Pythagorean Theorem II

A. Pythagorean Theorem IItime limit per test

3 seconds

Memory limit per test

256 megabytes

Input

Standard input

Output

Standard output

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 (
Two sides that meet at a right angle ).

The theorem can be written as an equation relating the lengths of the sidesA,BAndC,
Often called the Pythagorean equation:

A ^{2 cores + CoresB2 bytes = bytesC2}

WhereCRepresents the length of the hypotenuse, andAAndBRepresent
The lengths of the other two sides.

GivenN, Your task is to count how many right-angled triangles with side-lengthsA,BAndCThat
Satisfied an inequality 1 limit ≤ limitALimit ≤ limitBLimit ≤ limitCLimit ≤ limitN.

Input

The only line contains one integerN(1 digit ≤ DigitNLimit ≤ limit 10^{4)
We mentioned above.}

Output

Print a single integer-the answer to the problem.

Sample test (s) input

5

Output

1

Input

74

Output

35

Idea: 10 ^ 8MLE, two points + enumeration n. Enumerative n is a good posture. Do not look for map for the free time of table creation, or it will die very quickly.

In addition, Yu GE's direct root code is actually the simplest and most effective method.

#include <algorithm>#include <iostream>#include <iomanip>#include <cstdio>#include <map>#include <cstdlib>#include <cstring>typedef long long ll;#defineclr(a)memset((a),0,sizeof (a))#definerep(i,a,b)for(int i=(a);i<(int)(b);i++)#defineper(i,a,b)for(int i=((a)-1);i>=(int)(b);i--)#defineinf0x7ffffff#defineeps1e-6using 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' functionbool 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;}