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题目链接：

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 -41 1 1 1

Sample Output

390880

这个题直接暴力求x1,x2,x3，算出x4,判断一下，再加一个特判就可以过，不过用哈希来实现更简单，

#include 
   
    #include
    
     #include
     
      #define ll long longusing namespace std;const int maxn=1e6+7;int hash1[maxn],hash2[maxn];int main(int argc, char** argv) {	int a,b,c,d;	while(~scanf("%d%d%d%d",&a,&b,&c,&d)){		if((a>0&&b>0&&c>0&&d>0)||(a<0&&b<0&&d<0&&c<0)){			printf("0\n");			continue;		} 		int ans=0;		memset(hash1,0,sizeof(hash1));		memset(hash2,0,sizeof(hash2));		for(int i=1;i<=100;++i){			for(int j=1;j<=100;++j){				int x=a*i*i+b*j*j;				if(x>=0) hash1[x]++;				else hash2[-x]++;			}		}		for(int i=1;i<=100;++i){			for(int j=1;j<=100;++j){				int x=c*i*i+d*j*j;				if(x>0) ans+=hash2[x];				else ans+=hash1[-x];			}		}		printf("%d\n",ans*16);	}	return 0;}
     
    
   

 

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