Poj 2247 humble numbers

Humble numbers

Time limit:1000 ms		Memory limit:65536 K
Total submissions:9453		Accepted:4440

Find out the numbers of all the factors that are only 2, 3, 5, and 7, give n, and find the nth number. In this case, only 6000 is not used in the table.

Note that the output of 11th 12 13 is different from the output of 1 2 3, and.

#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