MarsOJ 上的题，是 B3637 最长上升子序列 的加强版，$n\leq 5\times 10^3,1\leq a_i\leq 10^9$，20 个点 WA 了一个，Read 4, Expected 5.

我试着跟二分求解 LIS 的代码对拍，但是拍了十几分钟还是没有出错。我的代码是哪里出了问题，谢谢！

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll N = 1e5 + 10;
ll n, m, a[N], b[N], c[N], t[N], f[N];
inline ll read() {
	ll x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') {x=(x<<3)+(x<<1)+(ch&15);ch=getchar();}
	return x*f;
}
ll lowbit(ll x) {
    return x & (-x);
}
void add(ll i, ll x) {
    ll pos = i;
    while (pos <= n) {
        t[pos] = max(t[pos], x);
        pos += lowbit(pos);
    }
}
ll ask(ll i) {
    ll res = 0;
    while (i) {
        res = max(res, t[i]);
        i -= lowbit(i);
    }
    return res;
}
ll query(ll x) {
	return lower_bound(b + 1, b + m + 1, x) - b;
}

int main() {
    n=read();
    for(ll i=1;i<=n;i++) {
        a[i] = read(); a[i] ++; b[i] = a[i];
    }
    sort(b + 1, b + n + 1);
    m = unique(b + 1, b + n + 1) - b;
    for (int i=1;i<=n;i++) {
    	c[i]=query(a[i]);
	}
    for(ll i=1;i<=n;i++) {
        f[i] = ask(c[i]-1) + 1;
        add(c[i], f[i]);
    }
    ll ans=0;
    for(ll i=1;i<=n;i++) {
        ans=max(ans,f[i]);
    }
    cout << ans << "\n";
	return 0;
}