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  • Fermat’s Chirstmas Theorem (素数打表的)

                                                                             Fermat’s Chirstmas Theorem
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

    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
    ",l, u, x, y );  
    	}  
    	return 0;  
    }  
    
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  • 原文地址:https://www.cnblogs.com/yspworld/p/3900513.html
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