Fermat’s Chirstmas Theorem (素数打表的)

                                                                         Fermat’s Chirstmas Theorem

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status Practice SDUTOJ 2093

Description

In a letter dated December 25, 1640; the great mathematician Pierre de Fermat wrote to Marin Mersenne that he just proved that an odd prime p is expressible as p = a2 + b2 if and only if p is expressible as p = 4c + 1. As usual, Fermat didn’t include the proof, and as far as we know, never

wrote it down. It wasn’t until 100 years later that no one other than Euler proved this theorem.

Input

Your program will be tested on one or more test cases. Each test case is specified on a separate input line that specifies two integers L, U where L ≤ U < 1, 000, 000

The last line of the input file includes a dummy test case with both L = U = −1.

Output

L U x y

where L and U are as specified in the input. x is the total number of primes within the interval [L, U ] (inclusive,) and y is the total number of primes (also within [L, U ]) that can be expressed as a sum of squares.

Sample Input

10 20

11 19

100 1000

-1 -1

Sample Output

10 20 4 2

11 19 4 2

100 1000 143 69

#include<stdio.h>  

#include<string.h>  



#define N 1000005



int prime[100005];  

int flag[1000005];  

int e;

  

void getP()  // 素数打表,找出素数存栈  

{  

	int i, j;  

	e = 0;  

	memset(flag, 0, sizeof(flag) ); //标记初始化

	

	for ( i=2; i<N; i++)  

	{  

		if ( flag[i]==0 ) 

		{

			prime[e++] = i; //进栈  

		}

		for ( j=0; j<e && i*prime[j]<N; j++ )  

		{  

			flag[ i * prime[j] ] = 1;  

		}  

	}  

}  



int main()  

{  

	int l,u,x,y;  



	getP(); 

	int i;

	while(scanf("%d %d",&l,&u))  

	{  

	  

		if(l==-1 && u==-1)  

            break;  

		x=0;  

		y=0;

		for( i=0; i<e; i++)  

		{  

			if(prime[i]>=l && prime[i]<=u )  

			{  

				x++;  

				if(prime[i]%4==1) 

				{

                    y++;  

				}

			}  

			if(prime[i]>u)  

                break;  

		}  

		if(l<=2 && u>=2)

		{

			y++;

		}

		printf("%d %d %d %d\n",l, u, x, y );  

	}  

	return 0;  

}