Hot！ 浙大计算机研究生复试上机考试—— 历年题目(链接) 全真模拟（链接） EquationsTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1387 Accepted Submission(s): 542Problem DescriptionConsider equations having the following form: a*x1^2+b*x2^2+c*x3^2+d*x4^2=0a, 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 Input1 2 3 -41 1 1 1 Sample Output390880
#include <iostream> using namespace std; int base[101]; int hash[2000009]; int main() { int a,b,c,d; int i,j,k; for(i=0;i<101;i++) base[i]=i*i; while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF) { if( (a>0 && b>0 && c>0 && d>0 )|| (a<0 && b<0 && c<0 && d<0) ) { printf("0/n"); continue; } else { memset(hash,0,sizeof(hash)); for(i=1;i<101;i++) for(j=1;j<101;j++) hash[base[i]*a+base[j]*b+1000000]++; int sum=0; for(i=1;i<101;i++) for(j=1;j<101;j++) sum+=hash[-(c*base[i]+d*base[j])+1000000]; cout<<sum*16<<endl; //X分正负，每个X有2个解，4个就有16种不同的解 } } }
