poj1840 hash

Eqs

题意：给出a1, a2, a3, a4, a5

求满足表达式 a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=0 的解(x1, x2, x3, x4, x5)的个数，其中xi范围是[-50, 50], xi != 0 (i = 1, 2, 3, 4, 5) 
分析：这题挺简单的，将解分成两组(x1, x2, x3) 和 (x4, x5)，先三重循环求出a1x1 ³+ a2x2 ³+ a3x3 ³他们的解，用开散列法对应到数组中，再用两重循环求出a4x4 ³+ a5x5 ³求得结果在hash表中查找然后判断就行了。

代码

    
      

     
       #include
     
       <
     
       stdio.h
     
       >
     
       
#include
     
       <
     
       string
     
       .h
     
       >
     
       
#include 
     
       <
     
       iostream
     
       >
     
       

     
       using
     
        
     
       namespace
     
        std;

     
       #define
     
        NN 1000020
     
       

     
       #define
     
        MAXHASH 100003 
     
       //
     
       这是一个素数
     
       

     
       int
     
        x1, x2, x3, x4, x5;

     
       int
     
        a1, a2, a3, a4, a5;

     
       int
     
        idx;

typedef 
     
       struct
     
        po{
 
     
       int
     
        sum;
 
     
       struct
     
        po 
     
       *
     
       nxt;

}PO;

PO f[NN];
PO 
     
       *
     
       head[MAXHASH];


     
       int
     
        cube(
     
       int
     
        x){
 
     
       return
     
        x 
     
       *
     
        x 
     
       *
     
        x;
}


     
       int
     
        hash(
     
       int
     
        x){
 
     
       int
     
        h 
     
       =
     
        x 
     
       %
     
        MAXHASH;
 
     
       if
     
       (h 
     
       <
     
        
     
       0
     
       ) h 
     
       +=
     
        MAXHASH;
 
     
       return
     
        h;
}

     
       int
     
        OK(
     
       int
     
        tmp){
 
     
       int
     
        ans 
     
       =
     
        
     
       0
     
       ;
 
     
       int
     
        h 
     
       =
     
        hash(tmp);
 
     
       for
     
        (PO 
     
       *
     
       p 
     
       =
     
        head[h]; p; p 
     
       =
     
        p
     
       ->
     
       nxt){
 
     
       if
     
       (p
     
       ->
     
       sum 
     
       ==
     
        tmp){
 ans
     
       ++
     
       ;
 }
 }
 
     
       return
     
        ans;
}


     
       int
     
        main() {
 
     
       int
     
        tmp, h;
 scanf(
     
       "
     
       %d%d%d%d%d
     
       "
     
       , 
     
       &
     
       a1, 
     
       &
     
       a2, 
     
       &
     
       a3, 
     
       &
     
       a4, 
     
       &
     
       a5);
 idx 
     
       =
     
        
     
       0
     
       ;
 memset(head, 
     
       0
     
       , 
     
       sizeof
     
       (head));
 
     
       for
     
        (x1 
     
       =
     
        
     
       -
     
       50
     
       ; x1 
     
       <=
     
        
     
       50
     
       ; x1
     
       ++
     
       ){
 
     
       if
     
       (x1 
     
       ==
     
        
     
       0
     
       ) 
     
       continue
     
       ;
 
     
       for
     
        (x2 
     
       =
     
        
     
       -
     
       50
     
       ; x2 
     
       <=
     
        
     
       50
     
       ; x2
     
       ++
     
       ){
 
     
       if
     
       (x2 
     
       ==
     
        
     
       0
     
       ) 
     
       continue
     
       ;
 
     
       for
     
        (x3 
     
       =
     
        
     
       -
     
       50
     
       ; x3 
     
       <=
     
        
     
       50
     
       ; x3
     
       ++
     
       ){
 
     
       if
     
       (x3 
     
       ==
     
        
     
       0
     
       ) 
     
       continue
     
       ;
 tmp 
     
       =
     
        a1 
     
       *
     
        cube(x1) 
     
       +
     
        a2 
     
       *
     
        cube(x2) 
     
       +
     
        a3 
     
       *
     
        cube(x3);
 h 
     
       =
     
        hash(tmp);
 f[idx].sum 
     
       =
     
        tmp;
 f[idx].nxt 
     
       =
     
        head[h];
 head[h] 
     
       =
     
        f 
     
       +
     
        idx
     
       ++
     
       ;
 }
 }
 }
 
     
       int
     
        res 
     
       =
     
        
     
       0
     
       ;
 
     
       for
     
        (x4 
     
       =
     
        
     
       -
     
       50
     
       ; x4 
     
       <=
     
        
     
       50
     
       ; x4
     
       ++
     
       ){
 
     
       if
     
       (x4 
     
       ==
     
        
     
       0
     
       ) 
     
       continue
     
       ;
 
     
       for
     
        (x5 
     
       =
     
        
     
       -
     
       50
     
       ; x5 
     
       <=
     
        
     
       50
     
       ; x5
     
       ++
     
       ){
 
     
       if
     
       (x5 
     
       ==
     
        
     
       0
     
       ) 
     
       continue
     
       ;
 tmp 
     
       =
     
        a4 
     
       *
     
        cube(x4) 
     
       +
     
        a5 
     
       *
     
        cube(x5);
 tmp 
     
       =
     
        
     
       -
     
       tmp;
 res 
     
       +=
     
        OK(tmp);
 }
 }
 printf(
     
       "
     
       %d\n
     
       "
     
       , res);
 
     
       return
     
        
     
       0
     
       ;
}