Problem:
Score : 300 points

Problem Statement

A programming competition site AtCode provides algorithmic problems.
Each problem is allocated a score based on its difficulty.
Currently, for each integer i between 1 and D (inclusive), there are p_i problems with a score of 100i points.
These p_1 + … + p_D problems are all of the problems available on AtCode.

A user of AtCode has a value called total score.
The total score of a user is the sum of the following two elements:

Base score: the sum of the scores of all problems solved by the user.

Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of c_i points, aside from the base score (1 ≤ i ≤ D).

Takahashi, who is the new user of AtCode, has not solved any problem.
His objective is to have a total score of G or more points.
At least how many problems does he need to solve for this objective?

Constraints

1 ≤ D ≤ 10

1 ≤ p_i ≤ 100

100 ≤ c_i ≤ 10^6

100 ≤ G

All values in input are integers.

c_i and G are all multiples of 100.

It is possible to have a total score of G or more points.

Input

Input is given from Standard Input in the following format:

D G
p_1 c_1
:
p_D c_D

Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

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")
}

struct question {
  point: u32,
  stock: u32,
  c_point: u32,
}

fn main() {
  let d = read::<u32>();
  let g = read::<u32>();
  let mut qs = vec![];
  let mut answers = vec![];

  for i in 0..d {
    let p = read::<u32>();
    let c = read::<u32>();
    qs.push(question {
      point: (i + 1) * 100,
      stock: p,
      c_point: c,
    })
  }

  let two: u32 = 2;
  for i in 0..two.pow(d) {
    let result: String = format!("{:0>keta$b}", i, keta = d as usize);

    let r_char: Vec<char> = result.chars().collect();

    let yes_r: Vec<&question> = qs
      .iter()
      .filter(|x| r_char[((x.point / 100) - 1) as usize] == '1')
      .collect();
    let no_r: Vec<&question> = qs
      .iter()
      .filter(|x| r_char[((x.point / 100) - 1) as usize] != '1')
      .collect();

    let mut b_p: u32 = yes_r.iter().map(|x| (x.point * x.stock) + x.c_point).sum();
    let mut cost: u32 = yes_r.iter().map(|x| x.stock).sum();

    let mut no_r_cp = no_r.clone();

    if b_p >= g {
      answers.push(cost)
    } else {
      no_r_cp.sort_by_key(|x| x.point);
      let max_q: &question = match no_r_cp.last() {
        Some(s) => s,
        None => continue,
      };
      for _ in 1..max_q.stock {
        b_p += max_q.point;
        cost += 1;

        if b_p >= g {
          answers.push(cost)
        }
      }
    }
  }

  println!("{:?}", answers.iter().min().unwrap());
}