203A Two Problems

简单题，要注意可能做出0、1、2题，枚举所有情况即可；

# include <cstdio>



int a, b, x, da, db, t;

int i, j;



void solve(void)

{

    if (!x || (a>=x && (a-x)%da==0 && (a-x)/da<=t-1) || (b>=x &&(b-x)%db==0&&(b-x)/db<=t-1))

    {

        printf("YES\n");

        return ;

    }

    for (i = t-1; i >= 0; --i)

    for (j = t-1; j >= 0; --j)

    {

        if(a+b-x == i*da+j*db) {printf("YES\n"); return;}

    }

    printf("NO\n");

}



int main()

{

    while (~scanf("%d%d%d%d%d%d", &x, &t, &a, &b, &da, &db))

    {

        solve();

    }



    return 0;

}