hdu4655Cut Pieces

http://acm.hdu.edu.cn/showproblem.php?pid=4655

先以最大的来算为 N*所有的排列数  再减掉重复的 重复的计算方法：取相邻的两个数的最小值再与它前面的组合数和后面的组合数相乘

注意负值

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define LL long long
#define N 1000100
#define mod 1000000007
LL s1[N],s2[N],a[N],b[N];
int main()
{
    int i,j,n,t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        LL s = 1;
        for(i = 1; i <= n ; i++)
        {
            scanf("%lld",&a[i]);
            b[i] = a[i];
            s  = (s*a[i])%mod;
        }
        sort(b+1,b+n+1);
        j=0;
        for(i = 1 ; i <= n ; i++)
        if(i%2!=0)
        a[i] = b[++j];
        for(i = n ; i >= 1 ; i--)
        if(i%2==0)
        a[i] = b[++j];
        s1[0] = 1;
        for(i = 1; i <= n ; i++)
        s1[i] = (s1[i-1]*a[i])%mod;
        s2[n+1] = 1;
        for(i = n ; i>= 1; i--)
        s2[i] = (s2[i+1]*a[i])%mod;
        LL minz;
        LL ans = 0;
        for(i = 1; i < n ; i++)
        {
            minz = min(a[i],a[i+1]);
            ans=(ans+((s1[i-1]*minz)%mod*s2[i+2])%mod)%mod;
        }
        ans = ((s*n)%mod-ans)%mod;
        if(ans<0)
        ans+=mod;
        cout<<ans<<endl;
    }
    return 0;
}