Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4745 Accepted Submission(s): 1890
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
给定a,b,c,d。a*x1^2+b*x2^2+c*x3^2+d*x4^2=0
其中x1~x4 在 [-100,100]区间内， a,b,c,d在[-50,50] 区间内。
求满足上面那个式子的所有解的个数。
这题也能用hash做，涨姿势了……
思路：将等式变形为a*x1^2+b*x2^2= -(c*x3^2+d*x4^2) 先用两重循环列举a,b的所有情况，将等式的左边结果存入hash表。再用两重循环列举c,d的所有情况，看看结果的相反数在不在hash表中。统计输出。
优化：x1^2与x1的正负号无关，x2--x4同理，循环时从1-100循环即可，最后答案乘以16；若abcd均大于或小于0，直接输出0.
#include <iostream>
#include <cstring>
using namespace std;
int a,b,c,d,pin[101],hash[2000010],i,j,sum;
int main()
{
for (i=1;i<=100;++i)
pin[i]=i*i;
while (cin>>a>>b>>c>>d){
if ((a>0 && b>0 && c>0 && d>0) || (a<0 && b<0 && c<0 && d<0)){
cout<<"0\n";
continue;
}
memset(hash,0,sizeof(hash)); sum=0;
for (i=1;i<=100;++i)
for (j=1;j<=100;++j)
hash[a*pin[i]+b*pin[j]+1000000]++; //列举等式左边的所有x的解和a，b的组合，并置偏移1000000，这样可以偏置为正值，1000000减之可以偏置为负值
for (i=1;i<=100;++i)
for (j=1;j<=100;++j)
sum+=hash[1000000-(c*pin[i]+d*pin[j])];
/*例如hash[1000001]=1时也就是a*pin[i]+b*pin[j]=1,如果对应的(c*pin[i]+d*pin[j])=-1则sum+=hash[1000001]这里sum+=1也就是这个等式成立的情况+1，若前面一个循环没有获得这次索引为(c*pin[i]+d*pin[j]) 的哈希表，则+0也就是等式没有成立，最后根据sum的数量，有四个变量的组合，共16种，所有最后sum*16*/
cout<<sum*16<<endl;
}
return 0;
}