HDU 1058 Humble Numbers(打表+暴力)

HDU 1058 Humble Numbers(打表+暴力)

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22427    Accepted Submission(s): 9800

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. Write a program to find and print the nth element in this sequence

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line saying "The nth humble number is number.". Depending on the value of n, the correct suffix "st", "nd", "rd", or "th" for the ordinal number nth has to be used like it is shown in the sample output.

题解：

找出能被2，3，5，7整除的数，即素因子。既是素数，又是因子。然后找出是第几个。。。。

AC代码：

暴力+打表

#include #define min(a,b) ((a) < (b) ? (a):(b))#define min4(a,b,c,d) min(min(a,b),min(c,d))int a[5850];int main(void){ int n = 1; int p2,p3,p5,p7; p2 = p3 = p5 = p7 = 1; a[1] = 1; while(a[n] < 2000000000) { a[++n] = min4(2 * a[p2],3 * a[p3], 5 * a[p5], 7 * a[p7]); if(a[n] == 2 * a[p2]) p2++; if(a[n] == 3 * a[p3]) p3++; if(a[n] == 5 * a[p5]) p5++; if(a[n] == 7 * a[p7]) p7++; } while(~scanf("%d",&n)) { if(n==0)break; printf("The %d",n); int ten = n/10%10; if(n%10 == 1 && ten != 1) printf("st"); else if(n%10 == 2 && ten != 1) printf("nd"); else if(n%10 == 3 && ten != 1) printf("rd"); else printf("th"); printf(" humble number is %d.\n",a[n]); } return 0;}

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。