https://www.acwing.com/problem/content/5442/

t数组就是他们的排序，如果有0个比i大，那么就是最大，排好序后写不等式

hi + (n) * ai > hi+1 + (n) * ai+1

 const int N = 1e6 + 10;
struct node
{
    int t, h, a;
}st[N];
int cmp(node x, node y)
{
    return x.t < y.t;
}
void solve() 
{
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> st[i].h;
    for (int i = 1; i <= n; i++) cin >> st[i].a;
    for (int i = 1; i <= n; i++) cin >> st[i].t;
    sort(st + 1, st + 1 + n,cmp);
    int l = 0, r = 1e18;
    int ok = 1;
    for (int i = 1; i <= n - 1; i++)
    {
        if (st[i].a < st[i + 1].a)
        {
            if (st[i].h <= st[i + 1].h)
            {
                ok = 0;
                break;
            }
        }
        if (st[i].a == st[i + 1].a)
        {
            if (st[i].h <= st[i + 1].h)
            {
                ok = 0;
                break;
            }
        }
        else
        {
            if (st[i].a > st[i + 1].a)
            {
                l = max(l, (int)floor(1.0*(st[i + 1].h - st[i].h) / (st[i].a - st[i + 1].a)) + 1);//下取整加一，做到了即使是整除也要加一
            }
            else
            {
                r = min(r, (int)ceil(1.0 * (st[i].h - st[i + 1].h) / (st[i + 1].a - st[i].a)) - 1);//上取整减一，做到了即使整除也要减一
            }
        }
    }
    if (l > r || ok == 0) cout << "-1" << '\n';
    else cout << l << '\n';
}

https://www.acwing.com/problem/content/description/4908/

不等式有两个未知数，而升级数越大用时越短，成单调性，所以二分相当于加多了一条方程（x+y=mid），变成了两个未知数两条方程，可解

const int N=1e6+10;
int a[N],b[N],c[N];
int n,x,y;
int cheak(int ok)
{
    int l=max((int)0,ok+1-y);
    int r=min(ok,x-1);
    for(int i=1;i<=n;i++)
    {
        int cnt=b[i]-a[i];
        //int ans=a[i]-b[i];
        int res=c[i]-a[i]*x-b[i]*y+b[i]*ok;
        if(cnt==0&&res<0) return 0;
        else if(cnt>0) r=min(r,(int)floor(1.0*res/cnt));
        else l=max(l,(int)ceil(1.0*res/cnt));
    }
    if(l<=r) return 1;
    else return 0;
}
void solve()
{
    cin>>n>>x>>y;
    for(int i=1;i<=n;i++) cin>>a[i]>>b[i]>>c[i];
    int l=0,r=x+y-2;
    int daan=0;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(cheak(mid)) daan=mid,r=mid-1;
        else l=mid+1;
    }
    cout<<daan<<endl;
}