杭电oj1496，代码解析
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
“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛
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LL | We have carefully selected several similar problems for you: 1280 1800 1264 2600 2522
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
#include
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<