hdu 1058 Humble Numbers

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1058

题目描述：

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.

题目大意：

一个数的质因子只有2、 3、 5 、 7我们把它称为Humble Numbers，现在叫你输出第n个Humble Numbers

题目分析：

因为Humble Numbers的质因子只有2、 3、 5 、 7，所以一个Humble Numbers乘以2、 3、 5 、 7一定还是一个Humble Numbers，这里我们用打表的方法解决这一题

AC代码：

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;long long int dp[5843];int main(){    dp[1]=1;    for(int i=2;i<=5842;i++)    {        dp[i]=dp[i-1]*2;        for(int j=1;j<i;j++)        {            if(dp[j]*7<=dp[i-1])                continue;            if(dp[j]*2>dp[i-1])                dp[i]=min(dp[i],dp[j]*2);            if(dp[j]*3>dp[i-1])                dp[i]=min(dp[i],dp[j]*3);            if(dp[j]*5>dp[i-1])                dp[i]=min(dp[i],dp[j]*5);            if(dp[j]*7>dp[i-1])                dp[i]=min(dp[i],dp[j]*7);        }    }    int n;    while(scanf("%d",&n)!=EOF,n)    {        if((n-1)%10==0&&(n-11)%100!=0)             printf("The %dst humble number is %lld.\n",n,dp[n]);        else if((n-2)%10==0&&(n-12)%100!=0)             printf("The %dnd humble number is %lld.\n",n,dp[n]);        else if((n-3)%10==0&&(n-13)%100!=0)             printf("The %drd humble number is %lld.\n",n,dp[n]);        else             printf("The %dth humble number is %lld.\n",n,dp[n]);    }    return 0;}