Table of contents

• Problem
• Explanation
• Time complexity
• Solution

# Problem

https://atcoder.jp/contests/abc085/tasks/abc085_c

Given 2 integers $N$ and $Y$, report if it's possible to make $Y$ yen with $N$ bills or not.
A bill is either a 10000 yen, a 5000 yen or a 1000 yen.
If it's possible, print any combination of them.

# Explanation

Letting the numbers of 10000 yen bill, 5000 yen bill and 1000 yen bill $a$, $b$ and $c$, you get 2 equations (1) and (2).
And as $a$, $b$ and $c$ are the number of bills, you get (3).

$\begin{cases} a + b + c = N (1) \\ 10000a + 5000b + 1000c = Y (2) \\ 0 \leq a, b, c \leq N (3) \end{cases}$

From (1) $c$ is unique for a combination of $a$ and $b$.

$c = N - a - b$

If you find any combination of $a$, $b$ and $c$ which satisfies (2) and (3), print it and done.

# Time complexity

$O(N^2)$

Double loop for making combinations of $a$ and $b$.

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define MAX_N 2001
#define MAX_Y 20000001

Int N, Y;

void input() {
  cin >> N >> Y;
}

void solve() {
  loop(a,0,N+1) {
    loop(b,0,N+1) {
      Int c = N - a - b;
      if (a + b > N || c < 0) continue;
      if (a * 10000 + b * 5000 + c * 1000 == Y) {
        cout << a << ' ' << b << ' ' << c << endl;
        return;
      }
    }
  }
  cout << "-1 -1 -1" << endl;
}

int main(void) {
  input();
  solve();
  return 0;
}