Problem:
Score : 300 points

Problem Statement

A railroad running from west to east in Atcoder Kingdom is now complete.

There are N stations on the railroad, numbered 1 through N from west to east.

Tomorrow, the opening ceremony of the railroad will take place.

On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.

The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.

Here, it is guaranteed that F_i divides S_i.

That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.

For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Constraints

1≤N≤500

1≤C_i≤100

1≤S_i≤10^5

1≤F_i≤10

S_i％F_i=0

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 S_1 F_1
:
C_{N-1} S_{N-1} F_{N-1}

Output

Print N lines. Assuming that we are at Station i (1≤i≤N) when the ceremony begins, if the earliest possible time we can reach Station N is x seconds after the ceremony begins, the i-th line should contain x.

Sample Input 1

3
6 5 1
1 10 1

Sample Output 1

12
11
0

We will travel from Station 1 as follows:

5 seconds after the beginning: take the train to Station 2.

11 seconds: arrive at Station 2.

11 seconds: take the train to Station 3.

12 seconds: arrive at Station 3.

We will travel from Station 2 as follows:

10 seconds: take the train to Station 3.

11 seconds: arrive at Station 3.

Note that we should print 0 for Station 3.

Sample Input 2

4
12 24 6
52 16 4
99 2 2

Sample Output 2

187
167
101
0

Sample Input 3

4
12 13 1
44 17 17
66 4096 64

Sample Output 3

4162
4162
4162
0

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

#![allow(unused_mut)]
#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(unused_imports)]
#![allow(unused_macros)]
#![allow(dead_code)]
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => {
        cmp::max($x, max!($($y),+))
    };
}

macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => {
        cmp::min($x, min!($($y),+))
    };
}

const MOD: usize = 1_000_000_007;
const MAX: usize = 510_000;

struct BinCoeff {
    fac: Vec<usize>,
    finv: Vec<usize>,
    inv: Vec<usize>,
}

impl BinCoeff {
    fn new() -> BinCoeff {
        let fac = vec![0; MAX];
        let finv = vec![0; MAX];
        let inv = vec![0; MAX];
        let mut bincoeff = BinCoeff { fac: fac, finv: finv, inv: inv };
        bincoeff.com_init();
        bincoeff
    }

    fn com_init(&mut self) {
        let fac = &mut self.fac;
        let finv = &mut self.finv;
        let inv = &mut self.inv;
        fac[0] = 1;
        fac[1] = 1;
        finv[0] = 1;
        finv[1] = 1;
        inv[1] = 1;
        for i in 2..MAX {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    fn bin_coeff(&mut self, n: usize, k: usize) -> usize {
        if k > n/2 {
            self.bin_coeff(n-k, k)
        } else {
            self.fac[n] * (self.finv[k] * self.finv[n - k] % MOD) % MOD
        }
    }
}

// must input n <= m
fn gcd(n: usize, m: usize) -> usize {
    if n == 0 {
        m
    } else {
        gcd(m%n, n)
    }
}

use std::cmp;
use std::collections::{HashSet, BinaryHeap, VecDeque, HashMap};
use std::iter;
fn main() {
    input! {
        N: usize,
        stations: [(usize, usize, usize); N-1],
    }

    for start in 0..N-1 {
        let ans = stations[start..].iter().fold(0, |acc, &(dist, first, interval)| {
            cmp::max(first, interval * ((acc+interval-1)/interval)) + dist
        });
        println!("{}", ans);
    }
    println!("0");
}