HDU 3835 R(N)
R(N)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2761 Accepted Submission(s):
1420
as x=A^2+B^2(A,B are integers). Take x=10 for example,
10=(-3)^2+1^2.
We
define R(N) (N is positive) to be the total number of variable presentation of
N. So R(1)=4, which consists of 1=1^2+0^2, 1=(-1)^2+0^2, 1=0^2+1^2,
1=0^2+(-1)^2.Given N, you are to calculate R(N).
one integer N(N<=10^9).
6
10
25
65
0
8
12
16
For the fourth test case, (A,B) can be (0,5), (0,-5), (5,0), (-5,0), (3,4), (3,-4), (-3,4), (-3,-4), (4,3) , (4,-3), (-4,3), (-4,-3)
include <iostream>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
if(n==)
{
cout<<<<endl;
continue;
}
int c=sqrt(n/);
int ans=;
int i,j;
for(i=;i<=c;i++)
{
j=sqrt(n-i*i);
if(i*i+j*j==n)
{
if(i==||j==||i==j)
ans=ans+;
else
ans=ans+;
}
}
printf("%d\n",ans);
}
return ;
}
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