Codeforces Round #744 (Div. 3)

A. Casimir's String Solitaire

思路：

分別記錄字母A, B, C的數量，顯然要滿足B的數量等於A的數量和C的數量的和才行

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
 
#define x first
#define y second
 
using namespace std;
 
typedef long long LL;
typedef pair<int, int>PII;
 
const int N = 100010;
 
int a[N];
vector<PII>index[100010];
 
int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        string s;
        cin >> s;
        int cnta = 0, cntb = 0, cntc = 0;
        for (int i = 0; i < s.size(); i++)
            if (s[i] == 'A') cnta++;
            else if (s[i] == 'B') cntb++;
            else if (s[i] == 'C') cntc++;
 
        if (cntb == cnta + cntc) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}

B. Shifting Sort

題意：

題意有點難懂，意思是給你規定了一種操作，問要怎么進行一些列的這種操作才能使原數組遞增

思路：

比賽的時候想了了土方法，類比冒泡排序從后往前枚舉，遇到一個逆序對（a[i] > a[j]且，i < j）就整體前移一個，由於這個題數據范圍只有50，所以O(n^3)完全ok的

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
 
#define x first
#define y second
 
using namespace std;
 
typedef long long LL;
typedef pair<int, int>PII;
 
const int N = 55;
 
vector<PII>index[100010];
 
int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        int a[N];
        for (int i = 1; i <= n; i++) cin >> a[i];
        int tmp[N], b[N];
        for (int i = 1; i <= n; i++) tmp[i] = a[i], b[i] = a[i];
        sort(tmp + 1, tmp + 1 + n);
        bool flag = true;
        for (int i = 1; i <= n; i++)
            if (tmp[i] != a[i])
                flag = false;
        if (flag)
        {
            cout << 0 << endl;
            continue;
        }
        //else
        int k = 0;
        //k是逆序對的數量
        for (int i = n; i >= 1; i--)
            for (int j = i - 1; j >= 1; j--)
                if (b[i] < b[j])
                {
                    int x = b[j];//整體往前一個
                    for (int p = j; p <= i - 1; p++)
                        b[p] = b[p + 1];
                    b[i] = x;
                    k++;
                }
 
        cout << k << endl;
        int l = 0, r = 0, d = 1;
        for (int i = n; i >= 1; i--)
            for (int j = i - 1; j >= 1; j--)
                if (a[i] < a[j])
                {
                    cout << j << ' ' << i << ' ' << 1 << endl;
                    int x = a[j];//整體往前一個
                    for (int p = j; p <= i - 1; p++)
                        a[p] = a[p + 1];
                    a[i] = x;
                }
 
    }
    return 0;
}

D. Productive Meeting

思路：

優先隊列，同時再開一個idx數組記錄兩個交談的人的下標，每次取出兩個隊頭元素，然后判斷是否大於1，大於一都自減一，然后再放回隊列，直到隊列中 元素只有一個或沒有，需要注意的是，我們剛開始往隊列中加元素的時候只存大於0的元素

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
 
#define x first
#define y second
 
using namespace std;
 
typedef long long LL;
typedef pair<int, int>PII;
 
const int N = 200010;
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        vector<PII>idx;
        idx.clear();
        priority_queue<PII>q;
        for (int i = 1; i <= n; i++)
        {
            int x;
            cin >> x;
            if(x) q.push({ x, i });
        }
 
        while (q.size() > 1)
        {
            auto a = q.top();
            q.pop();
            auto b = q.top();
            q.pop();
 
            idx.push_back({ a.second, b.second });
            if (a.first - 1 > 0) q.push({ a.first - 1, a.second });
            if (b.first - 1 > 0) q.push({ b.first - 1, b.second });
        }
 
        if (q.size()) q.pop();
        cout << idx.size() << endl;
 
        for (int i = 0; i < idx.size(); i++)
            cout << idx[i].first << ' ' << idx[i].second << endl;
        
    }
    return 0;
}

E1. Permutation Minimization by Deque

思路：

貪心 + 雙端隊列，每次比較和隊頭元素的大小，大於隊頭元素就插到隊尾，小於就插到隊首

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
 
#define x first
#define y second
 
using namespace std;
 
typedef long long LL;
typedef pair<int, int>PII;
 
const int N = 200010;
 
int a[N];
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        memset(a, 0, sizeof a);
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++) cin >> a[i];
        deque<int>q;
        q.push_back(a[1]);
        for (int i = 2; i <= n; i++)
            if (a[i] < q.front()) q.push_front(a[i]);
            else q.push_back(a[i]);
 
        for (auto x : q) cout << x << ' ';
        cout << endl;
 
    }
    return 0;
}

上了119分，還行吧