Combinatorics——HDUOJ 1492

原题：

• 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

  4
  12
  0

• Sample Output

  3
  6

解题思路：

题目大致意思就是：丑数有且只有四种因子：2，3，5，7。给你一个丑数，算出这个丑数有多少种约数。

• 题目既然已经说明了丑数分解之后最小的数为2、3、5、7，那么直接通过循环除法来得出可以分解出a个2，b个3，c个5，d个7.
• 最后通过（1+a）×（1+b）×（1+c）×（1+d）得出约数的个数

代码：

#include <stdio.h>int main(){    long long n;      short a, b, c, d;     while (scanf("%lld",&n)!=EOF&&n!=0)    {        a = b = c = d = 1;        while (n != 1 && n % 2 == 0){            a++; n /= 2;        }        while (n != 1 && n % 3 == 0){            b++; n /= 3;        }        while (n != 1 && n % 5 == 0){            c++; n /= 5;        }        while (n != 1 && n % 7 == 0){            d++; n /= 7;        }        printf("%d\n", a*b*c*d);    }}

总结：

• 求一个数的约数的个数：
  
  • 先分解质因数，得出每个因数对应出现的个数a、b……：例如12 = 2^2 × 3^1>>>a=2,b=1
  • 约数的个数等于（1+a）×（1+b）：例如f(12) = （1+2）×（1+1）= 6
• 求一个数的所有约数之和：
  
  • 先分解质因数，得X=a1^k1 × a2^k2× ……..× an^kn ：例如12 = 2^2 × 3^1>>>a1=2,k1=2，a2=3,k2=1
  • 约数的之和等于(1+a1+a1^2…a1^k1)×(1+a2+a2^2…a2^k2)×…..×(1+an+an^2…an^kn)：例如f(12) = （1+2+4）×（1+3）= 28