题目链接

题面：

题意：
求 1 j − i + 1 ∑ k = i j s k \frac{1}{j-i+1}\sum_{k=i}^js_k j−i+11​∑k=ij​sk​ 的期望，其中 1 ≤ i ≤ j ≤ n 1\le i\le j\le n 1≤i≤j≤n。

我们可以先求出 ∑ a l l 1 j − i + 1 ∑ k = i j s k \sum_{all}\frac{1}{j-i+1}\sum_{k=i}^js_k ∑all​j−i+11​∑k=ij​sk​ 然后再除以区间个数 n ∗ ( n + 1 ) / 2 n*(n+1)/2 n∗(n+1)/2 即可。

题解：
参考 另一篇类似的博客

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<ctime>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
#define fhead(x) for(int i=head[(x)];i;i=nt[i])
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const double alpha=0.75;
const int mod=1e9+7;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=400100;
const int maxm=100100;
const int maxp=100100;
const int up=100100;

ll inv[maxn],sum[maxn];

void init(void)
{
    inv[1]=1;
    for(int i=2;i<maxn;i++)
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
}

ll mypow(ll a,ll b)
{
    ll ans=1;
    a%=mod;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
int main(void)
{
    init();
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&sum[i]);
            sum[i]=(sum[i]+sum[i-1])%mod;
        }
        ll ans=0,res=0;
        for(int i=1;i<=n;i++)
        {
            res=(res+sum[n-i+1]-sum[i-1])%mod;
            ans=(ans+res*inv[i])%mod;
        }
        ans=(ans%mod+mod)%mod;
        printf("%lld\n",ans*mypow(1ll*n*(n+1)/2,mod-2)%mod);
    }
    return 0;
}