Codeforces 876B Divisiblity of Differences

B. Divisiblity of Differences
time limit per test
1 second
memory limit per test
512 megabytes
input
standard input
output
standard output
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

No

input

4 3 5
2 7 7 7

output

Yes
2 7 7 

또 간단한 문제야.
제목: n개수에서 k개수를 꺼내서 이 k개수 두 두 개의 차이를 m의 배수로 만든다.
개수마다%m, 나머지가 같으면 그 차이는 m의 배수가 맵다~~

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
int n,k,m,s,num,q[200200],a[200200],sum[200200];
int main()
{
	scanf("%d%d%d",&n,&k,&m);
	for(int i=1;i<=n;i++){scanf("%d",&a[i]);q[i]=a[i]%m;}
	for(int i=1;i<=n;i++) sum[q[i]]++;
	bool flag=false;
	int i;
	for(i=0;i=k){flag=true;break;}
	if(flag){
		puts("Yes");
		num=i;
		for(int i=1;i<=n;i++)
			if(q[i]==num){
				s++;
				if(s==k){printf("%d
",a[i]);break;}
				else printf("%d ",a[i]);
			}
	}else puts("No");
	return 0;
}