poj Equations

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Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6262 Accepted Submission(s): 2536

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.

就是让你求出有几种情况我们这里首先想到的当然是用一个4重循环了，但是4重肯定超时了,然后我们会想到一个三重循环就是用0-三重的数值看看其结果是不是在正常的范围内了。这样也不好有没有更简便的方法呢？那就是用两次两重循环大体意思是吧a*x1*x1+b*x2*x2看为一体，把剩下的看为一体，这样只要确定前面的，就可以知道后面的了，因为他们两个相加是一个确定的值啊！首先取前面的a*x1*x1+b*x2*x2看看他们所有的值并用哈希的方式储存起来，意思就是说我用f[102] = 2那就表示在x1x2有两种情况是等于102的，这样就是找找以后的了，，，具体看代码

Sample Input

1 2 3 -41 1 1 1
#include <iostream>#include <string.h>#include <algorithm>using namespace std;int f[2000005];int main(){    int a,b,c,d;    while(~scanf("%d %d %d %d",&a,&b,&c,&d))    {        int i,j;        if(a > 0&&b > 0&&c > 0&&d > 0)        {            printf("0\n");            continue;        }        if(a < 0&&b < 0&&c < 0&&d < 0)        {            printf("0\n");            continue;        }        memset(f,0,sizeof(f));        for(i = -100;i <= 100;i++)        {if(i == 0)            continue;            for(j = -100;j <= 100; j++)            {                if(j == 0)                    continue;                f[1000001+i*a*i+j*b*j]++;    //fang zhi yue jie jiu jia shang 1000001    
       }        }        int sum = 0;        for(i = -100;i <= 100;i++)        {            if(i == 0)            continue;            for(j = -100;j <= 100; j++)            {                if(j == 0)                    continue;                sum = f[1000001-i*c*i-j*d*j]+sum;            }        }        printf("%d\n",sum);    }}

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