POJ1411数论+优化

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

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

2 2
313 313
23 73
43 43
37 53

#include
#include
#include

using namespace std;

struct node
{
	bool flag;
	int num;
};

node loop[10005];
int prime[10005];
int number;
double a,b,m;

void init()
{
	int i,j;
	number=0;
	for(i=0;i<10005;i++)
	{
		loop[i].flag=false;
		loop[i].num=0;
	}
	for(i=2;i<10005;i++)
	{
		if(!loop[i].flag)
		{
			for(j=i*i;j<10005;j=j+i)
				loop[j].flag=true;
		}
	}
	for(i=2;i<10005;i++)
	{
		if(!loop[i].flag)
		{
			prime[number]=i;
			loop[i].num=number;
			number++;
		}
	}
}

bool judge(int p,int q)
{
	double x=p*1.0;
	double y=q*1.0;
	if(x*y>m)
		return false;
	if(x/y>1)
		return false;
	if(x/y=2;i--)
		{
			if(!loop[i].flag)
				break;
		}
		int index=loop[i].num;
		for(i=0;i<=index;i++)
		{
			for(j=i;j<=index;j++)
			{
				p=prime[i];
				q=prime[j];
				if(judge(p,q)&&p*q>max)
				{
					max=p*q;
					ansp=p;
					ansq=q;
				}
			}
		}
		printf("%d %d/n",ansp,ansq);
	}
	return 0;
}