poj2739

用素数筛，枚举所有区间，把加和，并把ans对应的位+1.

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      #include 
    
      <
    
      iostream
    
      >
    
      
#include 
    
      <
    
      cstdio
    
      >
    
      
#include 
    
      <
    
      cstdlib
    
      >
    
      
#include 
    
      <
    
      cstring
    
      >
    
      
#include 
    
      <
    
      cmath
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      #define
    
       maxn 10005
    
      


    
      bool
    
       
    
      is
    
      [maxn];

    
      int
    
       prm[maxn];

    
      int
    
       sum[maxn];

    
      int
    
       ans[maxn];


    
      int
    
       getprm(
    
      int
    
       n)
{
 
    
      int
    
       i, j, k 
    
      =
    
       
    
      0
    
      ;
 
    
      int
    
       s, e 
    
      =
    
       (
    
      int
    
      ) (sqrt(
    
      0.0
    
       
    
      +
    
       n) 
    
      +
    
       
    
      1
    
      );
 memset(
    
      is
    
      , 
    
      1
    
      , 
    
      sizeof
    
      (
    
      is
    
      ));
 prm[k
    
      ++
    
      ] 
    
      =
    
       
    
      2
    
      ;
 
    
      is
    
      [
    
      0
    
      ] 
    
      =
    
       
    
      is
    
      [
    
      1
    
      ] 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
       (i 
    
      =
    
       
    
      4
    
      ; i 
    
      <
    
       n; i 
    
      +=
    
       
    
      2
    
      )
 
    
      is
    
      [i] 
    
      =
    
       
    
      0
    
      ;
 
    
      for
    
       (i 
    
      =
    
       
    
      3
    
      ; i 
    
      <
    
       e; i 
    
      +=
    
       
    
      2
    
      )
 
    
      if
    
       (
    
      is
    
      [i])
 {
 prm[k
    
      ++
    
      ] 
    
      =
    
       i;
 
    
      for
    
       (s 
    
      =
    
       i 
    
      *
    
       
    
      2
    
      , j 
    
      =
    
       i 
    
      *
    
       i; j 
    
      <
    
       n; j 
    
      +=
    
       s)
 
    
      is
    
      [j] 
    
      =
    
       
    
      0
    
      ;
 
    
      //
    
       因为j是奇数，所以+奇数i后是偶数，不必处理！
    
      

    
       }
 
    
      for
    
       (; i 
    
      <
    
       n; i 
    
      +=
    
       
    
      2
    
      )
 
    
      if
    
       (
    
      is
    
      [i])
 prm[k
    
      ++
    
      ] 
    
      =
    
       i;
 
    
      return
    
       k; 
    
      //
    
       返回素数的个数
    
      

    
      }


    
      int
    
       main()
{
 
    
      //
    
      freopen("D:\\t.txt", "r", stdin);
    
      

    
       memset(ans, 
    
      0
    
      , 
    
      sizeof
    
      (ans));
 
    
      int
    
       tot 
    
      =
    
       getprm(
    
      10000
    
      );
 sum[
    
      0
    
      ] 
    
      =
    
       prm[
    
      0
    
      ];
 ans[sum[
    
      0
    
      ]] 
    
      =
    
       
    
      1
    
      ;
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      1
    
      ; i 
    
      <
    
       tot; i
    
      ++
    
      )
 {
 sum[i] 
    
      =
    
       sum[i 
    
      -
    
       
    
      1
    
      ] 
    
      +
    
       prm[i];
 
    
      if
    
       (sum[i] 
    
      <
    
       maxn)
 ans[sum[i]]
    
      ++
    
      ;
 }
 
    
      for
    
       (
    
      int
    
       i 
    
      =
    
       
    
      0
    
      ; i 
    
      <
    
       tot; i
    
      ++
    
      )
 
    
      for
    
       (
    
      int
    
       j 
    
      =
    
       
    
      0
    
      ; j 
    
      <
    
       i; j
    
      ++
    
      )
 
    
      if
    
       (sum[i] 
    
      -
    
       sum[j] 
    
      <
    
       maxn)
 ans[sum[i] 
    
      -
    
       sum[j]]
    
      ++
    
      ;
 
    
      int
    
       n;
 
    
      while
    
       (scanf(
    
      "
    
      %d
    
      "
    
      , 
    
      &
    
      n) 
    
      !=
    
       EOF 
    
      &&
    
       n 
    
      !=
    
       
    
      0
    
      )
 printf(
    
      "
    
      %d\n
    
      "
    
      , ans[n]);
 
    
      return
    
       
    
      0
    
      ;
}