Eqs POJ
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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.
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.
The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.
The output will contain on the first line the number of the solutions for the given equation.
37 29 41 43 47
654
#include<stdio.h>#include<math.h>#include<string.h>short jie[25000000];int main(){ int a1,a2,a3,a4,a5,i,j,t,k,sum; scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5); memset(jie,0,sizeof(jie)); for(i=-50;i<=50;i++) { for(j=-50;j<=50;j++) { if(i!=0&&j!=0) { t=i*i*i*a1+j*j*j*a2; if(t>0) jie[t]++; else { t=t+25000000; jie[t]++; } } } } sum=0; 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) { t=a3*i*i*i+a4*j*j*j+a5*k*k*k; t=-t; if(t>0&&t<25000000) sum=sum+jie[t]; if(t<=0&&t>-25000000) { t=t+25000000; sum=sum+jie[t]; } } } } } printf("%d\n",sum); return 0;
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