HDU 1058

雪黛依梦 — Sat, 11 Sep 2010 12:56:00 GMT

//重在理解方法：每次找到最小的 2 3 5  7 的因子数，之后利用所存数的下标的关系改变
#include
using namespace std;

int num[5843];  //存储前5842个丑数

int find_min ( int a, int b, int c, int d )
{
    int temp = a < b ? a : b;
    int index = c < d ? c : d;
    return temp < index ? temp : index;
}

void solve ( )
{
     int i1, i2, i3, i4, i;
     int h1, h2, h3, h4;
     i1 = i2 = i3 = i4 = 1;
     num[1] = 1;
     for (i = 2; i < 5843; i ++ )
     {
         h1 = num[i1] * 2;
         h2 = num[i2] * 3;
         h3 = num[i3] * 5;
         h4 = num[i4] * 7;
         int min = find_min ( h1, h2, h3, h4 );

         num[i] = min;


         //易错点：这里不可以用else if 因为ti中可能会有相同的最小值，如当：min = 6 时
         if ( min == h1 )
         i1 ++;

         if ( min == h2 )
         i2 ++;

         if ( min == h3 )
         i3 ++;

         if ( min == h4 )
         i4 ++;
     }
}

int main ()
{
    solve ();
    int n;
    while ( scanf ("%d", &n), n )
    {
          if ( n % 100 != 11 && n % 10 == 1 )
            printf ("The %dst humble number is %d.\n", n, num[n]);

          else if ( n % 100 != 12 && n % 10 == 2 )
            printf ("The %dnd humble number is %d.\n", n, num[n]);

          else if ( n % 100 != 13 && n % 10 == 3 )
            printf ("The %drd humble number is %d.\n", n, num[n]);

          else
            printf ("The %dth humble number is %d.\n", n, num[n]);
    }
    //system ("pause");
    return 0;
}

雪黛依梦 2010-09-11 20:56 发表评论

C++博客-雪黛依梦-随笔分类-数学题

HDU 1058