哈希算法---Eqs

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Consider equations having the following form:
a1x1 ³+ a2x2 ³+ a3x3 ³+ a4x4 ³+ a5x5 ³=0
The coefficients are given integers from the interval [-50,50].
It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}.

Determine how many solutions satisfy the given equation.

Input

The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

Output

The output will contain on the first line the number of the solutions for the given equation.

Sample Input

37 29 41 43 47

Sample Output

/*（a1*x1^3+a2*x2^3+a3*x3^3） =-(a4*x4^3+a5*x5^3)最大值是 2.325*10^7右边最大值1.25*10^7    当结果是负数时，数组下标不能为负数---故加上2.5*10^7 */#include<iostream>#include<cstring>using namespace std;char Hash[25000000];//int ==Memory Limit const int MOD=25000000;int main(){int sum,a1,a2,a3,a4,a5,num;int x1,x2,x3,x4,x5;while(scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5)!=EOF){memset(Hash,'0',sizeof(Hash));//当字符串时---memset(Hash,0,sizeof(Hash));与其不一样 num=0;for(x1=-50;x1<=50;x1++){if(x1==0)continue;for(x2=-50;x2<=50;x2++){if(x2==0)continue;for(x3=-50;x3<=50;x3++){if(x3==0)continue;sum=(a1*x1*x1*x1+a2*x2*x2*x2+a3*x3*x3*x3);if(sum<0)sum=sum+MOD;//此处 取绝对值是不对的 Hash[sum]++;}}}for(x4=-50;x4<=50;x4++){if(x4==0)continue;for(x5=-50;x5<=50;x5++){if(x5==0)continue;sum=-1*(a4*x4*x4*x4+a5*x5*x5*x5);if(sum<0)sum=sum+MOD;//此处 会出现等式成立 但x不同的情况  所以存在多种情况 num+=Hash[sum]-'0';}}cout<<num<<endl;}return 0;}

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