LA2565 Extraterrestrial Intelligence

最新推荐文章于 2023-04-06 07:30:00 发布

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本文介绍了一个寻找最佳图片尺寸的问题，给定一个整数m和一个分数a/b，目标是在宽度和高度均为素数的情况下，找到满足面积不超过m且宽度与高度比介于a/b和1之间的最大尺寸。

Calling Extraterrestrial Intelligence Again

Time limit: 3.000 seconds

A message from humans to extraterrestrial inteIigence was sent through the Arecibo radio telescope in Puerto Rico on the afternoon of Saturday November l6, l974. The message consisted of l679 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 l 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 haction a/b . It holds that m > 4 and 0 < a/b $\le$ 1 . You should find the pair of prime numbers p, q such that pq $\le$ m and a/b $\le$ p/q $\le$ 1 , and furthermore, the product pqtakes 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.

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 $\le$ 100000 and 1 $\le$ a $\le$ b $\le$ 1000 .

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.

Sample Input

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

Sample Output

這道題不難，本以為需要binary search質數表，但最後沒有用還是AC了。簡單而言此題就是告訴你m,a,b三個值，要滿足三個條件，即p*q<=m,0<a/b<=p/q<=1,p*q要最大。而m最大值為10萬，其實質數表建到5萬就夠了，因為p*q<=m,p至少為2，所以最大質數會在5萬以內。

#include <cstdio>
#include <math.h>
using namespace std;
int p[50000]; // to store all prime numbers, since m<=100000 and p*q<=m,so actually the size m/2 is enough
int find_prime()
{
    int i;
    int k = 1;
    p[0] = 2;  // first prime number
    for( i=3;i<50000;i++ )
    {
        int s = 0;
        for( int j=2;j<=sqrt(i);j++ )  // mod from 2 to sqrt(i) to check whether it is prime number 
        {
            if( i%j == 0 ) s = 1;
        }
        if( s == 0 )
        {
            p[k] = i;
            k++;
        } 
    }
    return k; // return index of array p
}
void ans( int m, int a, int b, int mm ) // mm is the index of p
{
    int pp, qq;
    int max = 0;
    for( int i=0;i<mm&&p[i]<=sqrt(m);i++) // since p*q<=m & p<=q, so max(p) = sqrt(m)
    {
        for( int j=i;j<mm;j++)  // q>=p
        {
            if( p[i]*b<p[j]*a ) break;
            if( p[i]*p[j]>m ) break;
            if( p[i]*p[j]>max )
            {
                max = p[i]*p[j];
                pp = p[i];
                qq = p[j];
            }
        }
    }
    printf("%d %d\n",pp,qq);
}
int main()
{
    int m, a, b;
    int mm = find_prime();
    while( scanf("%d%d%d",&m,&a,&b)!=EOF && (m+a+b)!=0 )
    {
        ans(m,a,b,mm);
    }
    return 0;
}

我這裡用的找質數的方法效率很低，即判斷一個數字i跟2~sqrt(i)每個數進行取餘數即判斷能否整除來決定是否為質數。但有更加高效的算法，以後我會補上討論。