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 t…

The number of divisors(约数) about Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2039    Accepted Submission(s): 1002 Problem Description A number whose only prime factors are 2,…

The number of divisors(约数) about Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2044    Accepted Submission(s): 1006 Problem DescriptionA number whose only prime factors are 2,3,…

The number of divisors(约数) about Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3033    Accepted Submission(s): 1465 Problem Description A number whose only prime factors are 2,3…

题目连接 The number of divisors(约数) about Humble Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3416 Accepted Submission(s): 1676 Problem Description A number whose only prime factors are 2,3…

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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1492 题目大意: 给出一个数,因子只有2 3 5 7,求这个数的因子个数 解题思路: 直接求出指数即可 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #include<…

The number of divisors(约数) about Humble Numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3337    Accepted Submission(s): 1632 Problem Description A number whose only prime factors are 2,3…

HDU - 1711 A - Number Sequence   Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... ,…

题目链接:传送门 题意: 求2004^x的全部约数的和. 分析: 由唯一分解定理可知 x=p1^a1*p2^a2*...*pn^an 那么其约数和 sum = (p1^0+p1^1^-+p1^a1)*-* (pn^0+pn^1^-+pn ) 代码例如以下: #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; con…