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最新推荐文章于 2021-04-23 11:24:47 发布

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本文介绍了一种算法问题，即从一个多集中选择大小为k的子集，使得子集中任意两数之差可被m整除。通过按余数分组的方式解决此问题，并提供了一段实现该算法的C++代码。

Divisiblity of Differences

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.

Examples

Input

3 2 3
1 8 4

Output

Yes
1 4

Input

3 3 3
1 8 4

Output

Input

4 3 5
2 7 7 7

Output

Yes
2 7 7
题意:给你n个数，要你找出一个大小为k的子集，要求子集中任意两个数的差的绝对值能被m整除。
思路:n的范围很大，暴力超时。而当两个数的差的绝对值能被m整除时，他们%m的值绝对相等。所以我们按照余数来建立子集，当子集大小大于k时有解。

#include <iostream>
#include <fstream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
#include <functional>
#define inf 0x7fffffff
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
const int MAX=1e5+10;
const double eps=1e-6;

int n,k,m;
int s[MAX];
vector<int>ans[MAX];

int main(){
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    #endif

    cin>>n>>k>>m;
    int flag=0;
    for(int i=0;i<n;i++){
        scanf("%d",&s[i]);
        ans[s[i]%m].push_back(s[i]);
    }
    for(int i=0;i<m;i++){
        if(ans[i].size()>=k){
            cout<<"Yes"<<endl;
            for(int j=0;j<k;j++){
                if(j==0)
                    cout<<ans[i][j];
                else
                    cout<<" "<<ans[i][j];
            }
            cout<<endl;
            flag++;
            break;
        }
    }
    if(!flag){
        cout<<"No"<<endl;
    }

    return 0;   
}