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链接：

https://www.nowcoder.com/acm/contest/139/F


来源：牛客网

题目描述

Given a1, a2, …, an, find

modulo (109+7).

输入描述:

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains an integer n.
The second line contains n integers a1, a2, ..., an.

输出描述:

For each test case, print an integer which denotes the result.

示例1

输入

复制

2
1 2
5
2 3 3 3 3
​

输出

复制

3
453

备注:

* 1 ≤ n ≤ 1000
* 1 ≤ ai ≤ 109
* The number of test cases does not exceed 10.

转载自

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long ll;

const int maxn=1010;
const int mod=1e9+7;
ll val[maxn],b[maxn];

/*
因为数字太大，根本不可能一项一项的求出来结果，就算是
ai--a(i+1)，也是不可能的
但是可以直接用 n-i+1项，直接 预估出来结果(后悔matlab没认真学)

*/


//拉格朗日插值模版
/****************************************  head  ****************************************/
namespace polysum {
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    const int D=2010;
    ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
    ll powmod(ll a,ll b){ll res=1;a%=mod;assert(b>=0);for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    ll calcn(int d,ll *a,ll n) { // a[0].. a[d]  a[n]
        if (n<=d) return a[n];
        p1[0]=p2[0]=1;
        rep(i,0,d+1) {
            ll t=(n-i+mod)%mod;
            p1[i+1]=p1[i]*t%mod;
        }
        rep(i,0,d+1) {
            ll t=(n-d+i+mod)%mod;
            p2[i+1]=p2[i]*t%mod;
        }
        ll ans=0;
        rep(i,0,d+1) {
            ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
            if ((d-i)&1) ans=(ans-t+mod)%mod;
            else ans=(ans+t)%mod;
        }
        return ans;
    }
    void init(int M) {
        f[0]=f[1]=g[0]=g[1]=1;
        rep(i,2,M+5) f[i]=f[i-1]*i%mod;
        g[M+4]=powmod(f[M+4],mod-2);
        per(i,1,M+4) g[i]=g[i+1]*(i+1)%mod;
    }
    //要求m次的方程，Y=a[0]...a[m]是 X=0...m对应的值，
    //预估sum[n-1]的值
    ll polysum(ll m,ll *a,ll n) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
        ll b[D];
        for(int i=0;i<=m;i++) b[i]=a[i];
        b[m+1]=calcn(m,b,m+1);
        rep(i,1,m+2) b[i]=(b[i-1]+b[i])%mod;
        return calcn(m+1,b,n-1);
    }
    ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
        if (R==1) return polysum(n,a,m);
        a[m+1]=calcn(m,a,m+1);
        ll r=powmod(R,mod-2),p3=0,p4=0,c,ans;
        h[0][0]=0;h[0][1]=1;
        rep(i,1,m+2) {
            h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
            h[i][1]=h[i-1][1]*r%mod;
        }
        rep(i,0,m+2) {
            ll t=g[i]*g[m+1-i]%mod;
            if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
            else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
        }
        c=powmod(p4,mod-2)*(mod-p3)%mod;
        rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
        rep(i,0,m+2) C[i]=h[i][0];
        ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
} // polysum::init();



ll pow_(ll base,int n,int mod){
    ll ans=1;
    while(n){
        if(n&1)ans=(ans*base)%mod;
        base=(base*base)%mod;
        n>>=1;
    }
    return ans;
}

int main(){
    int n;
    polysum::init(1010);

    while(scanf("%d",&n)==1){
        for(int i=1;i<=n;i++){
            scanf("%lld",&val[i]);
        }
        sort(val+1,val+n+1);
        val[0]=0;
        ll ans=0,now=1;

        for(int i=1;i<=n;i++){
            if(val[i]==val[i-1]){
                now=(now*val[i])%mod;
                continue;
            }
            b[0]=0;
            for(int j=1;j<=n-i+1;j++){
                b[j]=j*(pow_(j,n-i+1,mod)+mod-pow_(j-1,n-i+1,mod))%mod;
            }
            ll tmp=(polysum::polysum(n-i+1,b,val[i]+1)+mod-polysum::polysum(n-i+1,b,val[i-1]+1))%mod;
            ans=(ans+now*tmp%mod)%mod;
            now=now*val[i]%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}