Sum of squares#519
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Sum of squares (Editorial)

Since a, b \leq 10^9, then a^2 + b^2 \leq 2 \cdot 10^{18}. Therefore, to compute the result, the long long type should be used.

Algorithm implementation

Read the input data.

scanf("%lld %lld",&a,&b);

Compute and print the answer.

res = a * a + b * b;
printf("%lld\n",res);

Algorithm implementation — STL

#include <iostream>
using namespace std;

long long res, a, b;

int main(void)
{
  cin >> a >> b;
  res = a * a + b * b;
  cout << res << endl;
  return 0;
}
Java

Java implementation

import java.util.*;

public class Main
{
  public static void main(String []args)
  {
    Scanner con = new Scanner(System.in);
    long a = con.nextLong();
    long b = con.nextLong();
    long res = a*a + b*b;    
    System.out.println(res);
    con.close();
  }
}
Java

Python implementation

Read the input data.

a, b = map(int,input().split())

Compute and print the answer.

res = a**2 + b**2
print(res)

Go implementation

package main

import "fmt"

func main() {
  var a, b, res int64
  fmt.Scanf("%d %d", &a,&b)
  res = a * a + b * b
  fmt.Println(res)
}
Go