POJ 1595 Prime Cuts 素数打表+找中间数规律

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

Source

​​South Central USA 1996​​

算法分析：

题意：

给你一个数n，求出1~n的素数个数sum，sum为偶数中间开始打印2*m个，如果sum为奇数中间开始打印2*m-1，注意这题把一算作素数

实现：

题目不难，线性打表就ok，但是有许多坑，数组要开到1200，输出有个空行，还有一些细节处理。

代码实现：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
 
using namespace std;    
const long N = 1200;   
long prime[N] = {0},num_prime = 0;    
int isNotPrime[N] = {1, 1};   
int init()    
{     
      for(long i = 2 ; i < N ; i ++)       
        {            
    if(! isNotPrime[i])               
      prime[num_prime ++]=i;  
     //无论是否为素数都会下来     
    for(long j = 0 ; j < num_prime && i * prime[j] <  N ; j ++)
        {               
            isNotPrime[i * prime[j]] = 1;  
        if( !(i % prime[j] ) )  //遇到i最小的素数因子
          //关键处1                  
        break;           
    }        
  }        
  return 0;   
} 
int main()
{
  int n,m;
  init();
  while(~scanf("%d%d",&n,&m))
  {
    int sum=1;
    for(int i=0;i<=n;i++)
    {
      if(n>=prime[i])
        sum++;
      else
        break;
    }
      if(sum%2==0)
    {
      if(2*m>=sum)
      {
      
        printf("%d %d: 1",n,m);
        for(int i=0;i<sum-1;i++)
          printf(" %d",prime[i]);
        cout<<endl;
      }
      else
      {
      int i=(sum-2*m)/2+1-2;  //找到的规律，-2的原因跟prime匹配起来  
      printf("%d %d: %d",n,m,prime[i]);
      i++;
      for(int j=1;j<2*m;i++,j++)
      {
        printf(" %d",prime[i]);
        
      } 
      cout<<endl;
      }
    }
    else
    {
      if(2*m-1>=sum)
      {
        printf("%d %d: 1",n,m);
        for(int i=0;i<sum-1;i++)
          printf(" %d",prime[i]);
        cout<<endl;
      }
      else
      {
      int i=(sum-2*m+1)/2+1-2;  
      printf("%d %d: %d",n,m,prime[i]);
      i++;
      for(int j=1;j<2*m-1;i++,j++)
      {
        printf(" %d",prime[i]);
        
      } 
      cout<<endl;
      
      }
    }cout<<endl;
    //printf("%d\n",cnt);
  }
}