Codeforces1131B. Preparation for International Women's Day(同余)

https://codeforces.com/contest/1133/problem/B

n=7 k=3
1 2 2 3 2 4 5
注意数据2,3 判断k为偶数中间值if else
无需开数组 只需sum[t%k]即可

#include<iostream>
#include<fstream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<cmath>
#include<cctype>
using namespace std;

#define PI acos(-1.0)
#define fi first
#define se second
#define pb push_back
#define forn(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++) 
#define rep(i,a,n)  for(int i=a;i<n;i++)
#define IO                       \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int INF=0x3f3f3f3f;
const int maxn=2e5+5;
const int mod=1e9+7;

int gcd(int a,int b)
{
	return b!=0?gcd(b,a%b):a;

}
int sum[100];
int main()
{
	int n,k,t,ans=0;//ans max 2e5 
	cin>>n>>k;
	for(int i=1;i<=n;i++)
	{
		cin>>t;
		sum[t%k]++;
	} 
	//(0+k)%k==0
	ans+=sum[0]/2;	//sum[0]能成几对
	for(int i=1;i<=k/2;i++)
	{
		if(i*2==k)
		{
			ans+=sum[i]/2;//k=2 走else 中间值重复出现 
		}
		else
			ans+=min(sum[i],sum[k-i]);
	} 
	cout<<ans*2<<endl;
	return 0;
}