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swust oj 方程式(0300)
Description
Consider equations having the following form: a*x1*x1 + b*x2*x2 + c*x3*x3 + d*x4*x4 = 0 a, b, c, d are integers from the interval [-50,50] and any of them cannot be 0. It is consider a solution a system ( x1,x2,x3,x4 ) that verifies the equation, xi is an integer from [-100,100] and xi != 0, any i ∈{1,2,3,4}. Determine how many solutions satisfy the given equation.
Input
The input consists of several test cases. Each test case consists of a single line containing the 4 coefficients a, b, c, d, separated by one or more blanks.
Output
or each test case, output a single line containing the number of the solutions.
Sample InpUT
1 2 3 -4
1 1 1 1
Sample Output
39088
0
#include<stdio.h>struct pz{ int x;}book1[1000000],book2[1000000];int main(){ int a,b,c,d; while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){ if((a > 0 && b > 0 && c > 0 && d > 0) || (a < 0 && b < 0 && c < 0 && d < 0)) { printf("0\n"); continue; } int x; int i,j; for(i=1;i<=100;i++) for(j=1;j<=100;j++){ x=a*i*i+b*j*j; if(x>0) book1[x].x++; else book2[-x].x++; // printf("%d %d\n",book1[k-1],book2[k-1]); } int ans=0; for(i=1;i<=100;i++) for(j=1;j<=100;j++){ x=-(c*i*i+d*j*j); if(x>0) ans+=book1[x].x; else ans+=book2[-x].x; } printf("%d\n",ans*4*4); for(i=0;i<1000000;i++){ book1[i].x=0; book2[i].x=0; } } return 0;}
swust oj 方程式(0300)
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