AtCoder Beginner Contest 384 D

https://atcoder.jp/contests/abc384/tasks/abc384_d

繰り返し部分を除いて、数列Aの中か両端で和がSになる範囲を探せばいいですが、尺取り法でO(N)でできます。

// Repeated Sequence
#![allow(non_snake_case)]


//////////////////// library ////////////////////

fn read<T: std::str::FromStr>() -> T {
    let mut line = String::new();
    std::io::stdin().read_line(&mut line).ok();
    line.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace()
            .map(|e| e.parse().ok().unwrap()).collect()
}

fn YesNo(b: bool) -> String {
    return if b { "Yes" } else { "No" }.to_string()
}


//////////////////// process ////////////////////

fn read_input() -> (i64, Vec<i64>) {
    let v: Vec<i64> = read_vec();
    let S = v[1];
    let A: Vec<i64> = read_vec();
    (S, A)
}

fn accumulate(A: &Vec<i64>) -> Vec<i64> {
    let mut acc: Vec<i64> = vec![0];
    for &a in A.iter() {
        acc.push(acc[acc.len()-1] + a)
    }
    acc
}

fn is_equal_mid(S: i64, acc: &Vec<i64>) -> bool {
    let mut k = 0;
    let mut l = 0;
    while l < acc.len() {
        if acc[l] - acc[k] == S {
            return true
        }
        else if acc[l] - acc[k] < S {
            l += 1
        }
        else {
            k += 1
        }
    }
    false
}

fn is_equal_edge(S: i64, acc: &Vec<i64>) -> bool {
    is_equal_mid(*acc.iter().last().unwrap() - S, acc)
}

fn F(S: i64, A: Vec<i64>) -> bool {
    let acc = accumulate(&A);
    let sum = *acc.iter().last().unwrap();
    let r = S % sum;
    is_equal_mid(r, &acc) || is_equal_edge(r, &acc)
}

fn main() {
    let (S, A) = read_input();
    println!("{}", YesNo(F(S, A)))
}