제출 #742854

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
742854		Luicosas	순열 (APIO22_perm)	C++17	0 / 100	0 ms	212 KiB

perm

#include<bits/stdc++.h>

using namespace std;

#define ll long long
#define sz(x) (int)(x.size())
#define itr(x) x.begin(), x.end()
#ifdef LOC
#define debug(...) cerr << #__VA_ARGS__ << " : "; for(auto& dval : {__VA_ARGS__}) cerr << dval << " "; cerr << "\n";
#else
#define debug(...)
#endif

template<typename T> 
ostream& operator << (ostream& out, vector<T> vals) {
    for(auto& i : vals) {
        out << i << " ";
    }
    return out;
}

// i hypothesize that we can greedily append to a permutation the greatest number that will not make the increasing sequences > k and that this will terminate with at most 90 numbers
// we can get 2 ^ k - 1 with 1, 2, 3, 4, ... k
// n - (2 ^ k - 1) will not have a bit higher than k set
// 2 ^ k - 1 is what the numbers are 
// so after we get close we can use the 2 ^ k - 1 and we do this at most k times? 

// v*2 + 1, greatest will become (answer - 1) / 2
// v*2 - greatest
// + 1
// + 1 + smallest = 1 + 1 = 2
// + 1 + smallest + second smallest = 1 + 1 + 1 or 2 = 3 or 4

void mul2(vector<int>& vals) {
    vals[0] += 1;
    vals.push_back(sz(vals));
}

void add1(vector<int>& vals) {
    for(int& i : vals) {
        i++;
    }
    vals.push_back(1);
}

void add3(vector<int>& vals) {
    for(int& i : vals) {
        i += (i > 2);
    }
    vals.push_back(3);
}

vector<int> construct_permutation(ll k) {
    vector<int> bits;
    for(int i = 0, cnt = 0; i < 62; i++) {
        if((k >> i) & 1) {
            bits.resize(sz(bits) + cnt, 0);
            bits.push_back(1);
            cnt = 0;
        } else {
            cnt++;
        }
    }

    bool fadd1 = 0;
    bits.pop_back();
    vector<int> ret(1, 1);
    while(sz(bits) > 1) {
        int a = bits.back(); 
        bits.pop_back();
        int b = bits.back();
        bits.pop_back();
        
        if(a == 0 && b == 0) {
            mul2(ret);
            mul2(ret);
        }
        if(a == 0 && b == 1) {
            mul2(ret);
            mul2(ret);
            add1(ret);
            fadd1 = 1;
        }
        if(a == 1 && b == 0) {
            mul2(ret);
            add1(ret);
            mul2(ret);
            fadd1 = 1;
        }
        if(a == 1 && b == 1 && !fadd1) {
            mul2(ret);
            add1(ret);
            mul2(ret);
            add1(ret);
            fadd1 = 1;
        } else if(a == 1 && b == 1 && fadd1) {
            mul2(ret);
            mul2(ret);
            add3(ret);
        }
    }
    if(sz(bits) == 1 && bits[0] == 0) {
        mul2(ret);
    }
    if(sz(bits) == 1 && bits[0] == 1) {
        ret.push_back(sz(ret) + 1);
    }

    return ret;
}

#ifdef LOC
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    ll k;
    cin >> k;

    vector<int> val = construct_permutation(k);
    debug(val);

    vector<ll> dp(sz(val) + 1, 0);
    for(int& crn : val) {
        for(int i = 0; i < crn; i++) {
            dp[crn] += dp[i];
        }
        dp[crn] += 1;
    }
    debug(dp);
    
    ll incs = 0;
    for(ll& i : dp) {
        incs += i;
    }
    cerr << sz(val) << "\n";
    assert(sz(val) <= 90);
    assert(incs == k);
}
#endif

• Subtask 1: 0 / 10
• Subtask 2: 0 / 90