Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3853 Accepted Submission(s): 1551
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.
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.
End of file.
Output
For 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
Author
LL
Source
Recommend
LL
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=2000010; int hash[N]; int main(){ //freopen("input.txt","r",stdin); 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 && c<0 && d<0)){ printf("0\n"); continue; } memset(hash,0,sizeof(hash)); for(int i=1;i<=100;i++) for(int j=1;j<=100;j++) hash[a*i*i+b*j*j+1000000]++; int ans=0; for(int i=1;i<=100;i++) for(int j=1;j<=100;j++) ans+=hash[1000000-(c*i*i+d*j*j)]; printf("%d\n",ans*16); //每个解有正有负,结果有2^4种 } return 0; }