给每个数一个哈希值,做个异或前缀和,对于每个1的位置,正着做反着做,贡献加起来就可以了。
如果给个序列 3 1 2
因为你1是必要的,如果左边有3,那么会扩展到右边的2,而你右边的2只能扩展到1,所以贡献不重复。
// powered by c++11
// by Isaunoya
#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;
#define pii pair<int, int>
#define fir first
#define sec second
template <class T>
void cmax(T& x, const T& y) {
if (x < y) x = y;
}
template <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
template <class T>
void sort(vector<T>& v) {
sort(all(v));
}
template <class T>
void reverse(vector<T>& v) {
reverse(all(v));
}
template <class T>
void unique(vector<T>& v) {
sort(all(v)), v.erase(unique(all(v)), v.end());
}
void reverse(string& s) { reverse(s.begin(), s.end()); }
const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
char ch;
#ifndef __WIN64
char getchar() {
static char buf[io_size], *p1 = buf, *p2 = buf;
return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2) ? EOF : *p1++;
}
#endif
io_in& operator>>(char& c) {
for (c = getchar(); isspace(c); c = getchar())
;
return *this;
}
io_in& operator>>(string& s) {
for (s.clear(); isspace(ch = getchar());)
;
if (!~ch) return *this;
for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
;
return *this;
}
io_in& operator>>(char* str) {
char* cur = str;
while (*cur) *cur++ = 0;
for (cur = str; isspace(ch = getchar());)
;
if (!~ch) return *this;
for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
;
return *++cur = 0, *this;
}
template <class T>
void read(T& x) {
bool f = 0;
while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);
x = ~ch ? (ch ^ 48) : 0;
while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
x = f ? -x : x;
}
io_in& operator>>(int& x) { return read(x), *this; }
io_in& operator>>(ll& x) { return read(x), *this; }
io_in& operator>>(uint& x) { return read(x), *this; }
io_in& operator>>(ull& x) { return read(x), *this; }
io_in& operator>>(db& x) {
read(x);
bool f = x < 0;
x = f ? -x : x;
if (ch ^ '.') return *this;
double d = 0.1;
while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
return x = f ? -x : x, *this;
}
} in;
struct io_out {
char buf[io_size], *s = buf;
int pw[233], st[233];
io_out() {
set(7);
rep(i, pw[0] = 1, 9) pw[i] = pw[i - 1] * 10;
}
~io_out() { flush(); }
void io_chk() {
if (s - buf > io_limit) flush();
}
void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }
io_out& operator<<(char c) { return *s++ = c, *this; }
io_out& operator<<(string str) {
for (char c : str) *s++ = c;
return io_chk(), *this;
}
io_out& operator<<(char* str) {
char* cur = str;
while (*cur) *s++ = *cur++;
return io_chk(), *this;
}
template <class T>
void write(T x) {
if (x < 0) *s++ = '-', x = -x;
do {
st[++st[0]] = x % 10, x /= 10;
} while (x);
while (st[0]) *s++ = st[st[0]--] ^ 48;
}
io_out& operator<<(int x) { return write(x), io_chk(), *this; }
io_out& operator<<(ll x) { return write(x), io_chk(), *this; }
io_out& operator<<(uint x) { return write(x), io_chk(), *this; }
io_out& operator<<(ull x) { return write(x), io_chk(), *this; }
int len, lft, rig;
void set(int _length) { len = _length; }
io_out& operator<<(db x) {
bool f = x < 0;
x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
}
} out;
#define int long long
template <int sz, int mod>
struct math_t {
math_t() {
fac.resize(sz + 1), ifac.resize(sz + 1);
rep(i, fac[0] = 1, sz) fac[i] = fac[i - 1] * i % mod;
ifac[sz] = inv(fac[sz]);
Rep(i, sz - 1, 0) ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
vector<int> fac, ifac;
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 inv(int x) { return qpow(x, mod - 2); }
int C(int n, int m) {
if (n < 0 || m < 0 || n < m) return 0;
return fac[n] * ifac[m] % mod * ifac[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); }
int n;
const int maxn = 3e5 + 53;
int a[maxn];
ull hsh[maxn], pre[maxn], qwq[maxn];
ull rnd() { return 1ull * rand() * rand() * rand(); }
signed main() {
// code begin.
in >> n;
rep(i, 1, n) in >> a[i];
srand(19260817);
rep(i, 1, n) {
hsh[i] = rnd();
pre[i] = pre[i - 1] ^ hsh[i];
}
int ans = 0;
rep(i, 1, n) {
if (a[i] == 1) {
++ans;
}
}
rep(i, 1, n) qwq[i] = qwq[i - 1] ^ hsh[a[i]];
rep(i, 1, n) {
if (a[i] == 1) {
int ret = 0, mx = 1;
for (int j = i + 1; j <= n && a[j] ^ 1; j++) {
cmax(mx, a[j]);
if (j >= mx && j - mx + 1 <= i && pre[mx] == (qwq[j] ^ qwq[j - mx])) ++ret;
}
ans += ret;
}
}
reverse(a + 1, a + n + 1);
rep(i, 1, n) qwq[i] = qwq[i - 1] ^ hsh[a[i]];
rep(i, 1, n) {
if (a[i] == 1) {
int ret = 0, mx = 1;
for (int j = i + 1; j <= n && a[j] ^ 1; j++) {
cmax(mx, a[j]);
if (j >= mx && j - mx + 1 <= i && pre[mx] == (qwq[j] ^ qwq[j - mx])) ++ret;
}
ans += ret;
}
}
out << ans << '\n';
return 0;
// code end.
}