The number of divisors(約數) 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.

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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.

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Output For each test case, output its divisor number, one line per case.

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Sample Input 4 12 0

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Sample Output 3 6

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Author lcy

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Source “2006校園文化活動月”之“校慶杯”大學生程序設計競賽暨杭州電子科技大學第四屆大學生程序設計競賽

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Recommend LL 很簡單的題目。。 只要求得有多少個2，3，5，7 然后結果就是? (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; }

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