Codeforces Round #451 (Div. 2) Proper Nutrition 暴力枚举

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

NO

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;}