CodeForces

题意：给定n个数，从中选取k个数，使得任意两个数之差能被m整除，若能选出k个数，则输出，否则输出“No”。

分析：

1、若k个数之差都能被m整除，那么他们两两之间相差的是m的倍数，即他们对m取余的余数是相同的。

2、记录n个数对m取余的余数，计算出数量最多的余数ma。

3、ma>=k，才能选出，并输出即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int a[MAXN];
int yu[MAXN];
map<int, int> mp;
int main(){
    int n, k, m;
    scanf("%d%d%d", &n, &k, &m);
    for(int i = 0; i < n; ++i){
        scanf("%d", &a[i]);
    }
    for(int i = 0; i < n; ++i){
        yu[i] = a[i] % m;
        ++mp[yu[i]];
    }
    int ma = 0;
    int id = 0;
    for(map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it){
        if((*it).second > ma){
            ma = (*it).second;
            id = (*it).first;
        }
    }
    if(ma < k){
        printf("No
");
    }
    else{
        printf("Yes
");
        bool flag = true;
        int t = 0;
        for(int i = 0; i < n; ++i){
            if(yu[i] == id){
                if(flag) flag = false;
                else printf(" ");
                printf("%d", a[i]);
                ++t;
                if(t == k) break;
            }
        }
        printf("
");
    }
    return 0;
}