<cf>Two Problems

Codeforces比赛得分策略

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#each #input

本文探讨了在Codeforces特定比赛中，参赛者Valera如何通过解决两个随时间减少分数的问题来获得确切的分数x。文章详细解释了比赛机制，并提供了一个算法解决方案来验证目标分数是否可实现。

A boy Valera registered on site Codeforces as Valera, and wrote his first Codeforces Round #300. He boasted to a friend Arkady about winning as much as x points for his first contest. But Arkady did not believe his friend's words and decided to check whether Valera could have shown such a result.

He knows that the contest number 300 was unusual because there were only two problems. The contest lasted for t minutes, the minutes are numbered starting from zero. The first problem had the initial cost of a points, and every minute its cost reduced by d_a points. The second problem had the initial cost of b points, and every minute this cost reduced by d_b points. Thus, as soon as the zero minute of the contest is over, the first problem will cost a - d_a points, and the second problem will costb - d_b points. It is guaranteed that at any moment of the contest each problem has a non-negative cost.

Arkady asks you to find out whether Valera could have got exactly x points for this contest. You should assume that Valera could have solved any number of the offered problems. You should also assume that for each problem Valera made no more than one attempt, besides, he could have submitted both problems at the same minute of the contest, starting with minute 0 and ending with minute number t - 1. Please note that Valera can't submit a solution exactly t minutes after the start of the contest or later.

Input

The single line of the input contains six integers x, t, a, b, d_a, d_b (0 ≤ x ≤ 600; 1 ≤ t, a, b, d_a, d_b ≤ 300) — Valera's result, the contest's duration, the initial cost of the first problem, the initial cost of the second problem, the number of points that the first and the second problem lose per minute, correspondingly.

It is guaranteed that at each minute of the contest each problem has a non-negative cost, that is, a - i·d_a ≥ 0 and b - i· d_b ≥ 0 for all 0 ≤ i ≤ t - 1.

Output

If Valera could have earned exactly x points at a contest, print "YES", otherwise print "NO" (without the quotes).

Sample test(s)

input
30 5 20 20 3 5

output
YES

input
10 4 100 5 5 1

output
NO

Note

In the first sample Valera could have acted like this: he could have submitted the first problem at minute0 and the second problem — at minute 2. Then the first problem brings him 20 points and the second problem brings him 10 points, that in total gives the required 30 points.

这不是一道难题，但是一次AC还是不易的，因为有些细节问题很容易忽略，首先可以两题答错（x=0），也可以只答对一题，也可以两题皆对。

AC code

#include <iostream>
using namespace std;
int main()
{
    int x,t,a,b,da,db,i;
    double ta,tb;
    bool flag;
    while(cin>>x>>t>>a>>b>>da>>db)
    {
        flag=false;
        if(x==0) flag=true;
        for(i=0;i<t && !flag;i++)
        {
            int m=x-(a-i*da);
            if(m==0 || m==b)
            {
                flag=true;
            }
            else if(m>0 && m<b)
            {
                tb=1.0*(b-m)/db;
                if((int)tb==tb && tb>=0 && tb<t)
                {
                    flag=true;
                }
            }
        }
        for(i=0;i<t && !flag;i++)
        {
            int m=x-(b-i*db);
            if(m==0 || m==a)
            {
                flag=true;
            }
            else if(m>0 && m<a)
            {
                ta=1.0*(a-m)/da;
                if((int)ta==ta && ta>=0 && ta<t)
                {
                    flag=true;
                }
            }
        }
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}