Time Limit: 1000ms

Description

 Given N distinct elements, how many permutations we can get from all the possible subset of the elements?

Input

 The first line is an integer T that stands for the number of test cases.

Then T line follow and each line is a test case consisted of an integer N.

Constraints:

T is in the range of [0, 10000]

N is in the range of [0, 8000000]

Output

 For each case output the answer modulo 1000000007 in a single line.

Sample Input

 5

0

1

2

3

4

Sample Output

 0

1

4

15

64 
题意：给n个不同的数，求子集的排列数
分析：做道题竟然没想到是递推，天真的是考求组合数的知识。之后在看到大神讲解焕然大悟
以n为例：当子集是1个数时，有n种情况，子集个数是2，就有n*（n-1)，子集个数为三，就有n*(n-1)*(n*2)....直到为n时就是 n*(n-1)*(n-2)....1；其实也是在排列组合数
把s[n] = n + n*(n-1) + n*(n-1)*(n-2) + .....+n*(n-1)*(n-2)...*1;把n提出来就是s[n] = n*( 1+(n-1)+（n-1)*(n-2) +...... +(n-2)*...*1) = n*(1+s[n-1])

 #include <iostream>

 #include <cstring>

 #include <algorithm>

 #include <cstdio>

 using namespace std;

 const int mod = 1e9 + ;

 const int MAX =  + ;

 long long a[MAX];

 int main()

 {

     a[] = ;

     a[] = ;

     for(int i = ; i <= MAX; i++)

     {

         a[i] = i * ( a[i - ] + ) % mod;

     }

     int t;

     scanf("%d", &t);

     while(t--)

     {

         int num;

         scanf("%d", &num);

         printf("%lld\n", a[num]);

     }

     return ;

 }