ACM刷题之codeforce————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.

挺有意思的一道题。

题意：给你一列数，让你从中找出k个数，满足任意两两相减能够整除m。

如果暴力的话肯定是超时的，我们这里要用到取余的思想。

如果两个数相减后的结果能够被m整除，那么他们的余数肯定是相等的。（自己想想）

按照这个思想，就可以做下去了，扫一次，把数字的放到set里来判重复。

下面是ac代码：

#include<bits/stdc++.h>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=1e5+7;
int a[N];
int q[N];
set<int> si;

void outAns(int j, int maxi, int ke, int x) {
	
	printf("Yes\n%d", a[j]);
	int cnt = 1;
	for(int i = j + 1; i < maxi && cnt <x; i++) {
		if (q[i] == ke) {
			
			++cnt;
			printf(" %d", a[i]);	
		}

	}
	printf ("\n");
	return ;
}

int main()
{
	//freopen("f:/input.txt", "r", stdin);
	int b,c,d,i,j,k,m,n;
	cin>>n>>k>>m;
	//CLR(a);
	for (i = 0 ; i < n; i++) {
		scanf("%d", &a[i]);
		q[i] = a[i]%m;
	}
	
	int cnt=0;
	for (i = 0 ; i < n; i++) {
		if(si.find(q[i]) == si.end()) {
			cnt = 1;
			for(j = i + 1; j < n && cnt < k && n-j + 1>k-cnt; j++) {
				if (q[j] == q[i]) ++cnt;
			}
			if (cnt >= k) {
				outAns(i, n, q[i], k);
				return 0;
			}
			si.insert(q[i]);
		}
	}
	printf ("No\n");
}

ACM刷题之codeforce————Divisiblity of Differences

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