Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9366 Accepted Submission(s): 3892
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.
Sample Input
1 2 3 -4
1 1 1 1
Sample Output
39088
0
Author
LL
思路：将四重循环转成二重。
优化：可以从整数开始枚举，最后结果乘16（因为是平方，4个解的正负号组合数）。
Code：
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int AX = 1e2+6;
int s[AX];
map<LL,LL>mp;
int main(){
int a , b ,c ,d;
for( int i = 1 ; i <= 100 ; i++ ){
s[i] = i * i ;
}
while ( ~scanf("%d%d%d%d",&a,&b,&c,&d) ){
mp.clear();
if( a > 0 && b > 0 && c > 0 && d > 0 ){
printf("0\n");
continue;
}else if( a < 0 && b < 0 && c < 0 && d < 0 ){
printf("0\n");
continue;
}else{
for( int i = 1 ; i <= 100 ; i++ ){
for( int j = 1 ; j <= 100 ; j++ ){
LL tmp = a * s[i] + b * s[j];
mp[tmp] ++;
}
}
LL res = 0LL;
for( int i = 1 ; i <= 100 ; i++ ){
for( int j = 1 ; j <= 100 ; j++ ){
LL tmp = c * s[i] + d * s[j];
res += mp[0-tmp];
}
}
printf("%lld\n",res*16);
}
}
return 0 ;
}