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)


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")
}

use std::cmp::{max, min};
use std::collections::HashMap;

fn gcd(a: i64, b: i64) -> i64 {
  if a % b == 0 {
    b
  } else {
    gcd(b, a % b)
  }
}

fn lcm(a: i64, b: i64) -> i64 {
  a * b / gcd(a, b)
}

fn main() {
  let n = read::<i64>();
  let mut csf: Vec<Vec<i64>> = vec![];
  for _ in 0..n - 1 {
    let c = read::<i64>();
    let s = read::<i64>();
    let f = read::<i64>();
    csf.push(vec![c, s, f]);
  }

  for i in 0..n - 1 {
    let mut total = 0;
    for j in i..n - 1 {
      let j_size = j as usize;
      if total < csf[j_size][1] {
        total = csf[j_size][1];
        total += csf[j_size][0];
      } else {
        total += csf[j_size][2]
          - if (total % csf[j_size][2]) != 0 {
            (total % csf[j_size][2])
          } else {
            csf[j_size][2]
          };
        total += csf[j_size][0];
      }
    }
    println!("{}", total);
  }
  println!("{}", 0);
}