4644. 求和

https://www.acwing.com/problem/content/description/4647/

第十三届蓝桥杯省赛C++A/C组 , 第十三届蓝桥杯省赛JAVAA组

算法标签：推公式；前缀和

思路

• 推公式做法：
  $$(a_1+a_2+a_3+...+a_n{)}^2=a_1^2+a_2^2+a_3^2+...+a_n^2+2\times a_1\times a_2+2\times a_1\times a_3+...+2\times a_1\times a_n+2\times a_2\times a_3+...+2\times a_{n-2}\times a_{n-1}+2\times a_{n-2}\times a_n+2\times a_{n-1}\times a_n=a_1^2+a_2^2+a_3^2+...+a_n^2+2\times S$$

• 前缀和做法
  $$S=\sum _{i=1}^{n-1}{a}_i\times (s_n-s_i)$$

C++代码

思路1

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int N = 200010;
ll res, sum;
int n;
int main()
{
    cin >> n;
    for (int i = 0, t; i < n; i ++) cin >> t, res += t, sum += t * t;
    cout << (res * res - sum) / 2 << endl;
    return 0;
}

思路2

// (a + b + c)^2 = a^2 + b^2 + c^2  + 2ab + 2ac + 2bc
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int N = 200010;
ll res;
int sum[N], a[N];
int n;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> a[i], sum[i] = a[i] + sum[i-1];
    for (int i = 1; i <= n-1; i ++) res += (ll)a[i] * (sum[n] - sum[i]);
    cout << res << endl;
    return 0;
}