CF B. Divisiblity of Differences【water+WA9】

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

这是一道非常水的问题，但是因为我的考虑不周和一开始题意理解的偏差。
一共wrong answer了9发。心态直接爆炸，再加上A被hack，C也wrong answer。这场比赛下来，说实话的确是吓了一跳。
是简单题，就越来越不服，思路就弄不好，心态就越来越暴躁。
是应当好好再斟酌一下的。

代码：

#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define ll long long intconst ll maxn = 1e5 + 7;#define INF 0x3f3f3f3fstruct node{    ll val;    ll id;}b[maxn];bool cmp(node a, node b){    return a.val < b.val;}ll a[maxn];int main(){    ll n, k, m;    while (cin >> n >> k >> m)    {        for (int i = 1; i <= n; i++)        {            cin >> a[i];            b[i].val = a[i] % m;            b[i].id = i;        }        int flag = 0, ans = -1;        sort(b + 1, b + 1 + n, cmp);        b[0].val = INF;        for (int i = 2; i <= n; i++)        {            if (b[i].val == b[i - 1].val)            {                flag++;                if (b[i - 1].val != b[i - 2].val)                {                    flag++;                }                if (flag >= k)                {                    ans = b[i].val;                    break;                }            }            else            {                flag = 0;            }        }        if (ans == -1)        {            cout << "No" << endl;        }        else        {            cout << "Yes" << endl; int cnt = 0;            for (int i = 1; i <= n; i++)            {                if (cnt == k)                    break;                if (a[i] % m == ans&&cnt == 0)                {                    cout << a[i];                    cnt++;                }                else if (a[i] % m == ans&&cnt)                {                    cout << " " << a[i];                    cnt++;                }            }            cout << endl;        }    }    return 0;}