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; }