hdu1058 Humble Numbers 寻找丑数问题枚举+排序

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Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28683 Accepted Submission(s): 12611

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

1234111213212223100100058420

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.

#include<stdio.h>
#include<algorithm>
#include<cmath> 
using namespace std;
#define INF 0x7fffffff  //32位最大带符号整数

//丑数，就是不能被2,3,5,7以外的其他素数整除的数 
//对于任意一个丑数f[i]，它都是由它前面的丑数乘以2,3,5或者7得到的

int ans[10000];

void HN()
{
	__int64 t;//int64是有符号 64 位整数数据类型，相当于C++中的long long 
	int i,j,k,m,s,flag;
	s=0;flag=1;
	for(i=0;i<=12&&flag;i++)
	for(j=0;j<=14&&flag;j++)
	for(k=0;k<=20&&flag;k++)
	for(m=0;m<=31&&flag;m++)
	{
		t=(__int64)(pow(7.0,i)*pow(5.0,j)*pow(3.0,k)*pow(2.0,m));
		if(t<0||t>INF) 
		{
			break; 
			flag=0;
		}//之所以要判断t<0是为了防止t已经溢出 
		else 
		ans[s++]=t;
	}
	sort(ans,ans+s);
}

int main()
{    
	HN();//预处理 
	int n,i,answer,casenum;
	while(scanf("%d",&n)!=EOF&&n!=0)
	{
		answer=ans[n-1];//注意是n-1; 
	    if(n%10==1 && n%100!=11)
		printf("The %dst humble number is %d.\n",n,answer);
		else if(n%10==2 && n%100!=12)
		printf("The %dnd humble number is %d.\n",n,answer);
		else if(n%10==3 && n%100!=13)
		printf("The %drd humble number is %d.\n",n,answer);
		else 
		printf("The %dth humble number is %d.\n",n,answer);	
	}
	return 0;
}