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
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 using namespace std; 7 #define LL long long 8 #define N 1000100 9 #define mod 1000000007 10 LL s1[N],s2[N],a[N],b[N]; 11 int main() 12 { 13 int i,j,n,t; 14 cin>>t; 15 while(t--) 16 { 17 cin>>n; 18 LL s = 1; 19 for(i = 1; i <= n ; i++) 20 { 21 scanf("%lld",&a[i]); 22 b[i] = a[i]; 23 s = (s*a[i])%mod; 24 } 25 sort(b+1,b+n+1); 26 j=0; 27 for(i = 1 ; i <= n ; i++) 28 if(i%2!=0) 29 a[i] = b[++j]; 30 for(i = n ; i >= 1 ; i--) 31 if(i%2==0) 32 a[i] = b[++j]; 33 s1[0] = 1; 34 for(i = 1; i <= n ; i++) 35 s1[i] = (s1[i-1]*a[i])%mod; 36 s2[n+1] = 1; 37 for(i = n ; i>= 1; i--) 38 s2[i] = (s2[i+1]*a[i])%mod; 39 LL minz; 40 LL ans = 0; 41 for(i = 1; i < n ; i++) 42 { 43 minz = min(a[i],a[i+1]); 44 ans=(ans+((s1[i-1]*minz)%mod*s2[i+2])%mod)%mod; 45 } 46 ans = ((s*n)%mod-ans)%mod; 47 if(ans<0) 48 ans+=mod; 49 cout<<ans<<endl; 50 } 51 return 0; 52 }