Description
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])
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 const int mod = 1e9 + 7; 7 const int MAX = 8000000 + 10; 8 long long a[MAX]; 9 int main() 10 { 11 a[0] = 0; 12 a[1] = 1; 13 for(int i = 2; i <= MAX; i++) 14 { 15 a[i] = i * ( a[i - 1] + 1) % mod; 16 } 17 int t; 18 scanf("%d", &t); 19 while(t--) 20 { 21 int num; 22 scanf("%d", &num); 23 printf("%lld ", a[num]); 24 } 25 return 0; 26 }