Sum of Consecutive Prime Numbers--POJ 2739

1、题目类型：数论、打表。

2、解题思路：水题。

3、注意事项：优化dp[i][j]>10000，否则TLE。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     map
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     bool
   
      prime[
   
     10000
   
     ];

   
     int
   
      cnt,d[
   
     10000
   
     ],dp[
   
     1300
   
     ][
   
     1300
   
     ];
map
   
     <
   
     int
   
     ,
   
     int
   
     >
   
      M;


   
     void
   
      Table()
{
 
   
     int
   
      i,j;
 prime[
   
     1
   
     ]
   
     =
   
     1
   
     ;
 prime[
   
     2
   
     ]
   
     =
   
     prime[
   
     3
   
     ]
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     2
   
     ;i
   
     <
   
     100
   
     ;i
   
     ++
   
     )
 {
 
   
     for
   
     (j
   
     =
   
     2
   
     ;j
   
     *
   
     i
   
     <
   
     10000
   
     ;j
   
     ++
   
     )
 {
 prime[j
   
     *
   
     i]
   
     =
   
     true
   
     ;
 }
 }
}
 

   
     void
   
      Set()
{
 cnt
   
     =
   
     1
   
     ; 
 
   
     for
   
     (
   
     int
   
      i
   
     =
   
     1
   
     ;i
   
     <
   
     10000
   
     ;i
   
     ++
   
     )
 {
 
   
     if
   
     (
   
     !
   
     prime[i])
 d[cnt
   
     ++
   
     ]
   
     =
   
     i;
 }
}


   
     void
   
      Solve()
{
 
   
     int
   
      i,j;

 
   
     for
   
     (i
   
     =
   
     1
   
     ;i
   
     <
   
     cnt;i
   
     ++
   
     )
 {
 
   
     for
   
     (j
   
     =
   
     i;j
   
     <
   
     cnt;j
   
     ++
   
     )
 {
 dp[i][j]
   
     =
   
     dp[i][j
   
     -
   
     1
   
     ]
   
     +
   
     d[j];
 
   
     if
   
     (dp[i][j]
   
     >
   
     10000
   
     )
 
   
     break
   
     ;
 M[dp[i][j]]
   
     ++
   
     ;
 }
 }
}


   
     int
   
      main()
{
 Table();
 Set();
 Solve();
 
   
     int
   
      n;
 
   
     while
   
     (cin
   
     >>
   
     n
   
     &&
   
     n)
 cout
   
     <<
   
     M[n]
   
     <<
   
     endl;

}