Codeforces 876B Divisiblity of Differences 题解

B. Divisiblity of Differences
time limit per test1 second
memory limit per test512 megabytes
inputstandard input
outputstandard 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

给n个数，求能否选k个数膜m结果相同，若有输出这k个数，没有输出No。

记录将每个数膜m之后的数的个数，直接判断第一个大于k的数输出。复杂度O(n)

#include 

using namespace std;

int n,k,m,cnt;
int a[100005],num[100005],b[100005];

int main() {
    scanf("%d%d%d",&n,&k,&m);
    for(int i = 1; i <= n; i++) scanf("%d",&a[i]), b[i] = a[i]%m, num[b[i]]++;
    for(int i = 1; i <= n; i++) {
        if(num[b[i]] >= k) {
            printf("Yes\n");
            int tot = 0;
            for(int t = 1; t <= n; t++) {
                if(b[i] == b[t]) printf("%d ",a[t]), tot++;
                if(tot == k) break;
            }
            return 0;
        }
    }
    printf("No");
    return 0;
}