HDU 1492 The number of divisors(约数) about Humble Numbers
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HDU 1492 The number of divisors(约数) about Humble Numbers
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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): 1741????Accepted Submission(s): 852
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校園文化活動月”之“校慶杯”大學(xué)生程序設(shè)計競賽暨杭州電子科技大學(xué)第四屆大學(xué)生程序設(shè)計競賽 Recommend LL 排列組合求約數(shù)個數(shù)
#include <iostream> #include <map> #include <stdio.h> #include <math.h> #include <string.h> #include <stdlib.h> #include <algorithm> using namespace std; int main() {__int64 n;__int64 rc[4];int p[4]={2,3,5,7};while(scanf("%I64d",&n),n){for(int i=0;i<4;i++){rc[i]=1;if(n%p[i]==0)while(n%p[i]==0) {n/=p[i];rc[i]++;}}printf("%I64d\n",rc[0]*rc[1]*rc[2]*rc[3]);}return 0; }
轉(zhuǎn)載于:https://www.cnblogs.com/372465774y/archive/2012/10/20/2732386.html
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