The number of divisors(约数) about Humble Numbers

题目链接:?pid=1492

题意:给出一个只包含质因数2,3,5,7的数,求其约数的个数

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,根据乘法原理,则约数的个数为(n1+1)(n2+1)…(nn+1)

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<map>#include<queue>#include<cmath>#include<stack>#include<set>#include<vector>#include<algorithm>#define LL long long#define inf 1<<29#define s(a) scanf("%I64d",&a)#define CL(a,b) memset(a,b,sizeof(a))using namespace std;const int N=200005;__int64 n,m,a,b,c,d;int main(){while(s(n)&&n){a=1,b=1,c=1,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("%I64d\n",a*b*c*d);}return 0;}

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The number of divisors(约数) about Humble Numbers

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