P5161 WD与数列 [后缀自动机，线段树合并，启发式合并]

考虑选一对(A),(B)，使得 (A_i - B_i = A_j - B_j)，(|A| = |B|) 。
(A),(B) 不相交，否则就是 (sum LCP(i,j))。
然后转化成 (A_i - A_j = B_i - B_j) 然后令 (j = i - 1) 丢到 (SAM) 里。

我们考虑分类讨论，先考虑不相交的，答案直接加上。

ans1 表示 (endpos) 的数量。
ans2 表示 ([i]) 所对应的节点包含的子串的最后一个点 (i)。

ans1 = ans2 = 0; smt.qry(rt[u], x + 1 + len[u], n - 1, 1, n - 1);
ans += ans1 * len[u];
ans1 = ans2 = 0; smt.qry(rt[u], 1, x - len[u] - 1, 1, n - 1);
ans += ans1 * len[u];

考虑相交的部分。

ans1 = ans2 = 0; smt.qry(rt[u], x - len[u], x - 1, 1, n - 1);
ans += ans1 * (x - 1) - ans2;
ans1 = ans2 = 0; smt.qry(rt[u], x + 1, x + len[u], 1, n - 1);
ans += ans2 - ans1 * (x + 1);

这样就好了。

// clang-format off
// powered by c++11
// by Isaunoya
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#include <tr1/unordered_map>
#include<bits/stdc++.h>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define Rep(i,x,y) for(register int i=(x);i>=(y);--i)
using namespace std;using db=double;using ll=long long;
using uint=unsigned int;using ull=unsigned long long;
using pii=pair<int,int>;
#define Tp template
#define fir first
#define sec second
Tp<class T>void cmax(T&x,const T&y){if(x<y)x=y;}Tp<class T>void cmin(T&x,const T&y){if(x>y)x=y;}
#define all(v) v.begin(),v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T>void sort(vector<T>&v){sort(all(v));}Tp<class T>void reverse(vector<T>&v){reverse(all(v));}
Tp<class T>void unique(vector<T>&v){sort(all(v)),v.erase(unique(all(v)),v.end());}inline void reverse(string&s){reverse(s.begin(),s.end());}
const int SZ=1<<23|233;
struct FILEIN{char qwq[SZ],*S=qwq,*T=qwq,ch;
#ifdef __WIN64
#define GETC getchar
#else
inline char GETC(){return(S==T)&&(T=(S=qwq)+fread(qwq,1,SZ,stdin),S==T)?EOF:*S++;}
#endif
inline FILEIN&operator>>(char&c){while(isspace(c=GETC()));return*this;}inline FILEIN&operator>>(string&s){s.clear();while(isspace(ch=GETC()));if(!~ch)return*this;s=ch;while(!isspace(ch=GETC())&&~ch)s+=ch;return*this;}
inline FILEIN&operator>>(char*str){char*cur=str;while(*cur)*cur++=0;cur=str;while(isspace(ch=GETC()));if(!~ch)return*this;*cur=ch;while(!isspace(ch=GETC())&&~ch)*++cur=ch;*++cur=0;return*this;}
Tp<class T>inline void read(T&x){bool f=0;while((ch=GETC())<48&&~ch)f^=(ch==45);x=~ch?(ch^48):0;while((ch=GETC())>47)x=x*10+(ch^48);x=f?-x:x;}
inline FILEIN&operator>>(int&x){return read(x),*this;}inline FILEIN&operator>>(ll&x){return read(x),*this;}inline FILEIN&operator>>(uint&x){return read(x),*this;}inline FILEIN&operator>>(ull&x){return read(x),*this;}
inline FILEIN&operator>>(double&x){read(x);bool f=x<0;x=f?-x:x;if(ch^'.')return*this;double d=0.1;while((ch=GETC())>47)x+=d*(ch^48),d*=.1;return x=f?-x:x,*this;}
}in;
struct FILEOUT{const static int LIMIT=1<<22;char quq[SZ],ST[233];int sz,O,pw[233];
FILEOUT(){set(7);rep(i,pw[0]=1,9)pw[i]=pw[i-1]*10;}~FILEOUT(){flush();}
inline void flush(){fwrite(quq,1,O,stdout),fflush(stdout),O=0;}
inline FILEOUT&operator<<(char c){return quq[O++]=c,*this;}inline FILEOUT&operator<<(string str){if(O>LIMIT)flush();for(char c:str)quq[O++]=c;return*this;}
inline FILEOUT&operator<<(char*str){if(O>LIMIT)flush();char*cur=str;while(*cur)quq[O++]=(*cur++);return*this;}
Tp<class T>void write(T x){if(O>LIMIT)flush();if(x<0){quq[O++]=45;x=-x;}do{ST[++sz]=x%10^48;x/=10;}while(x);while(sz)quq[O++]=ST[sz--];}
inline FILEOUT&operator<<(int x){return write(x),*this;}inline FILEOUT&operator<<(ll x){return write(x),*this;}inline FILEOUT&operator<<(uint x){return write(x),*this;}inline FILEOUT&operator<<(ull x){return write(x),*this;}
int len,lft,rig;void set(int l){len=l;}inline FILEOUT&operator<<(double x){bool f=x<0;x=f?-x:x,lft=x,rig=1.*(x-lft)*pw[len];return write(f?-lft:lft),quq[O++]='.',write(rig),*this;}
}out;
#define int long long
struct Math{
vector<int>fac,inv;int mod;
void set(int n,int Mod){fac.resize(n+1),inv.resize(n+1),mod=Mod;rep(i,fac[0]=1,n)fac[i]=fac[i-1]*i%mod;inv[n]=qpow(fac[n],mod-2);Rep(i,n-1,0)inv[i]=inv[i+1]*(i+1)%mod;}
int qpow(int x,int y){int ans=1;for(;y;y>>=1,x=x*x%mod)if(y&1)ans=ans*x%mod;return ans;}int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*inv[m]%mod*inv[n-m]%mod;}
int gcd(int x,int y){return!y?x:gcd(y,x%y);}int lcm(int x,int y){return x*y/gcd(x,y);}
}math;
// clang-format on

