Codeforces 1175F The Number of Subpermutations

The Number of Subpermutations

想了想感觉可行解不多， 然后就暴力莽就完事了(FST我就删了这个博客 逃

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, a[N];
int maxPos[N];
int Map[N];

int Log[N];
struct ST {
    int dp[N][20]; int ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    inline LL query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
} rmq;

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    rmq.build(n, a, 1);
    int canGo = n;
    for(int i = n; i >= 1; i--) {
        if(Map[a[i]]) chkmin(canGo, Map[a[i]] - 1);
        Map[a[i]] = i;
        maxPos[i] = canGo;
    }
    LL ans = 0;
    for(int i = 1; i <= n; i++) {
        int now = i;
        while(now <= maxPos[i]) {
            if(rmq.query(i, now) == now - i + 1) {
                ans++;
                now++;
            } else {
                now = i + rmq.query(i, now) - 1;
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}

/*
*/