A. Casimir's String Solitaire

思路：

分别记录字母A, B, C的数量，显然要满足B的数量等于A的数量和C的数量的和才行

#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

题意：

题意有点难懂，意思是给你规定了一种操作，问要怎么进行一些列的这种操作才能使原数组递增

思路：

比赛的时候想了了土方法，类比冒泡排序从后往前枚举，遇到一个逆序对（a[i] > a[j]且，i < j）就整体前移一个，由于这个题数据范围只有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是逆序对的数量
        for (int i = n; i >= 1; i--)
            for (int j = i - 1; j >= 1; j--)
                if (b[i] < b[j])
                {
                    int x = b[j];//整体往前一个
                    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];//整体往前一个
                    for (int p = j; p <= i - 1; p++)
                        a[p] = a[p + 1];
                    a[i] = x;
                }
 
    }
    return 0;
}

D. Productive Meeting

思路：

优先队列，同时再开一个idx数组记录两个交谈的人的下标，每次取出两个队头元素，然后判断是否大于1，大于一都自减一，然后再放回队列，直到队列中 元素只有一个或没有，需要注意的是，我们刚开始往队列中加元素的时候只存大于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：

贪心 + 双端队列，每次比较和队头元素的大小，大于队头元素就插到队尾，小于就插到队首

#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，还行吧