本文共 1324 字，大约阅读时间需要 4 分钟。
Equations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2161 Accepted Submission(s): 839 Consider equations having the following form:
a*x1^2+b*x2^2+c*x3^2+d*x4^2=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.
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.
End of file.
For each test case, output a single line containing the number of the solutions.
#include<stdio.h>
#include<memory.h>
int pin[101];
int hash[2000003];
main()
{
int a,b,c,d,i,j,sum;
for(i=1;i<=100;i++)
{
pin[i]=i*i;
}
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;
}
sum=0;
memset(hash,0,sizeof(hash));
for(i=1;i<=100;i++)
{
for(j=1;j<=100;j++)
{
hash[a*pin[i]+b*pin[j]+1000000]++;
}
}
for(i=1;i<=100;i++)
{
for(j=1;j<=100;j++)
{
sum+=hash[-c*pin[i]-d*pin[j]+1000000];
}
}
printf("%d\n",sum*16);
}
}
转载地址：http://mttpi.baihongyu.com/