poj 1840
2015-08-03 23:51
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Eqs
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
Sample Output
Source
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 13967 | Accepted: 6858 |
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
#include<stdio.h> #include<string.h> #include<iostream> #include<algorithm> using namespace std; int hash[1000000]; int main(){ int a,b,c,d,e; while(cin>>a>>b>>c>>d>>e){ memset(hash,0,sizeof(hash)); int cnt=0; for(int i=-50;i<=50;i++){ if(i==0) continue; for(int j=-50;j<=50;j++){ if(j==0) continue; for(int k=-50;k<=50;k++){ if(k==0) continue; int temp=i*i*i*a+j*j*j*b+k*k*k*c; hash[cnt++]=temp; } } } int ans=0; sort(hash,hash+cnt); for(int i=-50;i<=50;i++){ if(i==0) continue; for(int j=-50;j<=50;j++){ if(j==0) continue; int temp2=i*i*i*e+j*j*j*d; ans+=upper_bound(hash,hash+cnt,temp2)-lower_bound(hash,hash+cnt,temp2); } } printf("%d\n",ans); } return 0; }
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