Codeforces Round 987 (Div. 2)

訓練記錄Codeforces Round 987 (Div. 2)

4/6

前四題都是簡單思維題

A. Penchick and Modern Monument

這個題目最貪心的做法是找到出現最多的數，保留種數字不變，其他按照題目要求改大小就行

// Problem: A. Penchick and Modern Monument
// Contest: Codeforces - Codeforces Round 987 (Div. 2)
// URL: https://codeforces.com/contest/2031/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//by codeforcer ——
//  ____  _   _  _   _  _   _   ____    _ 
// / ___|| | | || | | || | | | |___  \ | |
//| |    | |_| || |_| || |_| |   __) | | |
//| |    |  _  ||  _  ||  _  |  |__  < | |
//| |___ | | | || | | || | | |  ___) | | |
// \____||_| |_||_| |_||_| |_| |____ / |_|


#include<bits/stdc++.h>
using namespace std;

typedef int E;
typedef long long LL;
typedef pair<int, int> PII;
typedef tuple<int, int, int> PIII;
typedef tuple<LL, LL, LL> PLLL;
typedef pair<long long, long long> PLL;
typedef unsigned long long ULL;

#define endl '\n'
#define vec vector
#define pb push_back
#define pob pop_back
#define fir first
#define sec second
#define maxINT 0x3f3f3f3f
#define maxLL 0x3f3f3f3f3f3f3f3fLL
#define umap unordered_map
#define uset unordered_set
#define maxheap priority_queue<E, vector<E>, less<E>>
#define minheap priority_queue<E, vector<E>, greater<E>>

#define prvec(a)                			\
    for (int i = 0; i < (a).size(); i++) {	\
        cout << (a)[i] << " ";   			\
    }                            			\
    cout << endl;

#define debugvec(a, i, n)             		\
    cout << #a << ": ";               		\
    for (int k = (i); k <= (n); k++) { 		\
        cout << (a)[k] << " ";        		\
    }                                 		\
    cout << endl;							

LL gcd(LL a, LL b) { return (b) ? gcd(b, a % b) : a; }
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, y, x);y -= a / b * x;return gcd;}
LL qmi(LL a, LL b, LL mod) {LL res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}

const int N = 200010;

void solve() {
    int n;
    cin>>n;
    vec<int> a(n + 1);
    map<int,int> st;
    int mark = 0,cnt = 0;
    for(int i = 0;i < n;i ++)
    {
    	cin>>a[i];
    	st[a[i]] ++;
    	if(cnt < st[a[i]])
    	{
    		mark = a[i];
    		cnt = st[a[i]];
    	}
    }
    
    cout<<n - cnt<<endl;
    
    
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;cin >> t;for (int i = 0; i < t; i++)
        solve();

    return 0;
}

B. Penchick and Satay Sticks

考慮如果滿足答案的序列，經過移動後會變成怎樣的序列，比如1 2 3 4 5 ，透過題目給的移動方式，有1 3 2 4 5，經過手動模擬可以發現，一個數不能遠離應該在的位置超過2，例如3 1 2 4 5是不可能透過1 2 3 4 5移動過來的，所以記錄每個a[i]對於應該在的位置多原即可

// Problem: B. Penchick and Satay Sticks
// Contest: Codeforces - Codeforces Round 987 (Div. 2)
// URL: https://codeforces.com/contest/2031/problem/B
// Memory Limit: 256 MB
// Time Limit: 1500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//by codeforcer ——
//  ____  _   _  _   _  _   _   ____    _ 
// / ___|| | | || | | || | | | |___  \ | |
//| |    | |_| || |_| || |_| |   __) | | |
//| |    |  _  ||  _  ||  _  |  |__  < | |
//| |___ | | | || | | || | | |  ___) | | |
// \____||_| |_||_| |_||_| |_| |____ / |_|


#include<bits/stdc++.h>
using namespace std;

typedef int E;
typedef long long LL;
typedef pair<int, int> PII;
typedef tuple<int, int, int> PIII;
typedef tuple<LL, LL, LL> PLLL;
typedef pair<long long, long long> PLL;
typedef unsigned long long ULL;

#define endl '\n'
#define vec vector
#define pb push_back
#define pob pop_back
#define fir first
#define sec second
#define maxINT 0x3f3f3f3f
#define maxLL 0x3f3f3f3f3f3f3f3fLL
#define umap unordered_map
#define uset unordered_set
#define maxheap priority_queue<E, vector<E>, less<E>>
#define minheap priority_queue<E, vector<E>, greater<E>>

#define prvec(a)                			\
    for (int i = 0; i < (a).size(); i++) {	\
        cout << (a)[i] << " ";   			\
    }                            			\
    cout << endl;

#define debugvec(a, i, n)             		\
    cout << #a << ": ";               		\
    for (int k = (i); k <= (n); k++) { 		\
        cout << (a)[k] << " ";        		\
    }                                 		\
    cout << endl;							

