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 = ;
         for(i = ; i <= n ; i++)
         {
             scanf("%lld",&a[i]);
             b[i] = a[i];
             s  = (s*a[i])%mod;
         }
         sort(b+,b+n+);
         j=;
         for(i =  ; i <= n ; i++)
         if(i%!=)
         a[i] = b[++j];
         for(i = n ; i >=  ; i--)
         if(i%==)
         a[i] = b[++j];
         s1[] = ;
         for(i = ; i <= n ; i++)
         s1[i] = (s1[i-]*a[i])%mod;
         s2[n+] = ;
         for(i = n ; i>= ; i--)
         s2[i] = (s2[i+]*a[i])%mod;
         LL minz;
         LL ans = ;
         for(i = ; i < n ; i++)
         {
             minz = min(a[i],a[i+]);
             ans=(ans+((s1[i-]*minz)%mod*s2[i+])%mod)%mod;
         }
         ans = ((s*n)%mod-ans)%mod;
         if(ans<)
         ans+=mod;
         cout<<ans<<endl;
     }
     return ;
 }