题意:给出n个元素的序列s,1 ≤ i ≤ j ≤ n,i、j任意,求下列式子期望。
题解:数学
总的情况为(1+n)*n/2。
考虑相同长度区间,Sk的和有规律。
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<fstream>
#include<set>
#include<map>
#include<sstream>
#include<iomanip>
#define ll long long
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
int t, n, a[maxn], sum[maxn], inv[maxn], x[maxn];
ll fastpow(ll base, ll n, ll mod) {
ll ans = 1;
while (n) {
if (n & 1) ans *= base % mod, ans %= mod;
base *= base, base %= mod;
n >>= 1;
}
return ans % mod;
}
int add(ll x, ll y) {
ll temp = x + y;
if (temp > mod) return temp - mod;
else if (temp < 0) return temp + mod;
return temp;
}
int main() {
inv[1] = 1;
for (int i = 2; i < maxn; ++i) inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i++) sum[i] = add(sum[i - 1], a[i]);
int ans = 0;
for (int i = 1; i <= n; i++) {
x[i] = add(x[i - 1], sum[n - i + 1] - sum[i - 1]);
ans = add(ans, 1ll * x[i] * inv[i] % mod);
}
int y = fastpow(1ll * (1 + n) * n / 2 % mod, mod - 2, mod);
printf("%lld\n", 1ll * ans * y % mod);
}
return 0;
}