一.题目链接：

HDU-1058

二.题目大意：

规定因子（1 和 本身除外）只有  2,  3，5，7 的数为 Humble Number.

求第 n 个 Humble Number 的值.

三.分析：

题目给出：a[1] == 1.

然后求 2，3，5，7 的倍数，取最小的数，再让它的倍数++；

四.代码实现：

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-4
#define PI acos(-1.0)
#define ll long long int
using namespace std;

int a[6000];

void init()
{
    int x2, x3, x5, x7;
    x2 = x3 = x5 = x7 = 1;
    a[1] = 1;
    for(int i = 2; i < 5843; ++i)
    {
        a[i] = min(min(a[x2] * 2, a[x3] * 3), min(a[x5] * 5, a[x7] * 7));
        if(a[i] == a[x2] * 2)
            x2++;
        if(a[i] == a[x3] * 3)///注意这里不能是else if，因为两者可能相等.
            x3++;
        if(a[i] == a[x5] * 5)
            x5++;
        if(a[i] == a[x7] * 7)
            x7++;
    }
}

int main ()
{
    init();
    int n;
    while(scanf("%d", &n) && n)
    {
        if(n % 100 == 11 || n % 100 == 12 || n % 100 == 13)///蒟蒻格式盲区
            printf("The %dth humble number is %d.\n", n, a[n]);
        else if(n % 10 == 1)
            printf("The %dst humble number is %d.\n", n, a[n]);
        else if(n % 10 == 2)
            printf("The %dnd humble number is %d.\n", n, a[n]);
        else if(n % 10 == 3)
            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;
}