Codeforces Round #451 (Div. 2) Proper Nutrition 暴力枚举_vasya has n burles. one bottle of ber-cola costs a-优快云博客

本文介绍了一个简单的算法，用于解决二元一次不定方程的正整数解问题，即给定金额n和两种商品的价格a、b，判断是否能通过购买一定数量的商品使得总花费恰好等于n。通过枚举一种商品的数量并检查剩余金额是否为另一种商品价格的倍数，从而找到符合条件的解。

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Proper Nutrition

Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.

Find out if it’s possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.

In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it’s impossible.
Input

First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.

Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.

Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output

If Vasya can’t buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).

Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.

Any of numbers x and y can be equal 0.
Examples
Input

7
2
3

Output

YES
2 1

Input

100
25
10

Output

YES
0 10

Input

15
4
8

Output

Input

9960594
2551
2557

Output

YES
1951 1949

Note

In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.

In second example Vasya can spend exactly n burles multiple ways:

buy two bottles of Ber-Cola and five Bars bars;
buy four bottles of Ber-Cola and don’t buy Bars bars;
don’t buy Ber-Cola and buy 10 Bars bars.

In third example it’s impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.

题目的意思就是让你求二元一次不定方程c=ax+by的正整数根，我原以为要用扩展欧几里德来计算……没想到暴力也能过

#include<bits/stdc++.h>
using namespace std;
bool ok=false;
int main()
{
  long long n,a,b,x,y;
  cin>>n>>a>>b;
  for(int i=0;i<=n;i++)//枚举x
    { int c=(n-a*i)/b;
      if(c>=0&&(a*i+b*c)==n)
      {
        x=i;y=c;
        ok=true;
        break;
      }
    }
  if(ok)
    {
      cout<<"YES"<<endl;
      cout<<x<<endl;
      cout<<y<<endl;
    }
    else
    cout<<"NO"<<endl;

}