hdu1058.Humble Numbers

找出自然数中第 n 个质因子中只含 2，3，5，7 的数。

/*纯模拟*/
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX_N = 6000;

int n;
long long f[MAX_N];

void doit()
{
	f[1] = 1; f[2] = 2; f[3] = 3;
	for(int i = 4; i <= 5843; i ++){
		long long p1 = 0x7fffffff;
		for(int j = 1; j <= i - 1; j ++){
			if(f[j] * 2 > f[i - 1]) p1 = min(f[j] * 2, p1); 
			if(f[j] * 3 > f[i - 1]) p1 = min(f[j] * 3, p1);
			if(f[j] * 5 > f[i - 1]) p1 = min(f[j] * 5, p1);
			if(f[j] * 7 > f[i - 1]) p1 = min(f[j] * 7, p1);
		}
		f[i] = p1;
	}
}

int main()
{
	doit();
	while(scanf("%d", &n) != EOF){
		if(n == 0) break;
		if(n % 10 == 1 && n % 100 != 11) printf("The %dst humble number is %lld.\n", n, f[n]);
		else if(n % 10 == 2 && n % 100 != 12) printf("The %dnd humble number is %lld.\n", n, f[n]);
		else if(n % 10 == 3 && n % 100 != 13) printf("The %drd humble number is %lld.\n", n, f[n]);
		else printf("The %dth humble number is %lld.\n", n, f[n]);
	}
	return 0;
}