Problem Description

You are given an integer N. Find the number of the positive divisors of N!, modulo 109+7.

Constraints

• 1≤N≤103

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the positive divisors of N!, modulo 109+7.

Example

Sample Input 1

3

Sample Output 1

4
There are four divisors of 3! =6: 1, 2, 3 and 6. Thus, the output should be 4.

Sample Input 2

6

Sample Output 2

30

Sample Input 3

1000

Sample Output 3

972926972

題意：給出一個整數 n，求 n! 的約數個數

思路：

根據唯一分解定理，一個整數 n 可分解為：

那么，n 的約數個數為：

故而對于每個素因子，可以選擇 0~ai 個，根據乘法原理，直接統計即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000+5; const int dx[] = {0,0,-1,1,-1,-1,1,1}; const int dy[] = {-1,1,0,0,-1,1,-1,1}; using namespace std; int num[N]; int main() {int n;scanf("%d",&n);for(int i=2;i<=n;i++){int temp=i;for(int j=2;j<=temp;j++){while(temp%j==0){num[j]++;temp/=j;}}if(temp!=1)num[temp]++;}LL res=1;for(int i=1; i<=n; i++)if(num[i])res=(res*(num[i]+1))%MOD;printf("%lld\n",res);return 0; }

?

總結

以上是生活随笔為你收集整理的Factors of Factorial（AtCoder-2286）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。