LL gcd(LL a, LL b) { return (b) ? gcd(b, a % b) : a; }
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, y, x);y -= a / b * x;return gcd;}
LL qmi(LL a, LL b, LL mod) {LL res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}

const int N = 200010;

void solve() {
    int n;
    cin>>n;
	vec<int> a(n + 1);
	
	for(int i = 1;i <= n;i ++)cin>>a[i];
	
	for(int i = 1;i <= n;i ++)
	{
		if(abs(a[i] - i) > 1)
		{
			cout<<"NO"<<endl;
			return;
		}
	}
    cout<<"YES"<<endl;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;cin >> t;for (int i = 0; i < t; i++)
        solve();

    return 0;
}

C. Penchick and BBQ Buns

這個題目按給的數字n分成兩種情況：

• 偶數 一定有對應方式比如n = 6，序列可以是1 1 2 2 3 3，依次類推
• 奇數 有特定的方式能滿足這個情況，我們需要讓三個位置是相同的一個數，這裡以1舉例，放1的位置分別記作a b c，根據題目條件有 b - a = \(x^2\) ，c - b = \(y^2\)，\(x^2\) + \(y^2\) = \(c^2\) 看著是不是像勾股數，那麼最小的勾股數就是3 4 5了，9 + 16 = 25，
  按照題目先把1放在對應位置b[1] = 1,b[10] = 1,b[26] = 1，然後可以發現，1到10這個位置可以放偶數個數，但是10到26只能放奇數個數，看似一定不能滿足題目條件，
  我們可以透過一個數放在10到26之間一個數放在26右邊來讓10到26之間放下偶數個數，但是得滿足放在10到26之間的數和到大於26的這個數的位置相差是完全平方數，考慮最小的方法，首先1肯定不行，中間隔著一個a[26]，那就只有4了，也就是放一個數在a[23] 一個放在a[27]，剩餘沒放數的位置就是偶數個了。所以最小滿足條件的奇數是27，更大的奇數就只需要在27這個序列後面加兩個數就行

#include <bits/stdc++.h>
using namespace std;


const int N = 2e5 + 9;
int a[N] = {0, 1, 2, 2, 3, 3, 4, 4, 5, 5,  1,  6,  6,  7,  7,  8,  8,  9,  9, 10, 10, 11, 11, 13, 12, 12, 1,  13};
//           1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27
 
 
void solve()
{
    int n;cin >> n;
    if(n % 2)
    {
        if(n >= 27)
        {
            for(int i = 1;i <= 27; ++ i)cout << a[i] << ' ';
            for(int i = 28;i <= n; ++ i)cout << i / 2 << ' ';
            cout << '\n';
        }else{
            cout << -1 << '\n';
        }
    }else
    {
        for(int i = 1;i <= n; ++ i)cout << (i + 1) / 2 << ' ';
        cout << '\n';
    }
}
 

int main() 
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;
    cin >> t;
    while (t--) {
        solve();
    }

    return 0;
}

D. Penchick and Desert Rabbit

我們可以發現當從1到i的字首最大值大於i + 1到n的字尾最小值就可以讓i到達i + 1，也就是\(ans_i\) = \(ans_{i + 1}\)，反之， \(ans_i\) 只能等於字首最大值

// Problem: D. Penchick and Desert Rabbit
// Contest: Codeforces - Codeforces Round 987 (Div. 2)
// URL: https://codeforces.com/contest/2031/problem/D
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//by codeforcer ——
//  ____  _   _  _   _  _   _   ____    _ 
// / ___|| | | || | | || | | | |___  \ | |
//| |    | |_| || |_| || |_| |   __) | | |
//| |    |  _  ||  _  ||  _  |  |__  < | |
//| |___ | | | || | | || | | |  ___) | | |
// \____||_| |_||_| |_||_| |_| |____ / |_|


#include<bits/stdc++.h>
using namespace std;

typedef int E;
typedef long long LL;
typedef pair<int, int> PII;
typedef tuple<int, int, int> PIII;
typedef tuple<LL, LL, LL> PLLL;
typedef pair<long long, long long> PLL;
typedef unsigned long long ULL;

#define endl '\n'
#define vec vector
#define pb push_back
#define pob pop_back
#define fir first
#define sec second
#define maxINT 0x3f3f3f3f
#define maxLL 0x3f3f3f3f3f3f3f3fLL
#define umap unordered_map
#define uset unordered_set
#define maxheap priority_queue<E, vector<E>, less<E>>
#define minheap priority_queue<E, vector<E>, greater<E>>

#define prvec(a)                			\
    for (int i = 0; i < (a).size(); i++) {	\
        cout << (a)[i] << " ";   			\
    }                            			\
    cout << endl;

#define debugvec(a, i, n)             		\
    cout << #a << ": ";               		\
    for (int k = (i); k <= (n); k++) { 		\
        cout << (a)[k] << " ";        		\
    }                                 		\
    cout << endl;							

LL gcd(LL a, LL b) { return (b) ? gcd(b, a % b) : a; }
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, y, x);y -= a / b * x;return gcd;}
LL qmi(LL a, LL b, LL mod) {LL res = 1;while (b) {if (b & 1) res = res * a % mod;a = a * a % mod;b >>= 1;}return res;}

const int N = 200010;

void solve() {
	int n;
	cin>>n;
	vector<int> a(n + 1);
	vector<int> pre(n + 4),ed(n + 3,1e9 + 10);
	for(int i = 1;i <= n;i ++)
	{
		 cin>>a[i];
		 pre[i] = max(pre[i - 1],a[i]);
	}
	ed[n] = a[n];
	for(int i = n-1;i >= 1;i --)
	{
		ed[i] = min(ed[i + 1],a[i]);
	}
	vec<int> ans(n + 2);

	for(int i = n;i >= 1;i--)
	{
		if(pre[i] > ed[i+1])
		{
			ans[i] = ans[i + 1];
		}else
		{
			ans[i] = pre[i];
		}
	}
	
	for(int i = 1;i <= n;i ++)
	{
		cout<<ans[i]<<" ";
	}
	cout<<endl;
    
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t;cin >> t;for (int i = 0; i < t; i++)
        solve();

    return 0;
}