poj 2247 Humble Numbers

/* Name:poj 2247 Humble Numbers Author:Unimen Date: 24/04/11 16:51*//*解题报告：1、递推计算，并打表，注意计算质因数的方法 2、注意输出格式 */#include <iostream>using namespace std;int anResult[6000];int MinTwo(int a, int b){return a<b ? a : b;}int MinFour(int a, int b, int c, int d){int nTemp = MinTwo(a, b);nTemp = MinTwo(nTemp, c);nTemp = MinTwo(nTemp, d);return nTemp;}int main(){int p2, p3, p5, p7;p2 = p3 = p5 = p7 = 1;int n = 1;anResult [1] = 1;while(n <= 5842){anResult[++n] = MinFour(2*anResult[p2], 3*anResult[p3], 5*anResult[p5], 7*anResult[p7]);if(2*anResult[p2] == anResult[n])++p2;if(3*anResult[p3] == anResult[n])++p3;if(5*anResult[p5] == anResult[n])++p5;if(7*anResult[p7] == anResult[n])++p7;}while(cin>>n && n){int ten = n / 10 % 10;int nTemp = n;nTemp = n % 10;cout<<"The "<<n;if(1==nTemp && ten!=1)cout<<"st";else if(2==nTemp && ten!=1)cout<<"nd";else if(3==nTemp && ten!=1)cout<<"rd";else cout<<"th";cout<<" humble number is "<<anResult[n]<<"."<<endl;}return 0;}