着，每个人（编号定为1~n）有一个数字a（限定条件：a的范围为1~n，且每个数字只出现一次），

现在如果指定一个人为leader，下一个leader为从他的顺时针方向第一个人开始数 a .以此类推，现在给出一系列按顺序成为leader的人的编号，

问是否存在这样一种情况，能够在限定条件内确定每个人的数字，满足成为leader顺序的结果，如果有多种结果，输出一种，没有输出 -1。

解题思路：这其实就是一道约瑟夫环问题

 1 #include<stdio.h>
 2 #include<deque>
 3 #include<string.h>
 4 using namespace std;
 5 int main()
 6 {
 7     int n,k,i,j,b[100],a,c;
 8         deque<int> q;
 9     scanf("%d%d",&n,&k);
10     for(i=1;i<=n;i++)
11     {
12         q.push_back(i);
13     }
14     for(i=1;i<=k;i++)
15     {
16         scanf("%d",&b[i]);
17     }
18     for(i=1;i<=k;i++)
19     {
20         b[i]=b[i]%n;
21         for(j=1;j<=b[i];j++)
22         {
23             a=q.front();
24             q.pop_front();
25             q.push_back(a);
26         }
27         c=q.front();
28         printf("%d ",c);
29         q.pop_front();
30         n--;
31     }
32     return 0;
33 }