Submission #1296975

#	Time	Username	Problem	Language	Result	Execution time	Memory
1296975		dosts	Peru (RMI20_peru)	C++20	100 / 100	401 ms	128348 KiB

peru

#include "peru.h"
#include <bits/stdc++.h>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
#define int long long
#define pii pair<int,int>
#define vi vector<int>
#define ff first
#define ss second
#define sp << " " <<
#define all(x) x.begin(),x.end()
#define big(x) ((int)(x.size()))
using namespace std;
const int MOD = 1e9+7, LIM = 1e6+1, inf = 2e18;

const int N = 1e5+1;
int add(int x,int y) {
    if ((x + y) >= MOD) return x + y - MOD;
    return x + y;
}

int mult(int x,int y) {
    return (1LL * x * y) % MOD;
}

int expo(int x,int y) {
    if (!y) return 1;
    int e = expo(x, y >> 1);
    e = mult(e, e);
    if (y & 1) e = mult(e, x);
    return e;
}

struct ST {
    vi t;
    ST(){}
    ST(int nn) {
        t.assign(4*nn+1,inf);
    }
    void update(int node,int l,int r,int p,int v) {
        if (l == r) {
            t[node] = v;
            return;
        }
        int m = (l+r) >> 1;
        if (p <= m) update(2*node,l,m,p,v);
        else update(2*node+1,m+1,r,p,v);
        t[node] = min(t[2*node],t[2*node+1]);
    }

    int query(int node,int l,int r,int L,int R) {
        if (l > R || r < L) return inf;
        if (l >= L  && r <= R) return t[node];
        int m = (l+r) >> 1;
        return min(query(2*node,l,m,L,R),query(2*node+1,m+1,r,L,R));
    }
} st;

signed solve(signed n, signed k, signed* v){
    vi a(n+1);
    for (int i=1;i<=n;i++) a[i] = v[i-1];
    vi dp(n+1,inf);
    dp[0] = 0;
    a[0] = inf;
    int mx = 0;
    for (int j=1;j<k;j++) {
        mx = max(mx,a[j]);
        dp[j] = min(dp[j],mx);
    }
    ST st(n);
    deque<int> stk;
    for (int i=0;i<=n;i++) {
        while (!stk.empty() && a[stk.back()] <= a[i]) {
            st.update(1,1,n,stk.back(),inf);
            stk.pop_back();
        }
        if (!stk.empty()) st.update(1,1,n,i,a[i]+dp[stk.back()]);
        stk.push_back(i);
        while (stk.front() <= i-k) {
            stk.pop_front();
            if (!stk.empty()) st.update(1,1,n,stk.front(),inf);
        }
        if (i >= k) {
            dp[i] = st.query(1,1,n,1,n);
            dp[i] = min(dp[i],a[stk[0]]+dp[i-k]);
        }
    }
    int mlt = 1;
    int ans = 0;
    for (int i = n;i>=1;i--) {
        ans = add(ans,mult(dp[i]%MOD,mlt));
        mlt = mult(mlt,23);
    }
    return ans;
}

• Subtask 1: 18 / 18
• Subtask 2: 31 / 31
• Subtask 3: 51 / 51