HDU 1496 Equations

Equations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6441 Accepted Submission(s): 2603

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

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分析：

题意简单（类似“百钱买百鸡”问题）
传统方法是（几重循环）？复杂度？
上述方法如何判断两端相等？（查找？）
可行方案（两重循环＋Hash存储查找）
Hash表大小？

#include <iostream>#include <sstream>#include <iomanip>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <bitset>#include <string>#include <numeric>#include <algorithm>#include <functional>#include <iterator>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <cctype>#include <complex>#include <ctime>typedef long long LL;const double pi = acos(-1.0);const long long mod = 1e9 + 7;using namespace std;int x[2000005];int main(){    int a,b,c,d;    while(scanf("%d%d%d%d",&a,&b,&c,&d) != EOF)    {        if( (a > 0 && b > 0 && c > 0 && d > 0) ||  (a < 0 && b < 0 && c < 0 && d < 0) )        {            puts("0");            continue;        }        memset(x,0,sizeof(x));        for(int i = 1;i <= 100;i++)            for(int j = 1;j <= 100;j++)                x[a * i * i + b * j * j  + 1000000]++;        int sum = 0;        for(int i = 1;i <= 100;i++)            for(int j = 1;j <= 100;j++)                sum += x[ - c * i * i - d * j * j + 1000000];        printf("%d\n",sum * 16);    }    return 0;}