The number of divisors(約數(shù)) about Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1623????Accepted Submission(s): 789

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校園文化活動(dòng)月”之“校慶杯”大學(xué)生程序設(shè)計(jì)競(jìng)賽暨杭州電子科技大學(xué)第四屆大學(xué)生程序設(shè)計(jì)競(jìng)賽

?

Recommend LL 很簡(jiǎn)單的題目。。 只要求得有多少個(gè)2，3，5，7 然后結(jié)果就是? (p2-1)*(p3-1)*(p5-1)*(p7-1) #include<stdio.h> int main() {long long n;int p2,p3,p5,p7;while(scanf("%I64d",&n),n){p2=p3=p5=p7=0;while(n%2==0){n/=2;p2++;} while(n%3==0){n/=3;p3++;} while(n%5==0){n/=5;p5++;} while(n%7==0){n/=7;p7++;} printf("%d\n",(p2+1)*(p3+1)*(p5+1)*(p7+1));} return 0; }

?

總結(jié)

以上是生活随笔為你收集整理的HDU 1492 The number of divisors(约数) about Humble Numbers（数论，简单约数）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。