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2013
11-09

POJ 1411 Calling Extraterrestrial Intelligence Again [解题报告] Java

Calling Extraterrestrial Intelligence Again

问题描述 :

A message from humans to extraterrestrial intelligence was sent through the Arecibo radio telescope in Puerto Rico on the afternoon of Saturday November 16, 1974. The message consisted of 1679 bits and was meant to be translated to a rectangular picture with 23 x 73 pixels. Since both 23 and 73 are prime numbers, 23 x 73 is the unique possible size of the translated rectangular picture each edge of which is longer than 1 pixel. Of course, there was no guarantee that the receivers would try to translate the message to a rectangular picture. Even if they would, they might put the pixels into the rectangle incorrectly. The senders of the Arecibo message were optimistic.

We are planning a similar project. Your task in the project is to find the most suitable width and height of the translated rectangular picture. The term “most suitable” is defined as follows. An integer m greater than 4 is given. A positive fraction a/b less than or equal to 1 is also given. The area of the picture should not be greater than m. Both of the width and the height of the translated picture should be prime numbers. The ratio of the width to the height should not be less than a/b nor greater than 1. You should maximize the area of the picture under these constraints.

In other words, you will receive an integer m and a fraction a/b. It holds that m > 4 and 0 < a/b <= 1. You should find the pair of prime numbers p, q such that pq <= m and a/b <= p/q <= 1, and furthermore, the product pq takes the maximum value among such pairs of two prime numbers. You should report p and q as the "most suitable" width and height of the translated picture.

输入:

The input is a sequence of at most 2000 triplets of positive integers, delimited by a space character in between. Each line contains a single triplet. The sequence is followed by a triplet of zeros, 0 0 0, which indicates the end of the input and should not be treated as data to be processed.

The integers of each input triplet are the integer m, the numerator a, and the denominator b described above, in this order. You may assume 4 < m <= 100000 and 1 <= a <= b <= 1000.

输出:

The output is a sequence of pairs of positive integers. The i-th output pair corresponds to the i-th input triplet. The integers of each output pair are the width p and the height q described above, in this order.

Each output line contains a single pair. A space character is put between the integers as a delimiter. No other characters should appear in the output.

样例输入:

5 1 2
99999 999 999
1680 5 16
1970 1 1
2002 4 11
0 0 0

样例输出:

2 2
313 313
23 73
43 43
37 53

解题代码:

import java.util.*;

/**
 *
 * @author leo
 */
public class Main {

    /**
     * @param args the command line arguments
     */
    public static boolean prime(int n) {
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                return false;
            }
        }
        return true;

    }

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);
        int m, a, b;
        m = scan.nextInt();
        a = scan.nextInt();
        b = scan.nextInt();
        while (true) {
            if (a == 0 && b == 0 && m == 0) {
                break;
            }
            boolean flag = false;
            while (m > 0) {
                int t = (int) Math.sqrt((double) (m * a / b));
                if (t * t != m) {
                    t++;
                }
                int tmp = (int) Math.sqrt((double) m);
                for (int i = t; i <= tmp; i++) {
                    if (m % i == 0 && prime(i) && prime(m / i)) {
                        flag = true;
                        System.out.println(i + " " + m / i);
                        break;
                    }
                }
                if (flag) {

                    break;
                }
                m--;

            }

            m = scan.nextInt();
            a = scan.nextInt();
            b = scan.nextInt();
        }
        // TODO code application logic here
    }
}

  1. #include <cstdio>

    int main() {
    //answer must be odd
    int n, u, d;
    while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
    if(n<=u) { puts("1"); continue; }
    n-=u; u-=d; n+=u-1; n/=u;
    n<<=1, ++n;
    printf("%dn",n);
    }
    return 0;
    }

  2. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。