Eqs
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 10828 | Accepted: 5261 |
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.
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 47Sample Output
654Source
Romania OI 2002#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int a,b,c,d,e; int hash[2000000]; int main() { scanf("%d%d%d%d%d",&a,&b,&c,&d,&e); 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 tmp=i*i*i*a+j*j*j*b+k*k*k*c; hash[cnt++]=tmp; } } } sort(hash,hash+cnt); int ans=0; for(int i=-50;i<=50;i++) { if(i==0) continue; for(int j=-50;j<=50;j++) { if(j==0) continue; int tmp=i*i*i*d+j*j*j*e; ans+=upper_bound(hash,hash+cnt,-tmp)-lower_bound(hash,hash+cnt,-tmp); } } printf("%d ",ans); return 0; } |