hdu1492 The number of divisors(约数) about Humble Numbers(公约数个数问题)



Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, ... shows the first 20 humble numbers.

Now given a humble number, please write a program to calculate the number of divisors about this humble number.For examle, 4 is a humble,and it have 3 divisors(1,2,4);12 have 6 divisors.

 

Input

The input consists of multiple test cases. Each test case consists of one humble number n,and n is in the range of 64-bits signed integer. Input is terminated by a value of zero for n.

 

Output

For each test case, output its divisor number, one line per case.

 

Sample Input

4120

 

Sample Output

36
题意：给出一个只包含质因数2，3，5，7的数，求其公约数的个数
思路：如果n=p1^n1*p2^n2*...*pn^nn,其中p1,p2,...,pn表示质因数，n1,n2,...,nn表示相应质因数的指数，则公约数的个数为(n1+1)(n2+1)...(nn+1)
用java实现总是内存超过限制，改用c++实现，vs2013中用非安全的scanf会提示出错，代码中添加了忽略：#pragma warning(disable:4996)
代码如下
#include <cstdio>using namespace std;#pragma warning(disable:4996)typedef long long LL;LL f[4];LL n;bool input(){scanf("%lld", &n);if (n == 0) return false;return true;}void solve(){f[0] = 1;while (n % 2 == 0) {f[0]++;n /= 2;} f[1] = 1;while (n % 3 == 0) {f[1]++;n /= 3;}f[2] = 1;while (n % 5 == 0) {f[2]++;n /= 5;}f[3] = 1;while (n % 7 == 0) {f[3]++;n /= 7;}LL ans = f[0] * f[1] * f[2] * f[3];printf("%lld\n", ans);}int main(){#ifndef ONLINE_JUDGEfreopen("f:\\OJ\\uva_in.txt", "r", stdin);#endifwhile (input()) {solve();}return 0;}