HDU 1058 Humble Numbers

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int N=6000;

int a[N];

void Humble()
{
    int x2, x3, x5, x7, y2=1, y3=1, y5=1, y7=1, i;

    a[1] = 1;
    for (i = 2; i <= 5842; i++)
    {
        x2 = a[y2]*2;
        x3 = a[y3]*3;
        x5 = a[y5]*5;
        x7 = a[y7]*7;

        a[i] = min(x2, x3);
        a[i] = min(a[i], x5);
        a[i] = min(a[i], x7);

        if (x2 == a[i]) y2++; ///x2，x3，x4，x5的最小值作为a[i]，那么下标要加1，以免重复
        if (x3 == a[i]) y3++;
        if (x5 == a[i]) y5++;
        if (x7 == a[i]) y7++;
    }
}

int main ()
{
    int n;

    Humble();

    while (scanf("%d", &n), n)
    {
        if (n % 10 == 1 && n/10%10 != 1) printf("The %dst humble number is %d.\n", n, a[n]); ///注意英语中第几个数的输出，个位为1,2,3，但是十位不能是1,2,3，才能输出st，nd，rd，其他输出th
        else if (n % 10 == 2 && n/10%10 != 1) printf("The %dnd humble number is %d.\n", n, a[n]);
        else if (n % 10 == 3 && n/10%10 != 1) printf("The %drd humble number is %d.\n", n, a[n]);
        else printf("The %dth humble number is %d.\n", n, a[n]);
    }

    return 0;
}