Humble Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25735 Accepted Submission(s): 11344
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.
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.
本题思路很简单：就是用2，3，5，7循环来
求第i个f[i],第i个f[i]必定等于
前i-1个数中其中一个数与｛2，3，5，7｝中
其中一个的乘积，于是答案就出来了
#include<stdio.h>
#include<iostream>
using namespace std;
long long f[5843];
int main()
{
int prime[4]={2,3,5,7};
int i,j,k;
int n;
f[1]=1;
for(i=2;i<=5842;i++)
{
f[i]=2000000001;
for(j=0;j<4;j++)
{
for(k=i-1;k>=1;k--)
{
if(f[k]*prime[j]<=f[i-1])break;
if(f[k]*prime[j]<f[i])
f[i]=f[k]*prime[j];
}
}
}
while(cin>>n,n)
{
cout<<"The "<<n;
if(n==1&&n0!=11)cout<<"st ";
else if(n==2&&n0!=12)cout<<"nd ";
else if(n==3&&n0!=13)cout<<"rd ";
else cout<<"th ";
cout<<"humble number is "<<f[n]<<"."<<endl;
}
return 0;
}
