HDU1496：Equations（二分）

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8166    Accepted Submission(s): 3354

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

 

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

题意：给定a,b,c,d，求在定义域内有多少种解。

# include <stdio.h>
# include <math.h>
# include <algorithm>
using namespace std;
int x[10001], y[10001];
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 || a<0&&b<0&&c<0&&d<0)//全为负数的剪枝快1s多。
        {
            puts("0");
            continue;
        }
        int k=0;
        long long ans = 0;
        for(int i=1; i<=100; ++i)
            for(int j=1; j<=100; ++j)
            {
                x[k] = a*i*i+b*j*j;
                y[k++] = c*i*i+d*j*j;
            }
        sort(x, x+k);
        sort(y, y+k);
        for(int i=0; i<k; ++i)
        {
            int l = 0, r = k-1, mid;
            long long tmp = 0;
            while(l<r)
            {
                mid = (l+r)>>1;
                if(x[i]+y[mid]<0)
                    l = mid+1;
                else
                    r = mid;
            }
            if(x[i]+y[r] == 0)
            {
                ++tmp;
                for(int t=r-1; t>=0&&x[i]+y[t]==0; ++tmp,--t);
                for(int t=r+1; t<k&&x[i]+y[t]==0; ++tmp,++t);
                ans += tmp;
                while(i+1 < k && x[i+1]==x[i])
                {
                    ++i;
                    ans += tmp;
                }
            }
        }
        printf("%lld\n",ans<<4);//因为是x^2,所以有2*2*2*2种情况。
    }
    return 0;
}