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;
}