题解 自然数（留坑）

传送门

首先可以 \(O(nlogn)\) 预处理出每个位置在一开始时的mex值
然后枚举左端点，设法维护出每个右端点的答案
令下一次出现这个位置上数的位置为 \(nxt_i\) 于是删去一个点的贡献等价于将 \([i, nxt_i)\) 的答案与 \(a_i\) 取min
又因为我们需要查区间和，很容易想到用吉司机线段树维护
但不会打吉司机线段树，于是挂了
但这样想其实麻烦了，发现线段树上的答案是单调不降的，所以可以在区间 \([i, nxt_i)\) 内通过线段树上二分找到第一个大于 \(a_i\) 的点
然后区间覆盖即可
吉司机线段树先咕了

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 200010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int a[N];

namespace force{
	ll ans;
	int tem[N], len;
	int mex(int l, int r) {
		len=r-l+1;
		for (int i=l; i<=r; ++i) tem[i-l+1]=a[i];
		sort(tem+1, tem+len+1);
		// cout<<"tem: "; for (int i=1; i<=len; ++i) cout<<tem[i]<<' '; cout<<endl;
		for (int i=0; ; ++i) {
			int *p=lower_bound(tem+1, tem+len+1, i);
			if ((p-tem>len) || *p!=i) return i;
		}
	}
	void solve() {
		for (int i=1; i<=n; ++i)
			for (int j=i; j<=n; ++j)
				ans+=mex(i, j); //, cout<<"mex: "<<i<<' '<<j<<' '<<mex(i, j)<<endl;
		printf("%lld\n", ans);
		exit(0);
	}
}

namespace task1{
	ll ans;
	int ua[N], uni[N], usize, lst[N], val[N], nxt[N];
	priority_queue<int, vector<int>, greater<int>> q;
	void solve() {
		for (int i=1; i<=n; ++i) uni[++usize]=a[i];
		sort(uni+1, uni+usize+1);
		usize=unique(uni+1, uni+usize+1)-uni-1;
		for (int i=1; i<=n; ++i) ua[i]=lower_bound(uni+1, uni+usize+1, a[i])-uni;
		int lin=0;
		for (int i=1; i<=n; ++i) {
			q.push(a[i]);
			while (q.size() && q.top()<=lin) {
				if (q.top()==lin) ++lin;
				q.pop();
			}
			val[i]=lin;
		}
		// cout<<"val: "; for (int i=1; i<=n; ++i) cout<<val[i]<<' '; cout<<endl;
		// cout<<"ua: "; for (int i=1; i<=n; ++i) cout<<ua[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) lst[i]=n+1;
		for (int i=n; i; --i) {
			nxt[i]=lst[ua[i]];
			lst[ua[i]]=i;
		}
		// cout<<"nxt: "; for (int i=1; i<=n; ++i) cout<<nxt[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) {
			for (int j=i; j<=n; ++j) ans+=val[j];
			for (int j=i; j<nxt[i]; ++j) {
				val[j]=min(val[j], a[i]);
			}
			// cout<<"val: "; for (int i=1; i<=n; ++i) cout<<val[i]<<' '; cout<<endl;
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

namespace task{
	ll ans;
	int ua[N], uni[N], usize, lst[N], val[N], nxt[N];
	priority_queue<int, vector<int>, greater<int>> q;
	int tl[N<<2], tr[N<<2], maxn[N<<2], len[N<<2], tag[N<<2];
	ll sum[N<<2];
	#define tl(p) tl[p]
	#define tr(p) tr[p]
	#define sum(p) sum[p]
	#define maxn(p) maxn[p]
	#define len(p) len[p]
	#define tag(p) tag[p]
	#define pushup(p) sum(p)=sum(p<<1)+sum(p<<1|1), maxn(p)=max(maxn(p<<1), maxn(p<<1|1))
	void spread(int p) {
		if (tag(p)==-1) return ;
		sum(p<<1)=tag(p)*len(p<<1); maxn(p<<1)=tag(p<<1)=tag(p);
		sum(p<<1|1)=tag(p)*len(p<<1|1); maxn(p<<1|1)=tag(p<<1|1)=tag(p);
		tag(p)=-1;
	}
	void build(int p, int l, int r) {
		tl(p)=l; tr(p)=r; len(p)=r-l+1; tag(p)=-1;
		if (l==r) {sum(p)=maxn(p)=val[l]; return ;}
		int mid=(l+r)>>1;
		build(p<<1, l, mid);
		build(p<<1|1, mid+1, r);
		pushup(p);
	}
	void upd(int p, int l, int r, int val) {
		if (l<=tl(p)&&r>=tr(p)) {sum(p)=1ll*val*len(p); maxn(p)=val; tag(p)=val; return ;}
		spread(p);
		int mid=(tl(p)+tr(p))>>1;
		if (l<=mid) upd(p<<1, l, r, val);
		if (r>mid) upd(p<<1|1, l, r, val);
		pushup(p);
	}
	ll qsum(int p, int l, int r) {
		if (l<=tl(p)&&r>=tr(p)) return sum(p);
		spread(p);
		int mid=(tl(p)+tr(p))>>1; ll ans=0;
		if (l<=mid) ans+=qsum(p<<1, l, r);
		if (r>mid) ans+=qsum(p<<1|1, l, r);
		return ans;
	}
	int search(int p, int val) {
		if (tl(p)==tr(p)) return tl(p);
		spread(p);
		if (maxn(p<<1)>=val) return search(p<<1, val);
		else return search(p<<1|1, val);
	}
	void solve() {
		for (int i=1; i<=n; ++i) uni[++usize]=a[i];
		sort(uni+1, uni+usize+1);
		usize=unique(uni+1, uni+usize+1)-uni-1;
		for (int i=1; i<=n; ++i) ua[i]=lower_bound(uni+1, uni+usize+1, a[i])-uni;
		int lin=0;
		for (int i=1; i<=n; ++i) {
			q.push(a[i]);
			while (q.size() && q.top()<=lin) {
				if (q.top()==lin) ++lin;
				q.pop();
			}
			val[i]=lin;
		}
		// cout<<"val: "; for (int i=1; i<=n; ++i) cout<<val[i]<<' '; cout<<endl;
		// cout<<"ua: "; for (int i=1; i<=n; ++i) cout<<ua[i]<<' '; cout<<endl;
		for (int i=1; i<=n; ++i) lst[i]=n+1;
		for (int i=n; i; --i) {
			nxt[i]=lst[ua[i]];
			lst[ua[i]]=i;
		}
		// cout<<"nxt: "; for (int i=1; i<=n; ++i) cout<<nxt[i]<<' '; cout<<endl;
		build(1, 1, n);
		for (int i=1; i<=n; ++i) {
			ans+=qsum(1, i, n);
			// cout<<"+= "<<qsum(1, i, n)<<endl;
			if (maxn(1)>a[i]) {
				int pos=search(1, a[i]);
				// cout<<"pos: "<<pos<<endl;
				if (pos<nxt[i]) upd(1, pos, nxt[i]-1, a[i]);
			}
			// cout<<"val: "; for (int i=1; i<=n; ++i) cout<<val[i]<<' '; cout<<endl;
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

signed main()
{
	freopen("mex.in", "r", stdin);
	freopen("mex.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	task::solve();

	return 0;
}