CodeForces 876B-蒲公英云

CodeForces 876B

短命女 2022-06-07 10:48 297阅读 0赞

问题描述:

You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible.

Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.

Input

First line contains three integers n, k and m (2 ≤ k ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.

Second line contains n integers a1, a2, …, an (0 ≤ ai ≤ 109) — the numbers in the multiset.

Output

If it is not possible to select k numbers in the desired way, output «No» (without the quotes).

Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print k integers b1, b2, …, bk — the selected numbers. If there are multiple possible solutions, print any of them.

Example

Input

3 2 3
1 8 4

Output

Yes
1 4

Input

3 3 3
1 8 4

Output

No

Input

4 3 5
2 7 7 7

Output

Yes
2 7 7

题目题意:题目给我们一个含有n个数的数字集合，问我们能否找到一个含有k个数的数集，使得任意俩个数之差都可以被m整除。

题目分析:俩个之差能被m整除，那么这俩个对m取余，余数相等。

a1=k1*m+b a2=k2*m+b 则(a1-a2)%m=(k1-k2)*m %m =0

代码如下:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int maxn=1e5+100;
vector<int> vec[maxn];
int main()
{
    int n,k,m;
    scanf("%d%d%d",&n,&k,&m);
    for (int i=0;i<n;i++) {
        int a,b;
        scanf("%d",&a);
        b=a%m;
        vec[b].push_back(a);
    }
    bool flag=false;
    for (int i=0;i<m;i++) {
        if ((vec[i].size())>=k) {
                puts("Yes");
                flag=true;
            for (int j=0;j<k;j++) {
                if (j==k-1) printf("%d\n",vec[i][j]);
                else printf("%d ",vec[i][j]);
            }
        }
       if (flag) break;
    }
    if (!flag)
        puts("No");
    return 0;
}