Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5526 Accepted Submission(s): 2193
Problem Description
Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0.
It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}.
Determine how many solutions satisfy the given equation.
Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
End of file.
End of file.
Output
For each test case, output a single line containing the number of the solutions.
Sample Input
1 2 3 -4 1 1 1 1
Sample Output
39088 0
/*题解：
hash问题
*/
// =====================================================================================
//
// Filename: Equations.cpp
// Description: Equations
// Algorithm: hash+二重循环
// Status: RunTime:187ms RunMemory:8044K
// Version: Dev-C++ 4.9.9.1
// Created: 2014/10/9 22:32
// Revision: none
// Compiler: G++
// Author: Tip of the finger melody, 1466989448@qq.com
// Company: none
//
// =====================================================================================
#include<cstdio>
#include<cstring>
int p[10002],hash[2000002];
void solve()
{
int a,b,c,d,i,j,sum;
for(i=1; i<=100; i++)
p[i] = i*i;
while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
{
if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&c<0&&d<0))
{
printf("0\n");
continue;
}
memset(hash,0,sizeof(hash));
for(i=1; i<=100; i++)
for(j=1; j<=100; j++)
hash[p[i]*a+p[j]*b+1000000]++;
for(i=1,sum=0; i<=100; i++)
for(j=1; j<=100; j++)
sum += hash[-(p[i]*c+p[j]*d)+1000000];
printf("%d\n",sum*16);
}
}
int main()
{
solve();
return 0;
}