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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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