poj 1840 -- Eqs

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11865		Accepted: 5811

Description

Consider equations having the following form:
a1x1³+ a2x2³+ a3x3³+ a4x4³+ a5x5³=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.

Determine how many solutions satisfy the given equation.

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

654

思路：hash水过。

 1 /*======================================================================
 2  *           Author :   kevin
 3  *         Filename :   Eqs.cpp
 4  *       Creat time :   2014-07-24 17:49
 5  *      Description :
 6 ========================================================================*/
 7 #include <iostream>
 8 #include <algorithm>
 9 #include <cstdio>
10 #include <cstring>
11 #include <queue>
12 #include <cmath>
13 #define clr(a,b) memset(a,b,sizeof(a))
14 #define M 12500000
15 using namespace std;
16 short int ha[2*M+5];
17 int main(int argc,char *argv[])
18 {
19     int a,b,c,d,e;
20     while(scanf("%d%d%d%d%d",&a,&b,&c,&d,&e)!=EOF){
21         clr(ha,0);
22         for(int i = -50; i <= 50; i++){
23             if(!i) continue;
24             for(int j = -50; j <= 50; j++){
25                 if(!j) continue;
26                 int s = a*i*i*i+b*j*j*j;
27                 ha[M-s]++;
28             }
29         }
30         int cnt = 0;
31         for(int i = -50; i <= 50; i++){
32             if(!i) continue;
33             for(int j = -50; j <= 50; j++){
34                 if(!j) continue;
35                 for(int k = -50; k <= 50; k++){
36                     if(!k) continue;
37                     int s = c*i*i*i+d*j*j*j+e*k*k*k;
38                     s = M+s;
39                     if(s < 2*M && s > 0)
40                         cnt += ha[s];
41                 }
42             }
43         }
44         printf("%d
",cnt);
45     }
46     return 0;
47 }

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