POJ 1840 Eqs

 

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 15010		Accepted: 7366

 

Description

Consider equations having the following form: 
a1x13+ a2x23+ a3x33+ a4x43+ a5x53=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

Source

Romania OI 2002

 

按要求模拟即可

hash是个神奇的东西

下方代码注释部分是先三层循环后二层，正文部分是先二层循环后三层。两者都是正解，但是由于list插入比读取慢,先二层更快

/*
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<list>
using namespace std;
const int mxn=14997;
list<int>ha[mxn*2];
list<int>::iterator it;
int x,x1,x2,x3,x4,x5;
int ans=0;
int main(){
scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);
int i,j,k;
for(i=-50;i<=50;i++)
for(j=-50;j<=50;j++)
for(k=-50;k<=50;k++){
if(i==0||j==0||k==0)continue;
x=i*i*i*x1+j*j*j*x2+k*k*k*x3;
ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数
}
for(i=-50;i<=50;i++)
for(j=-50;j<=50;j++){
if(i==0|j==0)continue;
x=-(i*i*i*x4+j*j*j*x5);
//检查hash
for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){
if(*it==x)ans++;
}
}
printf("%d",ans);
return 0;

}
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<list>
using namespace std;
const int mxn=14997;
list<int>ha[mxn*2];
list<int>::iterator it;
int x,x1,x2,x3,x4,x5;
int ans=0;
int main(){
scanf("%d%d%d%d%d",&x1,&x2,&x3,&x4,&x5);
int i,j,k;
for(i=-50;i<=50;i++)
for(j=-50;j<=50;j++)
{
if(i==0||j==0)continue;
x=i*i*i*x1+j*j*j*x2;
ha[x%mxn+mxn].push_back(x);//hash //x%mxn+mxn保证hash完以后是正数
}
for(i=-50;i<=50;i++)
for(j=-50;j<=50;j++)
for(k=-50;k<=50;k++){
if(i==0|j==0||k==0)continue;
x=-(i*i*i*x3+j*j*j*x4+k*k*k*x5);
//检查hash
for(it=ha[x%mxn+mxn].begin();it!=ha[x%mxn+mxn].end();it++ ){
if(*it==x)ans++;
}
}
printf("%d",ans);
return 0;

}

 

niconiconi