hdu 1496 hash表求表达式的解的个数 2中hash方法

Equations

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3356    Accepted Submission(s): 1347

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

 

#include
#include
int hash[2000000];
int main()
{
     int a,b,c,d,i,j,ans;
	 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))
		 {
			 printf("0\n");continue;
		 }
		 ans=0;
		 memset(hash,0,sizeof(hash));
		 for(i=1;i<=100;i++)
			 for(j=1;j<=100;j++)
				 hash[1000000+a*i*i+b*j*j]++;//若a b 为正 则 c d其中有负数 那么用100000一减结果和这里的小标一样
		 for(i=1;i<=100;i++)
			 for(j=1;j<=100;j++)
				 ans+=hash[1000000-c*i*i-d*j*j];
	     printf("%d\n",ans*16);//由于每个xi都可能为正或为负 因为是平方 所以其实有16倍我们找到的答案
	 }
	 return 0;
}

下面这个方法 我认为应该是更加一般的方法 使用范围更加广泛一些 

/*
*  hash+数学，很好的题 
*  对整数求hash，采用除余法，及线性探测解决冲突 
*  注意：devc++中不能定义全局变量count，它和库函数中的函数名同名了 
*/
#include 
#include 
#include 
using namespace std;
const int M = 175447;
int counts[M];
int result[M];
int temp[101];
int hashInt(int s)
{
    int k = s % M;
    if (k < 0) k += M;
    while (counts[k] && result[k]!=s) k = (k + 1) % M;//如果当前对应不为空 且对应的不是s 则继续寻找
    return k;
}

int solve(int a, int b, int c, int d) {
    if (a>0&&b>0&&c>0&&d>0 || a<0&&b<0&&c<0&&d<0) return 0;
    memset(counts, 0, sizeof(counts));
    int s, p, i, j;
    for (i=1; i<101; ++i) {
        for (j=1; j<101; ++j) {
            s = a * temp[i] + b * temp[j];
            p = hashInt(s);
            result[p] = s;//
            ++counts[p];
        }
    }
    int ans = 0;
    for (i=1; i<101; ++i) {
        for (j=1; j<101; ++j) {
            s = -(c * temp[i] + d * temp[j]);
            p = hashInt(s);
            ans += counts[p];
        }
    }
    return ans;
}

int main() {
    int a, b, c, d;
    for (int i=1; i<101; ++i) temp[i] = i * i;
    while (scanf("%d%d%d%d", &a, &b, &c, &d) != EOF) {
        int ans = solve(a, b, c, d);
        printf ("%d\n", ans<<4);
    }
    return 0;
}