hdu- 1058 Humble Numbers

题目链接：hdu-1058

思路：可四个循环暴力，此处说下dp;让每个数都乘以2,3,5,7； 四个指针，分别指向已经乘过2,3,5,7的数的下一个，然后让这四个数分别乘以2,3,5,7取最小值，注意11，12,13的英文字母结尾都是th

#include <iostream>
#include <string>
using namespace std;
int num[5845];
int min(int a,int b,int c,int d) //求四者中最小值
{
a=a>b?b:a;
c=c>d?d:c;
return a>c?c:a;
}

int main()
{
int n;
int i1=1,i2=1,i3=1,i4=1;//四个数，分别记录2，3,5,7,的个数
num[1]=1;
for(int i=2;i<=5842;i++)
{
num[i]=min(num[i1]*2,num[i2]*3,num[i3]*5,num[i4]*7);//让每个数分别乘2,3,5,7，取四者中最小值

if(num[i]==num[i1]*2)i1++;
if(num[i]==num[i2]*3)i2++;
if(num[i]==num[i3]*5)i3++;
if(num[i]==num[i4]*7)i4++;
}

while(cin>>n&&n)
{
if(n % 10 == 1 && n % 100 != 11)
printf("The %dst humble number is %lld.\n",n ,num
);
else if(n % 10 == 2 && n % 100 != 12)
printf("The %dnd humble number is %lld.\n",n ,num
);
else if(n % 10 == 3 && n % 100 != 13)
printf("The %drd humble number is %lld.\n",n ,num
);
else
printf("The %dth humble number is %lld.\n",n ,num
);
}
return 0;
}