Codeforces Round #544 (Div. 3) Codeforces Round #544 (Div. 3)

International Women's Day is coming soon! Polycarp is preparing for the holiday.
There are
n
n
candy boxes in the shop for sale. The
i
i
-th box contains
d
i
di
candies.
Polycarp wants to prepare the maximum number of gifts for
k
k
girls. Each gift will consist of exactly two boxes. The girls should be able to share each gift equally, so the total amount of candies in a gift (in a pair of boxes) should be divisible by
k
k
. In other words, two boxes
i
i
and
j
j
(
i≠j
i≠j
) can be combined as a gift if
d
i
+
d
j
di+dj
is divisible by
k
k
.
How many boxes will Polycarp be able to give? Of course, each box can be a part of no more than one gift. Polycarp cannot use boxes "partially" or redistribute candies between them.
Input
The first line of the input contains two integers
n
n
and
k
k
(
1≤n≤2⋅
10
5
,1≤k≤100
1≤n≤2⋅105,1≤k≤100
) — the number the boxes and the number the girls.
The second line of the input contains
n
n
integers
d
1
,
d
2
,…,
d
n
d1,d2,…,dn
(
1≤
d
i
≤
10
9
1≤di≤109
), where
d
i
di
is the number of candies in the
i
i
-th box.
Output
Print one integer — the maximum number of the boxes Polycarp can give as gifts.
Examples
Input
Copy
7 2
1 2 2 3 2 4 10
Output
Copy
6
Input
Copy
8 2
1 2 2 3 2 4 6 10
Output
Copy
8
Input
Copy
7 3
1 2 2 3 2 4 5
Output
Copy
4

题解:题意给n个数和k,从中取两个数,使得两个数之和能被k整除,每个数只能取一次.求最多能使用多少数.

我的思路:先求出每个数除以k的余数.两个数的余数之和等于k就是符合条件的.例如,n为余数是1的数的个数和m是余数是k-1的数的个数,那么有min(n,m)2个数符合条件.a是余数是0的数的个数,符合条件的数为a/22;

#include <bits/stdc++.h>
const int N=2e5+5;
typedef long long ll;
using namespace std;
//int a[N];
int b[105];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	int tmp;
	for(int i=1;i<=n;i++){
		scanf("%d",&tmp);
		b[tmp%k]++;
		//printf("%d=%d\n",i,tmp%k);
	}
	int ans=0;
	tmp=k/2;
	if(k%2==0) tmp=k/2-1,ans+=b[k/2]/2*2;
	for(int i=1;i<=tmp;i++){
		ans+=2*min(b[i],b[k-i]);
		//cout<<"ans="<<ans<<endl;
	}
	printf("%d\n",ans+b[0]/2*2);
	return 0;

}