SCU 4424（求子集排列数）

A - A

Time Limit:0MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status Practice SCU 4424

Description

 Time Limit: 1000ms

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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 ;
 }