杭电 1492 The number of divisors(约数) about Humble Numbers

The number of divisors(约数) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1947    Accepted Submission(s): 959

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

4 12 0

 

 

Sample Output

3 6

 

 

Author

lcy

 

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

 

Recommend

LL

 

    题意简述： 由质数2,3,5,7相乘(每个数的数目不限)所得到的数，称为“humble number”，该题给定一个humble number，求它一共有多少约数？

 

   解题思路：先求出该数在只用2,3,5,7表示时2,3,5,7的数目,分别用b[0],b[1],b[2],b[3]表示，然后（b[0]+1）*(b[1]+1)*(b[2]+1)*(b[3]+1)即为所求该数约数的个数！为什么呢？数学中的组合问题，分步来做！第一步，从b[0]个2中选择若干个2，可以是，0,1,2，，，b[0]个，总共b[0]+1种情况，然后从b[1]个3中，选择若干个3，以此类推。。。

#include <stdio.h>
#include <string.h>
int main()
{
    int b[4], i, cnt;
    __int64 n, a[4] = {2,3,5,7};
    while( (scanf("%I64d",&n)!=EOF) && n )
    {
        cnt = 1;
        memset(b,0,sizeof(b));
        for( i = 0; i < 4; i++ )
            while( !(n % a[i]) )
            {
                   b[i]++;
                   n = n / a[i];
            }
    for( i = 0; i < 4; i++ )
         cnt *= (b[i] + 1);
    printf( "%d\n", cnt );
    }
    return 0;
}