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前几天做了一个简单题

思路

通过读题我们很容易看出他就是让我们求这么一个东西：

\(\displaystyle\sum_{i = 1}^n \sum_{j = i + 1}^{n} (a_i - a_j)^2\)

我们把后边的完全平方公式展开就是:

\(\displaystyle\sum_{i = 1}^n \sum_{j = i + 1}^{n} a_i^2 -2a_ia_J + a_j^2\)

显然我们还可以把 \(a_i^2\)提出来

\(\displaystyle\sum_{i = 1}^n (n - i)a_i^2-2a_i\sum_{j = i + 1}^{n}a_J + \sum_{j = i + 1}^{n}a_j^2\)

显然后边的\(a_j\) 和\(a_j^2\)我们可以提前用前缀和处理

我们用sum[i]表示对a[i]数组的前缀和，用sum2表示对\(a[i]^2\)的前缀和

那么原来的式子就能化成:

\(\displaystyle\sum_{i = 1}^n (n - i)a_i^2-2a_i*(sum[n] - sum[i])+ (sum2[n]-sum2[i])\)

最后的时候注意取膜就行了

code

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

#define ll long long
#define N 500010
#define M 1010

using namespace std;
const int mod = 1000000007;
int n;
ll a[N], sum[N], sum2[N];

ll read() {
	ll s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

int main() {
	n = read();
	for (int i = 1; i <= n; i++) {
		a[i] = read();
		sum[i] = (sum[i - 1] + a[i]) % mod;
		sum2[i] = (sum2[i - 1] + (a[i] * a[i]) % mod) % mod;
	}
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		ll x1 = ((n - i) * ((a[i] * a[i]) % mod)) % mod;
		ll x2 = ((2 * a[i] % mod) * ((sum[n] - sum[i] + mod) % mod)) % mod;
		ll x3 = (sum2[n] - sum2[i] + mod) % mod;
		ll x = (x1 - x2 + x3 + mod) % mod;
		ans = (ans + x) % mod;
	}
	cout << ans;
}

原文地址：https://www.cnblogs.com/zzz-hhh/p/12774262.html