HDU 1058 Humble Numbers

Question：

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.

Sample Input

1

2

3

4

11

12

13

21

22

23

100

1000

5842

0

Sample Output

The 1st humble number is 1.

The 2nd humble number is 2.

The 3rd humble number is 3.

The 4th humble number is 4.

The 11th humble number is 12.

The 12th humble number is 14.

The 13th humble number is 15.

The 21st humble number is 28.

The 22nd humble number is 30.

The 23rd humble number is 32.

The 100th humble number is 450.

The 1000th humble number is 385875.

The 5842nd humble number is 2000000000.

题目大意：一个数的素因子中只含2，3，5，7的数叫做humble number，求出第几个数humble number是多少。

解题思路：用数组来保存这些数，先把1放入，分别用，2，3，5，7乘以a[1],选取4个答案中最小的一个并保存，再将选取的这个数是由哪一个数乘以a[1]得到，将其乘以a数组的后一位（例如：第一个选取2，则变为，2*a[2],3*a[1],5*a[1],7*a[1],以此类推）

(http://acm.hust.edu.cn/vjudge/contest/125402#problem/F)

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long LL;
LL a[5845];
int main()
{
int t2=1,t3=1,t5=1,t7=1;
a[1]=1;
int  h=1;
while(h<=5843)
{
LL tp;
tp=min(min(2*a[t2],3*a[t3]),min(5*a[t5],7*a[t7]));
a[++h]=tp;
if(tp==2*a[t2])
t2++;
if(tp==3*a[t3])
t3++;
if(tp==5*a[t5])
t5++;
if(tp==7*a[t7])
t7++;
}
int n,ncase=0;
while(~scanf("%d",&n),n)
{
printf("The %d",n);
if(n%10==1&&n%100!=11)     //输出格式有点复杂
printf("st ");
else if(n%10==2&&n%100!=12)
printf("nd ");
else if(n%10==3&&n%100!=13)
printf("rd ");
else printf("th ");
printf("humble number is %lld.\n",a
);
}
}

体会：做此类题开始想不到方法，但要善于变通