1842-上三角形矩阵的和-代数式变形，模拟

 1 

#include
 2 using namespace std;
 3 long long sum[100005] = { 0 };
 4 long long square[100005] = { 0 };
 5 long long ans[10] = { 0 };
 6 vector <int> vec;
 7 int main() {
 8     int t=1;//数据总数
 9     scanf("%d", &t);
10     int o = t, k = 0;
11     while (t > 0) {
12         long long temp,  numtmp = 0, SUM = 0;
13         int n;
14         scanf("%d", &n);
15         for (int i = 0; i < n; i++) {
16             scanf("%lld", &temp);
17             vec.push_back(temp);
18         }//输入
19         for (int i = 0; i < n; i++) {
20             SUM += vec[i];
21         }
22         for (int i = 0; i < n; i++) {
23             ans[k] +=vec[i]*vec[i] ;
24         }
25         ans[k] += SUM * SUM;
26         ans[k] = ans[k] / 2;
27         /*for (int i = 0; i < vec.size() ; i++)
28             cout << vec[i] << " ";*/
29         t--; k++;
30         vec.clear();
31     }
32     for (int j = 0; j < o; j++) {
33         printf("%lld\n", ans[j]);
34     }
35     return 0;
36 }