Humble Numbers（优先队列）

Humble Numbers

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 22   Accepted Submission(s) : 3

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.

 

Source

University of Ulm Local Contest 1996

转载来自  http://blog.csdn.net/qianchangdiyin/article/details/50787676这个博客

解释如下

# include<iostream># include<queue># include<cstdio># define MAX 5842using namespace std;__int64 humble_number[MAX+1];struct cmp1{bool operator()(__int64 &a,__int64 &b)    {        return a>b;//最小值优先    }};int main(){    int i=1,n;    humble_number[0]=0;    priority_queue<__int64,vector<__int64>,cmp1> q;    while(i<=MAX)    {        q.push(1);        while(!q.empty())        {            if(i>MAX)                break;            __int64 j=q.top();            q.pop();            if(j!=humble_number[i-1])            {                humble_number[i++]=j;                q.push(j*2);                q.push(j*3);                q.push(j*5);                q.push(j*7);            }        }    }        //这边是控制输入的问题     while(cin>>n)    {        if(n==0)            break;        else        {            if(n%10==1&&n%100!=11)                printf("The %dst humble number is %d.\n",n,humble_number[n]);            else if(n%10==2&&n%100!=12)                printf("The %dnd humble number is %d.\n",n,humble_number[n]);            else if(n%10==3&&n%100!=13)                printf("The %drd humble number is %d.\n",n,humble_number[n]);            else                printf("The %dth humble number is %d.\n",n,humble_number[n]);        }    }    return 0;}