ANS=0;
if x*x+x*x==n ANS+=4;//( x,x) (-x,-x) (-x,x) (x,-x)
if x*x+0*0==n ANS+=4;//(x,0) (0,x) (-x,0) (0,-x)
if a*a+b*b==n ANS+=8;
View Code
1 #include<stdio.h> 2 #include<math.h> 3 #include<string.h> 4 const int maxn = 10005; 5 int a[ maxn ]; 6 int main(){ 7 memset( a,0,sizeof(a)); 8 a[0]=1; 9 for( int i=1;i<maxn;i++ ){ 10 int tmp=i/2; 11 int now=1; 12 while( now*now<tmp ){ 13 int sum=i-now*now; 14 int tp=sqrt( sum*1.0 ); 15 if( tp*tp==sum ) a[ i ]+=8; 16 now++; 17 } 18 if( (int)sqrt(i*1.0)*(int)sqrt(i*1.0)==i ) a[ i ]+=4; 19 if( (int)sqrt(i*1.0)*(int)sqrt(i*1.0)*2==i ) a[ i ]+=4; 20 } 21 int n; 22 while( scanf("%d",&n)==1 ){ 23 if( n<maxn ){ 24 printf("%d\n",a[ n ]); 25 continue; 26 } 27 int ans=0; 28 int i=n; 29 int tmp=i/2; 30 int now=1; 31 while( now*now<tmp ){ 32 int sum=i-now*now; 33 int tp=sqrt( sum*1.0 ); 34 if( tp*tp==sum ) ans+=8; 35 now++; 36 } 37 if( (int)sqrt(i*1.0)*(int)sqrt(i*1.0)==i ) ans+=4; 38 if( (int)sqrt(i*1.0)*(int)sqrt(i*1.0)*2==i ) ans+=4; 39 printf("%d\n",ans); 40 } 41 return 0; 42 }