int n;
const int maxn = 6e5 + 56;
int a[maxn], rt[maxn];
int ans1, ans2;

struct SegMentTree {
	int cnt;
	
	SegMentTree() {
		cnt = 0;
	}
	
	int ls[maxn << 5], rs[maxn << 5];
	int c[maxn << 5], v[maxn << 5];
	
	void upd(int &p , int l , int r , int x) {
		if(! p) 
			p = ++ cnt;
		c[p] ++, v[p] += x;
		if(l == r) return;
		
		int mid = l + r >> 1;
		if(x <= mid)
			upd(ls[p], l, mid, x);
		else
			upd(rs[p], mid + 1, r, x);
	}
	
	int merge(int x, int y, int l, int r) {
		if(! x || ! y) return x | y;
		if(l == r) {
			c[x] += c[y];
			v[x] += v[y];
			return x;
		}
		int mid = l + r >> 1;
		ls[x] = merge(ls[x], ls[y], l, mid);
		rs[x] = merge(rs[x], rs[y], mid + 1, r);
		c[x] = c[ls[x]] + c[rs[x]];
		v[x] = v[ls[x]] + v[rs[x]];
		return x;
	}
	
	void qry(int p, int a, int b, int l, int r) {
		if(a > b) return;
		if(! p) return;
		if(a <= l && r <= b) {
			ans1 += c[p];
			ans2 += v[p];
			return;
		}
		int mid = l + r >> 1;
		if(a <= mid)
			qry(ls[p], a, b, l , mid);
		if(b > mid)
			qry(rs[p], a, b, mid + 1, r);
	}
} smt;

int ans = 0;
struct suffix_auto_maton {
	int las, cnt;
	
	suffix_auto_maton() { las = cnt = 1; }
	
	unordered_map <int , int> ch[maxn];
	int fa[maxn], len[maxn];
	vector < int > qwq[maxn];
	
	void ins(int c , int id) {
		int p = las, np = las = ++ cnt;
		len[np] = len[p] + 1;
		qwq[np].push_back(id);
		smt.upd(rt[np], 1, n - 1, id);
		for(; p && !ch[p][c]; p = fa[p])
			ch[p][c] = np;
		if(! p) {
			fa[np] = 1;
		} else {
			int q = ch[p][c];
			if(len[q] == len[p] + 1) {
				fa[np] = q;
			} else {
				int nq = ++ cnt;
				ch[nq] = ch[q];
				fa[nq] = fa[q], fa[q] = fa[np] = nq;
				len[nq] = len[p] + 1;
				for(; p && ch[p][c] == q; p = fa[p])
					ch[p][c] = nq;
			}
		}
	}
		
	vector < int > g[maxn];
	
	void build() {
		rep(i, 2, cnt) g[fa[i]].pb(i);
	}
	
	void merge(int u , int v) {
		if(sz(qwq[u]) < sz(qwq[v])) {
			swap(qwq[u], qwq[v]);
			swap(rt[u], rt[v]);
		}
		
		for(auto x : qwq[v]) {
			ans1 = ans2 = 0; smt.qry(rt[u], x + 1 + len[u], n - 1, 1, n - 1);
			ans += ans1 * len[u];
			ans1 = ans2 = 0; smt.qry(rt[u], 1, x - len[u] - 1, 1, n - 1);
			ans += ans1 * len[u];
			ans1 = ans2 = 0; smt.qry(rt[u], x - len[u], x - 1, 1, n - 1);
			ans += ans1 * (x - 1) - ans2;
			ans1 = ans2 = 0; smt.qry(rt[u], x + 1, x + len[u], 1, n - 1);
			ans += ans2 - ans1 * (x + 1);
			qwq[u].pb(x);
		}
		
		rt[u] = smt.merge(rt[u], rt[v], 1 , n - 1);
	}
	
	void dfs(int u) {
		for(int v : g[u]) {
			dfs(v);
			merge(u , v);
		}
	}
	
} sam;

signed main(){
	//code begin.
	in >> n;
	rep(i , 1 , n) in >> a[i];
	rep(i , 1 , n - 1)
		sam.ins(a[i + 1] - a[i], i);
	sam.build(), sam.dfs(1);
	ans += (n - 1) * n / 2;
	out << ans << '
';
	return 0;
	//code end.
}