codeforces 876B Divisiblity of Differences

codeforces 876B 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 No Input 4 3 5 2 7 7 7 Output Yes 2 7 7 题意要求我们找出能否在这n个数中找出k个数他们两两差可以整除m 那么我们以 m为分界线 然后按照m的余数划分 如果余数相同的那么我们一定满足两两之间差整除m

#include#include#define N 110000using namespace std;int cha[N],a[N],n,k,m,ans,last,cnt[N],top;int main(){ //freopen("cf.in","r",stdin); scanf("%d%d%d",&n,&k,&m); for (int i=1;i<=n;++i){ scanf("%d",&a[i]);int mm=a[i]%m; cnt[mm]++; } for (int i=0;i=k){ printf("Yes\n"); for (int j=1;j<=n;++j) if(a[j]%m==i) { printf("%d,a[j]);ans++;if (ans==k) break; }return 0; }printf("No"); return 0;}

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