HDU 1319 Prime Cuts & ZOJ 1312 Prime Cuts

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分类专栏：数组打表杭电素数位运算

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本文详细介绍了如何使用筛选法来找出一个给定范围内的素数，并通过一系列步骤将其分割为指定个数的中间素数。对于奇数个素数的情况，输出中间2*c-1个素数；对于偶数个素数，则输出中间2*c个素数。此外，文章还特别指出当c超过素数的数量时，应输出所有范围内的素数。

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/*
中文题目素数分割
中文翻译-大意

素数是一个统计数字(1、2、3、…),平均只能被1和本身整除。在这个问题你要编写一个程序,将减少一些数量的素数从列表中(并包括)之间的素数1和N。

给你一个整数n和整数c，当在n的范围内有奇数个素数时，输出中间2*c-1个素数，如果总的素数个数是偶数个的话，输出中间部分的2*c个素数
解题思路：先用筛选法将素数判断出来，在用另外一个数组将素数存起来。之后再在草稿纸中画出左右的界限，输出就可以了。
难点详解：当c超过在n范围内的素数时，将n内所有的素数全部输出。另外，本题是将 1 看成素数的，要注意一下题意
关键点：素数打表，理解题意
解题人：lingnichong
解题时间： 2014-08-22 01:10:43
解题体会：搞了半天前面要空格啊！害的我小心翼翼的。本题是将1看成素数的
*/

Prime Cuts

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1960 Accepted Submission(s): 833

Problem 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

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

Source

South Central USA 1996

#include<stdio.h>
#include<string.h>
#define MAXN 1100
int a[MAXN],b[MAXN];
int main()
{
	int i,j;
	int n,c,t,k,s;
	int l,r;
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	for(i=2;i<MAXN/2;i++)
	{
		for(j=2*i;j<MAXN;j+=i)
			if(a[j] != -1)
				a[j] = -1;
	}
	
	while(~scanf("%d%d",&n,&c))
	{
		k=t=0;
		for(i = 1; i <= n; i++)
		{
			if(a[i]==0)
			{
				b[k++]=i;
				t++;
			}
				
		}
		 printf("%d %d:",n,c);
		 if(c > t)
		 {
		 //	printf("1");
		 	for(i=0;i <= t-1; i++)
		 		printf(" %d",b[i]);
		 	printf("\n");
		 }
		 else
		 {
		 	if(t&1)
		 	{
		 		l=k/2-(c-1); r=k/2+(c-1);
		 		//printf("%d",b[l]);
		 		for(i=l;i<=r;i++)
		 			printf(" %d",b[i]);
		 		printf("\n");
		 	}
		 	else
		 	{
		 		l=k/2-c; r=k/2+c-1;
		 	//	printf("%d",b[l]);
		 		for(i=l;i<=r;i++)
		 			printf(" %d",b[i]);
		 		printf("\n");
		 	}
		 	
		 }
		 printf("\n");
	}
	return 0;
}