Pythagorean Theorem II

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In mathematics, the Pythagorean theorem — is a relation in Euclidean geometry among the three sides of a right-angled triangle. In terms of areas, it states:

In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:

a² + b² = c²

where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides.

$\text{[math]}$

Given n, your task is to count how many right-angled triangles with side-lengths a, b and c that satisfied an inequality 1 ≤ a ≤ b ≤ c ≤ n.

Input

The only line contains one integer n (1 ≤ n ≤ 10⁴) as we mentioned above.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input
5

output
1

input
74

output
35

思路：10^8MLE，二分+枚举n。枚举n是一个好姿势，打表的好姿势。打表的空余时间别找map，or会死的很有节奏。

另外，羽哥的直接开根号其实就是最简单有效的方法。

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <map>
#include <cstdlib>
#include <cstring>
typedef long long ll;
#define	clr(a)		memset((a),0,sizeof (a))
#define	rep(i,a,b)	for(int i=(a);i<(int)(b);i++)
#define	per(i,a,b)	for(int i=((a)-1);i>=(int)(b);i--)
#define	inf	0x7ffffff
#define	eps			1e-6
using namespace std;
int mm[10005];
int num[10005];
/*bool find(int a,int b,int  key){
    int mid=(a+b)/2;  
    if(mm[mid]==key){
       return 1;            
    } 
    if(a==b-1)return 0;//cause I sent the mid ，not the mid-1 or mid+1。 
    if(mm[mid]>key){
      return find(a,mid,key);     
    }    
    else{
      return find(mid,b,key);     
    }
}*/
// the more adequate 'find' function
bool find(int a,int b,int  key){
    if(a>b)return 0;
    int mid=(a+b)/2;  
    if(mm[mid]==key){
       return 1;            
    } 
    if(mm[mid]>key){
      return find(a,mid-1,key);     
    }    
    else{
      return find(mid+1,b,key);     
    }
}
int main(){  
  for(int i=1;i<=10000;i++){
      mm[i]=i*i;    
  }
  for(int i=10000;i>=1;i--){
     for(int j=i-1;j>=1;j--){
        int tt=i*i-j*j;
        if(tt>j*j)break;
        if(find(1,10000,tt)){        
            num[i]++;                   
        }
            
     }        
  }
  int n;
  scanf("%d",&n);
  int sum=0;
  for(int i=1;i<=n;i++){
      sum+=num[i];       
  }
  printf("%d\n",sum); 
  //system("pause");
  return 0;
}