https://atcoder.jp/contests/math-and-algorithm/tasks/abc204_d
2つに分けてトータル時間が半分に近くなればよいですが、どの時間が実現可能かはDPで簡単に分かるので、半分以上で半分に近い実現可能な時間を探せばよいです。
// Cooking #![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() } //////////////////// process //////////////////// fn read_input() -> Vec<i32> { let _N: usize = read(); let T = read_vec(); T } fn initialize_dp(T: &Vec<i32>) -> Vec<bool> { let S = T.iter().sum::<i32>() as usize; let mut dp: Vec<bool> = vec![false; S+1]; dp[0] = true; dp } fn update_dp(dp: Vec<bool>, t: i32) -> Vec<bool> { let mut new_dp = dp.to_vec(); for (i, _) in dp.into_iter().enumerate().filter(|(_, b)| *b) { new_dp[i+(t as usize)] = true } new_dp } fn f(T: Vec<i32>) -> usize { let half = (T.iter().sum::<i32>() as usize + 1) / 2; let mut dp = initialize_dp(&T); for t in T.into_iter() { dp = update_dp(dp, t) } (half..dp.len()).filter(|&i| dp[i]).next().unwrap() } fn main() { let T = read_input(); println!("{}", f(T)) }