AtCoder 2286 C

题目链接：http://arc067.contest.atcoder.jp/tasks/arc067_a?lang=en

C - Factors of Factorial

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Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

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

Constraints

• 1≤N≤103

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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.

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Sample Input 1

Copy

3

Sample Output 1

Copy

4

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

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Sample Input 2

Copy

6

Sample Output 2

Copy

30

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Sample Input 3

Copy

1000

Sample Output 3

Copy

972926972

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Submit

题意：求n!的因子数

解析：n = 2^x1 + 3^x2 + 5^x3 + 7^x4 + 11^x5............ 就是一个数素因子分解，n的因子就是 sum =  (x1+1) * (x2+1) * (x3+1)  * ...............

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<map>#include<cmath>#define N 1009using namespace std;const int INF = 0x3f3f3f3f;const long long mod = 1e9 + 7;int fun(int n){    int f[N];    memset(f, 0, sizeof(f));    long long ans = 1;    for(int i = 2; i <= n; i++)    {        int k = i;        for(int j = 2; j <= k; j++)        {            while(k % j == 0)            {                f[j]++;                k /= j;            }        }        if(k) f[k]++;    }    for(int i = 2; i <= n; i++) if(f[i]) ans = ans * (f[i] + 1) % mod;    return ans % mod;}int main(){    int n;    scanf("%d", &n);    cout << fun(n) << endl;    return 0;}