HDU-1496 Equations

最新推荐文章于 2018-08-11 23:39:48 发布

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本文介绍了一种计算特定形式四元二次方程解数量的方法。通过预先计算平方数并利用哈希表进行快速匹配，实现了高效求解。输入为四个整数系数，输出为该方程的有效整数解的数量。

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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.

InputThe 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.OutputFor 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

使用哈希快速查找，注意最后答案要乘以16

#include <cstdio>
#include <cstring>

#define MAX 2000001

int visit[MAX];
int num[MAX];

int main() {
    int a, b, c, d;
    for( int i = 1; i <= 100; i++ ) num[i] = i * i;
    while( scanf( "%d%d%d%d", &a, &b, &c, &d ) != EOF ) {
        memset( visit, 0, sizeof( visit ) );
        for( int i = 1; i <= 100; i++ ) {
            for( int j = 1; j <= 100; j++ ) {
                visit[a * num[i] + b * num[j] + 1000000]++;
            }
        }
        int sum = 0;
        for( int i = 1; i <= 100; i++ ) {
            for( int j = 1; j <= 100; j++ ) {
                sum = sum + visit[-( c * num[i] + d * num[j] ) + 1000000];
            }
        }
        printf( "%d\n", sum * 16 );
    }
    return 0;
}