【原】 POJ 2739 Sum of Consecutive Prime Numbers 筛素数+积累数组 解题报告

 

http://poj.org/problem?id=2739

 

方法：
埃氏筛法得到N以下的所有素数，复杂度n*lglgn
利用积累数组计算数组中连续元素的和

 

Description

Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations 5 + 7 + 11 + 13 + 17 and 53. The integer 41 has three representations 2+3+5+7+11+13, 11+13+17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime 
numbers, so neither 7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. 
Your mission is to write a program that reports the number of representations for the given positive integer.

Input

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10 000, inclusive. The end of the input is indicated by a zero.

Output

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be inserted in the output.

Sample Input

2

3

17

41

20

666

12

53

0

Sample Output

1

1

2

3

0

0

1

2

 

   1: #include <iostream>

   2: #include <fstream>

   3:  

   4: using namespace std ;

   5:  

   6: const int N = 10001 ;

   7: const int NC = 1500 ;

   8:  

   9: char prime[N] ;

  10: __int64 cumarr[NC];

  11:  

  12: void GetCumArr()

  13: {

  14:     int i,j;

  15:  

  16:     //筛得素数 n*lglgn

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

  18:         prime[i]=1 ;

  19:     for( i=2 ; i<N ; )

  20:     {

  21:         for( j=i*i ; j<N ; j+=i )

  22:             prime[j]=0 ;

  23:         do ++i;

  24:         while( i<N && prime[i]==0 );

  25:     }

  26:     //得到积累数组

  27:     cumarr[0]=0 ;  //注意第一个数需要为0

  28:     cumarr[1]=2 ;

  29:     for( i=3,j=2 ; i<N ; ++i )

  30:     {

  31:         if(prime[i]==1)

  32:         {

  33:             cumarr[j] = cumarr[j-1]+i ;

  34:             ++j ;

  35:         }

  36:     }

  37: }

  38:  

  39:  

  40: void run2739()

  41: {

  42:     ifstream in("in.txt");

  43:     GetCumArr();

  44:     int num ;

  45:     int i,j ;

  46:     int up,down ;

  47:     int count ;

  48:     while( in>>num && num!=0 )

  49:     {

  50:         count = 0 ;

  51:         //枚举

  52:         for( i=0 ; i<NC ; ++i )

  53:         {

  54:             for( int j=i ; j<NC ; ++j )

  55:             {

  56:                 if( cumarr[j]-cumarr[i] == num )

  57:                     ++count ;

  58:                 else if( cumarr[j]-cumarr[i] > num )

  59:                     break ;

  60:             }

  61:         }

  62:         cout<<count<<endl ;

  63:     }

  64: }