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1064 - 丑数

通过次数

0

提交次数

0

时间限制 : 1 秒 内存限制 : 32 MB

如果一个数的素因子只包含2，3，5或7，那么我们把这种数叫做丑数。序列1，2，3，4，5，6，7，8，9，10，12，14，15，16，18，20，21，24，25，27...展示了前20个丑数。
请你编程寻找这个序列中的第n个元素。

题目输入

输入包含多组测试数据。每组输入为一个整数n（1<=n<=5842），当n=0时，输入结束。

题目输出

对于每组输入，输出一行“The nth humble number is number.”。里面的n由输入中的n值替换，“st”，“nd”，“rd”和“th”这些序数结尾的用法参照输出样例。

输入/输出样例

输入格式

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

输出格式

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.

C语言解答

#include<stdio.h>

int min(int a,int b)
{
	if(a<b)
		return a;
	else 
		return b;
}
int main()
{
	int n;
	int a[5843];
	int j=1;
	int x2=1,x3=1,x5=1,x7=1;
	a[1]=1;
	for(int i=2;i<=5842;i++)
	{
		a[i]=min(min(2*a[x2],3*a[x3]),min(5*a[x5],7*a[x7]));
		if(a[i]==2*a[x2]) x2++;
		if(a[i]==3*a[x3]) x3++;
		if(a[i]==5*a[x5]) x5++;
		if(a[i]==7*a[x7]) x7++;
	}
	while(scanf("%d",&n)==1,n)
	{
		if(n%100==11||n%100==12||n%100==13)
			printf("The %dth humble number is %d.\n",n,a[n]);
		else if(n%10==1)
			printf("The %dst humble number is %d.\n",n,a[n]);
		else if(n%10==2)
			printf("The %dnd humble number is %d.\n",n,a[n]);
		else if(n%10==3)
			printf("The %drd humble number is %d.\n",n,a[n]);
		else 
			printf("The %dth humble number is %d.\n",n,a[n]);
	}
	return 0;
}

C++解答

#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
	int n=2,a[5843],i,p2,p3,p5,p7,ten;
	p2=p3=p5=p7=1;
	a[1]=1;
	while(n<=5842)
	{
		a[n]=min(min(2*a[p2],3*a[p3]),min(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++;
		n++;
	}
	while(scanf("%d",&n)!=EOF,n)
	{
		ten=n/10%10;
		printf("The %d",n);
		if(ten!=1&&n%10==1)
			printf("st");
		else if(ten!=1&&n%10==2)
			printf("nd");
		else if(ten!=1&&n%10==3)
			printf("rd");
		else
			printf("th");
		printf(" humble number is %d.\n",a[n]);
	}
	return 0;
}

Java解答

import java.io.File;
import java.io.FileOutputStream;
import java.math.*;
import java.util.Arrays;
import java.util.Scanner;


public class Main {
    public static boolean ok(int x){
    	while(x%2==0) x/=2;
    	while(x%3==0) x/=3;
    	while(x%5==0) x/=5;
    	while(x%7==0) x/=7;
    	return x==1;
    }
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in=new Scanner(System.in);
		 
		
              

		int n;
		int cas=1;
		int a[]=new int[5843];
		
		
		int b[]=new int[10];
		//System.out.println(b.charAt(0));
		
		while(in.hasNext()){
			n=in.nextInt();
			if(n==0) break;
			b[2]=b[3]=b[5]=b[7]=0;
			int t;
			t=n%10;
			a[0]=1;
			for(int i=1;i<n;i++){
				while(a[b[2]]*2<=a[i-1]) b[2]++;
				while(a[b[3]]*3<=a[i-1]) b[3]++;
				while(a[b[5]]*5<=a[i-1]) b[5]++;
				while(a[b[7]]*7<=a[i-1]) b[7]++;
				a[i]=Math.min(Math.min(a[b[2]]*2, a[b[3]]*3),Math.min(a[b[5]]*5, a[b[7]]*7));
				
			}
			
			
			 System.out.printf("The %d%s humble number is %d.\n",n,(t==1 && (n<10 || (n/10)%10!=1))?"st":((t==2 && (n<10 || (n/10)%10!=1))?"nd":((t==3 && (n<10 || (n/10)%10!=1))?"rd":"th")),a[n-1]);
			
		}
	}

}
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