HDOJ 1496 Equations hash

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.

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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题目大意：

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.

HDU ACM的PPT上有两种解法。第一种用了大数组，第二种用了小数组+冲突处理。

由于小数组的初始化时间少，第二种解法更优。

#include <iostream>
#include <cstring>
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]++;
for (i=1;i<=100;++i)
for (j=1;j<=100;++j)
sum+=hash[1000000-(c*pin[i]+d*pin[j])];
cout<<sum*16<<endl;
}
return 0;
}

#include <iostream>
#include <cstring>
using namespace std;
const int MAX=50021;
int f[MAX],g[MAX];
int a,b,c,d,pin[101],i,j,sum,s,p;

int hash(int k){
int t=k%MAX;
if (t<0) t+=MAX;
while (f[t]!=0 && g[t]!=k)
t=(t+1)%MAX;
return t;
}

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(f,0,sizeof(f)); sum=0;
for (i=1;i<=100;++i)
for (j=1;j<=100;++j){
s=a*pin[i]+b*pin[j];
p=hash(s);
g[p]=s;
f[p]++;
}
for (i=1;i<=100;++i)
for (j=1;j<=100;++j){
s=-(c*pin[i]+d*pin[j]);
p=hash(s);
sum+=f[p];
}
cout<<(sum<<4)<<endl;
}
return 0;
}

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