HDU 1496 Equations（hash）

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6652    Accepted Submission(s): 2695

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.

 

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

 

Author

LL

 

Source

“2006校園文化活動月”之“校慶杯”大學生程序設計競賽暨杭州電子科技大學第四屆大學生程序設計競賽

 

Recommend

LL

 

題解：雙向枚舉，hash。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#define M 1000005

using namespace std;

vector<int>z,f;
int a,b,c,d;
int h[2*M];

int H(int x) {
    return x+1000000;
}

void solve() {
    if(a>0&&b>0&&c>0&&d>0
            ||(a<0&&b<0&&d<0&&c<0)) {
        puts("0");
        return;
    }
    memset(h,0,sizeof h);
    for(int i=1; i<=100; i++) {
        for(int j=1; j<=100; j++) {
            h[H(a*i*i+b*j*j)]++;
        }
    }
    int ans=0;
    for(int i=1; i<=100; i++) {
        for(int j=1; j<=100; j++) {
            int s=H(-i*i*c-d*j*j);
            ans+=h[s];
        }
    }
    printf("%d\n",ans*16);
}

int main() {
    while(~scanf("%d%d%d%d",&a,&b,&c,&d)) {
        solve();
    }
}