humble numbers

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. <br><br>Write a program to find and print the nth element in this sequence<br>

 

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.<br>

 

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.<br>

 

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

思路：
分别将已有的数乘以 2 3 5 7 找到其中最小的那个，用四个位置变量分别记录 乘以 2 3 5 7的位置！

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#include <iomanip>#define maxn 5843#define MAXN 100005#define mod 1000000007#define INF 0x3f3f3f3f#define exp 1e-6 #define pi acos(-1.0)using namespace std;int humble[5843];void sethumble(){    int pos1=1;    int pos2=1;    int pos3=1;    int pos4=1;    int a,b,c,d;    humble[1]=1;    for(int i=2; i<=maxn; i++)    {        humble[i]=min(min(a=humble[pos1]*2,b=humble[pos2]*3),min(c=humble[pos3]*5,d=humble[pos4]*7));        if(humble[i]==a)        pos1++;        if(humble[i]==b)        pos2++;        if(humble[i]==c)        pos3++;        if(humble[i]==d)        pos4++;    }} int main(){    //freopen("D:\\a.txt","r",stdin);    ios::sync_with_stdio(false);    int i,j,k;    int n;    sethumble();    while(cin>>n,n)    {        cout<<"The "<<n;        if(n%100!=12&&n%10==2)        cout<<"nd";        else if(n%100!=11&&n%10==1)        cout<<"st";        else if(n%100!=13&&n%10==3)        cout<<"rd";        else        cout<<"th";        cout<<" humble number is "<<humble[n]<<"."<<endl;             }    return 0;}