CodeForces 701A Cards

每一组的和为$frac{{2×sumlimits_{i = 1}^n {a[i]} }}{n}$,然后暴力配对就可以了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-8;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar(); x = 0;while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar();  }
}

const int maxn=110;
int a[maxn],sum,n,f[maxn];

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),sum=sum+a[i];
    sum=sum/(n/2);
    for(int i=1;i<=n;i++)
    {
        if(f[i]==1) continue; f[i]=1;
        for(int j=1;j<=n;j++)
        {
            if(f[j]==1) continue;
            if(a[i]+a[j]==sum)
            {
                printf("%d %d
",i,j); f[j]=1;
                break;
            }
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/zufezzt/p/5802519.html