poj 2247 Humble Numbers

标签：poj

题目大意：找出所有因子中只有2,3,5,7的数，给出n，求出第n个这样   只有6000不到打表

注意第11  12  13 的输出与 1   2   3    几十一   几十二   几十三 的输出不同  

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

long humble[6000];

int main()
{
    int i, n2, n3, n5, n7;
    humble[1] = 1;
    n2 = 1;
    n3 = 1;
    n5 = 1;
    n7 = 1;
    for (i = 2; i <= 5900; i++)
    {
        humble[i] = min(min(humble[n2]*2, humble[n3]*3), min(humble[n5]*5, humble[n7]*7));
        if (humble[i] == humble[n2]*2)
            n2++;
        if (humble[i] == humble[n3]*3)
            n3++;
        if (humble[i] == humble[n5]*5)
            n5++;
        if (humble[i] == humble[n7]*7)
            n7++;
    }
    
    int num;
    while (cin >> num && num)
    {
          int tmp1, tmp2;
          tmp1 = num % 10;
          tmp2 = num % 100;
          if (tmp1 == 1)
          {
             if (tmp2 == 11)
                 cout << "The " << num << "th humble number is " << humble[num] << "." << endl;
             else 
                 cout << "The " << num << "st humble number is " << humble[num] << "." << endl;
          }
          else if (tmp1 == 2)
          {
             if (tmp2 == 12)
                 cout << "The " << num << "th humble number is " << humble[num] << "." << endl;
             else
                 cout << "The " << num << "nd humble number is " << humble[num] << "." << endl;
          }
          else if (tmp1 == 3)
          {
             if (tmp2 == 13)
                 cout << "The " << num << "th humble number is " << humble[num] << "." << endl;
             else
                 cout << "The " << num << "rd humble number is " << humble[num] << "." << endl;
          }
          else 
             cout << "The " << num << "th humble number is " << humble[num] << "." << endl;
    }
    
    system("pause");
}

poj 2247 Humble Numbers

标签：poj

原文地址：http://blog.csdn.net/fanerxiaoqinnian/article/details/38050875