Problem:
Score : 100 points

Problem Statement

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Constraints

1≦N≦100

Input

The input is given from Standard Input in the following format:

N

Output

Print the necessary number of candies in total.

Sample Input 1

3

Sample Output 1

6

The answer is 1+2+3=6.

Sample Input 2

10

Sample Output 2

55

The sum of the integers from 1 to 10 is 55.

Sample Input 3

1

Sample Output 3

1

Only one child. The answer is 1 in this case.

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn main() {
    let n: u32 = read();
    let mut ans = 0;
    for i in 0..n {
        ans += i+1;
    }
    println!("{}", ans);
}

// 以下関数

use std::io::*;
use std::str::FromStr;
pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub const MOD: u64 = 1_000_000_007;

pub fn is_prime(n: i32) -> bool {
    if n < 2 {
        return false;
    } else if n == 2 {
        return true;
    }
    let mut i = 2;
    while i * i < n {
        if n % i == 0 {
            return false;
        } else {
            i += 1;
            continue;
        }
    }
    true
}

pub fn lcm(mut n: i32, mut m: i32) -> i32 {
    n = if n > m { m } else { n };
    m = if n > m { n } else { m };
    let mut i = n;
    while i % m != 0 {
        i += n;
    }
    i
}

pub fn abs(n: i32) -> i32 {
    if n >= 0 {
        n
    } else {
        -n
    }
}

pub fn max(n: i32, m: i32) -> i32 {
    if n >= m {
        n
    } else {
        m
    }
}

pub fn min(n: i32, m: i32) -> i32 {
    if n <= m {
        n
    } else {
        m
    }
}

pub fn factorial(n: u32) -> u32 {
    if n == 0 {
        return 1;
    }
    let mut r = n;
    for i in 0..n - 1 {
        r *= n - i - 1;
    }
    r
}

pub fn combination(n: u32, m: u32) -> u32 {
    factorial(n) / factorial(m) / factorial(n - m)
}