hdu 1496 二分或者哈希

http://acm.hdu.edu.cn/showproblem.php?pid=1496

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

题目很简单，只贴代码：

二分：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL sum[400007];
LL ans,k;
void erfen(LL t)
{
int l=0,r=k;
int mid;
while(l<=r)
{
mid=(l+r)/2;
if(sum[mid]==t)
{
ans++;
//printf("%d\n",ans);
for(int x=mid-1; x>=1&&sum[x]==t; x--)
{
if(sum[x]==t)
ans++;
}
for(int x=mid+1; x<k&&sum[x]==t; x++)
{
if(sum[x]==t)
ans++;
}
return;
}
else if(sum[mid]<t)
l=mid+1;
else
r=mid-1;
}
}
int main()
{
int a,b,c,d;
while(~scanf("%d%d%d%d",&a,&b,&c,&d))
{
k=0;
if(a>0&&b>0&&c>0&&d>0)
{
printf("0\n");
continue;
}
for(int i=1; i<=100; i++)
for(int j=1; j<=100; j++)
sum[k++]=a*i*i+b*j*j;
sort(sum,sum+k);
ans=0;
for(int i=1; i<=100; i++)
for(int j=1; j<=100; j++)
{
LL t=-(c*i*i+d*j*j);
erfen(t);
}
printf("%I64d\n",ans*16);
}
return 0;
}

哈希：
#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn=1000005;
int vis[2000005];
int main()
{
int a,b,c,d;
while(~scanf("%d%d%d%d",&a,&b,&c,&d))
{
if(a>0&&b>0&&c>0&&d>0)
{
printf("0\n");
continue;
}
int ans=0;
memset(vis,0,sizeof(vis));
for(int i=1;i<=100;i++)
for(int j=1;j<=100;j++)
vis[i*i*a+b*j*j+maxn]++;
for(int i=1;i<=100;i++)
for(int j=1;j<=100;j++)
ans+=vis[-(c*i*i+d*j*j)+maxn];
printf("%d\n",ans*16);
}
return 0;
}