HDU

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27359 Accepted Submission(s): 12057

Problem Description

A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, … shows the first 20 humble numbers.

Write a program to find and print the nth element in this sequence

Input

The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

Output

For each test case, print one line saying “The nth humble number is number.”. Depending on the value of n, the correct suffix “st”, “nd”, “rd”, or “th” for the ordinal number nth has to be used like it is shown in the sample output.

Sample Input

1
2
3
4
11
12
13
21
22
23
100
1000
5842
0

Sample Output

The 1st humble number is 1.
The 2nd humble number is 2.
The 3rd humble number is 3.
The 4th humble number is 4.
The 11th humble number is 12.
The 12th humble number is 14.
The 13th humble number is 15.
The 21st humble number is 28.
The 22nd humble number is 30.
The 23rd humble number is 32.
The 100th humble number is 450.
The 1000th humble number is 385875.
The 5842nd humble number is 2000000000.

Source

University of Ulm Local Contest 1996

题意： 和POJ里有个丑数的定义差不多，那个是三个，这个是四个，分别是{2,3,5,7}，让你用这四个数来构造一些数，要求这些数分解的因子必须是这四个数的其中，然后问你第 n 个数是多少

分析： 因为样例中给了一个最大范围是2e9，在中间构造时很可能爆int，所以记得要开long long ，因为范围才五千多，直接暴力模拟即可，因为a[i]肯定比a[i-1]大，从a1,a2...ai−1分别乘上2，3，5，7，在符合条件下去一个最小值就是a[i]啦

还有：输出时要注意下，那个11,12,13，真坑呀

参考代码

#include<bits/stdc++.h>#define ll long longusing namespace std;const int N = 1e5 + 10, INF = 2e9 + 7;ll s[N];void init(){    ll a[5] = {2,3,5,7};    s[1] = 1;    for(int i = 2;i < 5844;i++){        s[i] = INF;        for(int j = 0;j < 4;j++)            for(int k = 1;k < i;k++)                if(a[j]*s[k] > s[i-1])                    s[i] = min(s[i],a[j]*s[k]);    }}int main(){    ios_base::sync_with_stdio(0);    init();    int n;    while(cin>>n,n){        if(n%10 == 1 && (n/10)%10 != 1) printf("The %dst humble number is %d.\n",n,s[n]);        else if(n%10 == 2 && (n/10)%10 != 1) printf("The %dnd humble number is %d.\n",n,s[n]);        else if(n%10 == 3 && (n/10)%10 != 1) printf("The %drd humble number is %d.\n",n,s[n]);        else printf("The %dth humble number is %d.\n",n,s[n]);    }    return 0;}

• 如有错误或遗漏，请私聊下UP，thx