首页 > ACM题库 > HDU-杭电 > HDU 1058 Humble Numbers-动态规划-[解题报告] C++
2013
11-26

HDU 1058 Humble Numbers-动态规划-[解题报告] C++

Humble Numbers

问题描述 :

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

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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.

题目意思很好懂。

暴力的想法是在已知的丑数中选出最小的,保存之,然后乘以2,3,5,7保存起来,这里要注意去重。不得不说其英文输出很坑爹。

代码如下:

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <map>
#include <iostream>
#define MAXN 5850
using namespace std;

int n, a[4] = {2, 3, 5 ,7};
long long ans[MAXN];  


priority_queue<long long, vector<long long>, greater<long long> > q;

map<long long,bool> mp;

void pre()
{ 
    int pos = 1, cnt = 0;
    q.push(pos); 
    mp[1] = 1;
    while (1) {
        long long pos = q.top();
        ans[++cnt] = pos;
        if (cnt >= 5842) {
            break;
        }
        q.pop();
        for (int i = 0; i < 4; ++i) {
            if (!mp.count(pos*a[i])) {
                q.push(pos*a[i]);
                mp[pos*a[i]] = 1;
            }
        }
    }
}

int main()
{
    pre();
    while (scanf("%d", &n), n) {
        if (n % 10 == 1 && n%100 != 11) {
            printf("The %dst humble number is ", n);
        }
        else if (n % 10 == 2 && n%100!= 12) {
            printf("The %dnd humble number is ", n);
        }
        else if (n % 10 == 3 && n%100 != 13) {
            printf("The %drd humble number is ", n);
        }
        else { 
            printf("The %dth humble number is ", n);
        } 
        cout << ans[n] << "." << endl; 
    }
    return 0;
}

 

还有一种思想就是动态规划的思想,也是看了大牛的代码后知道的。定义4个伪指针p1, p2, p3, p4指向数组下标,分别表示的意思是在已知的丑数中,能够与2,3,5,7相乘的最小的数的下标。比如说已知第一个数是1,那么能够和2相乘的最小数就是1(p1指向其数组下标)了。同理,在已知2是由1*2得到后,就不能再用1*2来生成2了,因为这样没有意思,此时就p1就指向了2所在的下标了。每次新增加的丑数就是 Min(num[p1]*2, num[p2]*3, num[p3]*5, num[p4]*7) 了。

代码如下:

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

int dp[5850];

inline int Min(int a, int b, int c, int d)
{
    return min(a, min(b, min(c, d)));
}

void DP()
{
    int p1 = 1, p2 = 1, p3 = 1, p4 = 1;
    dp[1] = 1;
    for (int i = 2; i <= 5842; ++i) {
        dp[i] = Min(dp[p1]*2, dp[p2]*3, dp[p3]*5, dp[p4]*7);
        if (dp[i] == dp[p1]*2) {
            ++p1;
        }
        if (dp[i] == dp[p2]*3) {
            ++p2;
        }
        if (dp[i] == dp[p3]*5) {
            ++p3;
        }
        if (dp[i] == dp[p4]*7) {
            ++p4;
        }
    }
}

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