2020 CCPC Wannafly Winter Camp Day6 Div.1&2 N 合并！

题目链接：https://ac.nowcoder.com/acm/contest/4137/N

其实看到数据量 2000应该要想到时间复杂度基本就On2

题解：不管怎么合并，答案都一样的， 所有数两两乘积的和

代码

#include<iostream>
#include<string>
#include <cstdlib>
#include <algorithm>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include <iomanip>

// #pragma comment(linker, "/STACK:1024000000,1024000000")
// #define pi acos(-1)
// #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x7f7f7f7f //2139062143
#define INF1 0x3f3f3f3f //1061109567
#define INF2 2147483647
#define llINF 9223372036854775807
#define pi 3.141592653589793//23846264338327950254
#define pb push_back
#define ll long long
#define debug cout << "debug
";
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
#define CNM ios::sync_with_stdio(false);cin.tie(NULL);
#define scai(x) scanf("%d", &x)
#define sca2i(x, y) scanf("%d %d", &x, &y)
#define scaf(x) scanf("%lf", &x)
#define sca2f(x, y) scanf("%lf %lf", &x, &y)
#define For(m,n) for (int i = m;  i < n; i++)

inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
#define local
#ifdef local
#endif

#define MAX 10233
#define LCH(i) ((i) << 1)
#define RCH(i) ((i) << 1 | 1)
const int N = 2000 + 5;
int a[N];

int main()
{
    CNM
    // freopen("in.txt","r",stdin);
    ll ans = 0;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = i + 1; j < n; j++)
        {
            ans += a[i]*a[j];
        }
    }
    cout << ans << endl;
    return 0;
}