Codeforces Round #451 (Div. 2) B. Proper Nutrition【枚举/扩展欧几里得/给你n问有没有两个非负整数x,y满足x·a + y·b = n】

B. Proper Nutrition

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 andx·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.

【题意】：给你n问有没有两个非负整数x，y满足x·a + y·b = n。

【分析】：略带技巧的枚举。非负整数x，y上面做文章。

【代码】：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,a,b;
    cin>>n>>a>>b;
    for(int i=;i*a<=n;++i)//自己这个地方是i<=n,错了，可能出现到负数的时候%b==0的情况
    {
        if((n-i*a)%b==)
        {
            int j=(n-i*a)/b;
            puts("YES");
            printf("%d %d\n",i,j);
            exit();
        }
    }
    puts("NO");
    return ;
}

枚举，整数性判断

#include <bits/stdc++.h>

using namespace std;
typedef long long int ll;
ll ex_gcd(ll a,ll b,ll &x,ll &y)
{
    if(b==)
    {
        x=;
        y=;
        return a;
    }
   ll r=ex_gcd(b,a%b,x,y);
    ll t=x;
    x=y;
    y=t-a/b*y;
    return r;
}
bool jie(ll a,ll b,ll c,ll &x,ll &y)
{
    ll r=ex_gcd(a,b,x,y);
    if(c%r!=) return ;
    ll zz=c/r;
    x*=zz;
    y*=zz;
    if(x<&&y<) return ;
     ll bb=b/r;
     ll aa=a/r;
    if(x<&&y>)
    {
        ll t=-*x/bb;
        if(x+bb*t<) t++;
        x=x+bb*t;
        y=y-aa*t;
        if(y>=) return ;
        else return ;
    }
   else if(x>&&y<)
    {
        ll t=y/aa;
        if(y-aa*t<) t--;

        x=x+bb*t;
        y=y-aa*t;
        if(x>=) return ;
        else return ;
    }
    return ;
}
ll a,b,n,x,y;
int main()
{
    while(~scanf("%lld%lld%lld",&n,&a,&b))
    {
        x=;
        y=;
        if(jie(a,b,n,x,y))
        {
            printf("YES\n");
            printf("%lld %lld\n",x,y);
        }
        else
        {
            printf("NO\n");
        }
    }
    return ;
}

拓展欧几里得