Sum of integers on the interval (Editorial)

If the sum is computed using a loop, then in the worst case (when a = -2 \cdot 10^9 and b = 2 \cdot 10^9) up to 4 \cdot 10^9 iterations would be required, which would lead to a time limit exceeded error. Therefore, we use the formula for the sum of an arithmetic progression. The first term is $a$, the last term is $b$, and the number of terms is b - a + 1. Then the required sum is computed using the formula:

It should also be noted that the resulting value may not fit into the int type, so the long long type must be used for the computations.

Algorithm implementation

Read the input data.

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

Compute and print the answer.

res = (a + b) * (b - a + 1) / 2;
printf("%lld\n",res);

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 + b) * (b - a + 1) / 2;
    System.out.println(res);
  }
}

Python implementation

Read the input data.

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

Compute and print the answer.

res = (a + b) * (b - a + 1) // 2
print(res)