InputThe 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.
OutputFor each test case, output its divisor number, one line per case.
Sample Input
4
12
0
Sample Output
3
6
题意：给出一个丑数n，这个数必定可以分解成多个2、3、5、7的形式
求：n的因子数
思路：唯一分解定理求因子个数，唯一不同的是唯一分解定理需要记录每个质数的指数，而该题则已经确定是2、3、5、7了，所以只要对这四个数进行计算指数再相互之间+1相乘即可
1 #include<stdio.h> 2 #include<iostream> 3 #include<string.h> 4 using namespace std; 5 typedef long long ll; 6 7 //ll n; 8 // 9 //ll prime() 10 //{ 11 // ll ans=0; 12 // for(ll i=1;i<=n;i++) 13 // { 14 // if(n%i==0) 15 // ans++; 16 // } 17 // return ans; 18 //} 19 20 int mark[4]; 21 int main() 22 { 23 ll n; 24 while(~scanf("%lld",&n)&&n) 25 { 26 // printf("%lld\n",prime()); 27 memset(mark,0,sizeof(mark));//每个数字的指数 28 while(n&&n%2==0) 29 { 30 mark[0]++; 31 n=n/2; 32 // cout<<n<<endl; 33 } 34 // cout<<"***"<<mark[0]<<endl; 35 while(n&&n%3==0) 36 { 37 mark[1]++; 38 n=n/3; 39 } 40 while(n&&n%5==0) 41 { 42 mark[2]++; 43 n=n/5; 44 } 45 while(n&&n%7==0) 46 { 47 mark[3]++; 48 n=n/7; 49 } 50 cout<<(1+mark[0])*(1+mark[1])*(1+mark[2])*(1+mark[3])<<endl; 51 } 52 return 0; 53 }
HDU-1492-The number of divisors(约数) about Humble Numbers -求因子总数+唯一分解定理的变形
原文：https://www.cnblogs.com/OFSHK/p/11342683.html