#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
using namespace std;
//分别枚举两边,a*x1*x1+b*x2*x2=-(c*x3*x3+d*x4*x4;
const int maxn=2000000+100;//50*100*100*2
const int BASE=maxn/2;
int val[maxn];
int main()
{
int a,b,c,d,i,j,t;
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(val,0,sizeof(val));
//不放此处可能会超时
for(i=1; i<=100; i++)
{
for(j=1; j<=100; j++)
{
t=a*i*i+b*j*j;
val[BASE-t]++;
}
}
int coun=0;
for(i=1; i<=100; i++)
{
for(j=1; j<=100; j++)
{
t=c*i*i+d*j*j;
coun+=val[BASE+t];//为啥是BASE,想想就知道了
//如果-(a*i*i+b*j*j)=c*i*i+d*j*j;BASE-a*x1*x1+b*x2*x2=BAse+c*i*i+d*j*j
}
}
printf("%d\n",coun*16);//每个解,都有正有负,每个位置有俩,2*2*2*2;
}
return 0;
}
E - Equations HDU - 1496
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转载自blog.csdn.net/zhangzhenjunaixuxin/article/details/81434642
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