poj Prime Cuts 1595 （素数打表 选择性素数输出）

Prime Cuts

Description

A prime number is a counting number (1, 2, 3, ...) that is evenly divisible only by 1 and itself. In this problem you are to write a program that will cut some number of prime numbers from the list of prime numbers between (and including) 1 and N. Your program will read in a number N; determine the list of prime numbers between 1 and N; and print the C*2 prime numbers from the center of the list if there are an even number of prime numbers or (C*2)-1 prime numbers from the center of the list if there are an odd number of prime numbers in the list.

Input

Each input set will be on a line by itself and will consist of 2 numbers. The first number (1 <= N <= 1000) is the maximum number in the complete list of prime numbers between 1 and N. The second number (1 <= C <= N) defines the C*2 prime numbers to be printed from the center of the list if the length of the list is even; or the (C*2)-1 numbers to be printed from the center of the list if the length of the list is odd.

Output

For each input set, you should print the number N beginning in column 1 followed by a space, then by the number C, then by a colon (:), and then by the center numbers from the list of prime numbers as defined above. If the size of the center list exceeds the limits of the list of prime numbers between 1 and N, the list of prime numbers between 1 and N (inclusive) should be printed. Each number from the center of the list should be preceded by exactly one blank. Each line of output should be followed by a blank line. Hence, your output should follow the exact format shown in the sample output.

Sample Input

21 2
18 2
18 18
100 7

Sample Output

21 2: 5 7 11

18 2: 3 5 7 11

18 18: 1 2 3 5 7 11 13 17

100 7: 13 17 19 23 29 31 37 41 43 47 53 59 61 67

//题意：

给出两个数n、m.先算出1--n之间的素数个数num，若num为奇数则输出2*m-1个素数。否则输出2*m个素数。输出的这些素数都是从中间往两边扩散的。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 1010
using namespace std;
int p[N];
bool pp[N];
int pnum;
int getp()//素数打表 
{
	int i,j;
	memset(pp,false,sizeof(pp));
	for(i=2;i<N;i++)
	{
		if(!pp[i])
			p[pnum++]=i;
		for(j=0;j<pnum&&p[j]*i<N;j++)
		{
			pp[p[j]*i]=true;
			if(i%p[j]==0)
				break;
		}
	}
	pp[0]=true;
}
int main()
{
	int n,m,num,cnt;
	int i,j;
	getp();
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		num=0;			
		for(i=1;i<=n;i++)
		{
			if(!pp[i])
				num++;
		}
		if(num&1)
			cnt=2*m-1;
		else
			cnt=2*m;
		printf("%d %d: ",n,m);
		if(cnt>=num)
		{
			printf("1");//在这里1也是素数，而打表里没有1. 
			for(i=0;i<num-1;i++)
				printf(" %d",p[i]);
				printf("\n\n");
		}
		else
		{
			if(num&1)
			{
				printf("%d",p[num/2-m]);
				for(i=num/2-m+1;i<num/2+m-1;i++)
					printf(" %d",p[i]);
					printf("\n\n");
			}
			else
			{
				printf("%d",p[num/2-m-1]);
				for(i=num/2-m;i<num/2+m-1;i++)
					printf(" %d",p[i]);
					printf("\n\n");
			}
		}
	}
	return 0;
}

//另一种打表：：

#include<stdio.h>
#include<string.h>
#define N 1010
int p[N];
bool pp[N];
int pnum;
int getp()
{
	int i,j;
	memset(pp,false,sizeof(pp));
	p[pnum++]=1;
	for(i=2;i<N;i++)
	{
		if(!pp[i])
		{
			for(j=i*2;j<N;j+=i)
				pp[j]=true;
			p[pnum++]=i;
		}
	}
} 
int main()
{
	int n,m,num,cnt;
	int i,j;
	getp();
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		num=0;			
		for(i=1;i<=n;i++)
		{
			if(!pp[i])
				num++;
		}
		if(num&1)
			cnt=2*m-1;
		else
			cnt=2*m;
		printf("%d %d: ",n,m);
		if(cnt>=num)
		{
			printf("%d",p[0]);//在这里1也是素数，而打表里没有1. 
			for(i=1;i<num;i++)
				printf(" %d",p[i]);
				printf("\n\n");
		}
		else
		{
			if(num&1)
			{
				printf("%d",p[num/2-m+1]);
				for(i=num/2-m+2;i<=num/2+m-1;i++)
					printf(" %d",p[i]);
					printf("\n\n");
			}
			else
			{
				printf("%d",p[num/2-m]);
				for(i=num/2-m+1;i<num/2+m;i++)
					printf(" %d",p[i]);
					printf("\n\n");
			}
		}
	}
	return 0;
}