Counting-out Rhymecodeforces

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 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.

题目大意：给出一个序列，该序列首位相连。每次删除若干步的点，该点后的一点为新的起始点。

解体思路：每次删除的点为t=(t+v[i]) % (n - i)。在删除每一个元素后重新进行排列。可使用vector，简化重新排列。

#include <iostream> #include <vector> #include <cstdio> using namespace std; int main() { int n, k, t = 0; cin >> n >> k; vector<int> a(n); int v[110], ans[110]; for(int i = 0; i < n; ++i) { a[i] = i + 1; } for(int i = 0; i < k; ++i) { cin >> v[i]; } for(int i = 0; i < k; ++i) { t = (t + v[i]) % (n - i); ans[i] = a[t]; a.erase(a.begin() + t); } for(int i = 0; i < k; ++i) { cout << ans[i] << " "; } cout << endl; return 0; }