POJ 1840 Eqs

 

 

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 9567		Accepted: 4710

Description

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

654

题解：简单哈希。数组定义成char型。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MOD=19999999;
int ans[111];
char hashtb[MOD];

int main()
{
	int n,cnt=0;
	int a1,a2,a3,a4,a5;
	for(int i=-50;i<=50;i++){
		if(i==0)continue;
		ans[cnt++]=i*i*i;
	}
	while(scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5)!=EOF){
		memset(hashtb,0,sizeof(hashtb));
		for(int i=0;i<100;i++){
			for(int j=0;j<100;j++)
				for(int k=0;k<100;k++){
					int temp=a1*ans[i]+a2*ans[j]+a3*ans[k];
					if(temp<0) temp+=MOD;
					temp=temp%MOD;
					hashtb[temp]++;
				}
		}
		cnt=0;
		for(int i=0;i<100;i++){
			for(int j=0;j<100;j++){
				int temp=-(a4*ans[i]+a5*ans[j]);
				if(temp<0) temp+=MOD;
				temp=temp%MOD;
				cnt+=hashtb[temp];
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}