题目学习——HDUOJ-1492

The number of divisors(约数) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

 

Author

lcy

暴力扫TLE

#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define LL long long intint main(){    LL n;    while(scanf("%lld",&n),n){        LL c=0;        long double m=sqrt(n);        for(LL i=1;i<m;++i)            if(!(n%i))++c;        c*=2;        LL M=m;        if(M==m)++c;        cout<<c<<endl;    }    return 0;}

#include <iostream>#include <cstdio>#include <cmath>using namespace std;#define LL long long intint main(){    LL n;    int factor[4]={2,3,5,7};    while(scanf("%lld",&n),n){        int index[4]={0,0,0,0};        for(int i=0;i<4;++i){            for(;n>1&&!(n%factor[i]);n=n/factor[i])                ++index[i];        }        LL c=1;        for(int i=0;i<4;++i)            c*=(1+index[i]);        cout<<c<<endl;    }    return 0;}

 因子分解 指数组合数 

input n=2^a*3^b*5^c*7^d

output (1+a)*(1+b)*(1+c)*(1+d)