hdu 1058 Humble Numbers（dp）

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看的题解。。。不知道这些方法都是怎么想出来的，好厉害啊，估计我是想不出来的，用b2,b3,b5,b7来分别记录

从第几位数开始还没有乘对应的2,3,5,7然后比较，小的数排上就ok了。。。注意有的时候是、会出现乘积相同，

所以都得加一，比如2*3和3*2，都得让b2,b3加一。这道题还涉及到了英语上的知识，就是只有11,12,13是序数

词时加th，1,2,3,分别是first，second，third。

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<vector>#include<set>#include<string>#include<algorithm>using namespace std;int main(){int a[5850];int temp,n;a[1] = 1;int b2,b3,b5,b7;b2 = b3 = b5 = b7 = 1;int m = 2;while(m<5843){temp = min(2*a[b2],min(3*a[b3],min(5*a[b5],7*a[b7])));a[m++] = temp;if(temp == 2*a[b2])b2++;if(temp == 3*a[b3])b3++;if(temp == 5*a[b5])b5++;if(temp == 7*a[b7])b7++;} while(cin >> n,n){int x = a[n];//printf("The 1st humble number is 1.")if(n%100==11 || n%100 == 12 || n%100 == 13)printf("The %dth humble number is %d.\n",n,x);else if(n%10 == 1)printf("The %dst humble number is %d.\n",n,x);else if(n%10 == 2)printf("The %dnd humble number is %d.\n",n,x);else if(n%10 == 3)printf("The %drd humble number is %d.\n",n,x);elseprintf("The %dth humble number is %d.\n",n,x);}return 0;}

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