STL用法

1.vector数组从尾部插入，尾部删除。相关用法代码如下：

        //1.尾部插入及删除数字
    vec2.push_back(1);  //尾部插入元素
        vec2.pop_back()     //删除尾部元素
 
    //2.使用下标访问元素，
    cout << vec2[0] << endl; //记住下标是从0开始的。
 
    //3.使用迭代器访问元素.
    vector<int>::iterator it;
    for (it = vec2.begin(); it != vec2.end(); it++)
        cout << *it << endl;
 
    //4.插入元素：    
    vec2.insert(vec2.begin() + i, a); //在第i + 1个元素前面插入a;
 
    
    //5.删除元素：    
    vec2.erase(vec2.begin() + 2); //删除第3个元素
 
    vec2.erase(vec2.begin() + i, vec2.end() + j); //删除区间[i, j - 1]; 区间从0开始
 
    //6.求数组大小:
    vec2.size();
 
    //7.清空 : 
    vec2.clear();

例题：Problem - B - Codeforces

该题利用两个vector分别存入小写和大写字母的下标，并且出现b和B的时候删除，最后按照下标从小到大输出

#include<bits/stdc++.h>
using namespace std;
#define ll long long
 
const int N = 210,inf = 1e9;
int n,m; 
int _;
 
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >>_;
    while(_ --)
    {
      string s;
      cin >> s;
      int n = s.size();
      vector<int> a,b; //用vector数组来存储大小写字母的下标
      for (int i = 0; i < n; i ++ )
      {
          if(s[i] == 'b')
          {
              if(a.size()) a.pop_back(); //从尾部删除
          }else if(s[i] == 'B'){
              if(b.size()) b.pop_back();
          }else if(s[i] >= 'a' && s[i] <= 'z'){
              a.push_back(i);//尾部插入
          }else {
              b.push_back(i);
          }
      }
      
      int k1 = 0,k2 = 0; //定义两个指针
      while(k1 < a.size() || k2 < b.size()){
          if(k1 < a.size() &&(k2 >= b.size() || a[k1] < b[k2])) {//比较下标哪个小输出哪个
              cout << s[a[k1]] ;
              k1 ++;
          }else 
          {
              cout << s[b[k2]];
              k2 ++;
          }
      }
      cout << endl;
    }
    return 0;
}

 2.vector + 哈希映射 Problem - 1890A - Codeforces

    #include <bits/stdc++.h>
    using namespace std;
    
    bool ok(string s) {
      for (size_t i = 1; i < s.length(); ++i)
        if (s[i] == s[i - 1])
          return false;
      return true;
    }
    
    
    string s;
    void solve() {
      int n;
      cin >> n;
      cin >> s;
      int cnt0 = 0, cnt1 = 0;
      for (int i = 0; i < s.length(); ++i) {
        cnt0 += s[i] == '0';
        cnt1 += s[i] == '1';
      }
      if (cnt0 != cnt1) {
        cout << -1 << std::endl;
        return;
      }
      vector<int> z;
      deque<char> q;
      for (int i = 0; i < s.length(); ++i)
        q.push_back(s[i]);
      
      int d = 0;
      while (!q.empty()) {
        if (q.front() == q.back()) {
          if (q.front() == '0') {
            q.push_back('0');
            q.push_back('1');
            z.push_back(n - d);
          } else {
            q.push_front('1');
            q.push_front('0');
            z.push_back(0 + d);
          }
          n += 2;
        }
        while (!q.empty() && q.front() != q.back()) {
          q.pop_back();
          q.pop_front();
          ++d;
        }
      }
    
      cout << z.size() << endl;
      for (int i = 0; i < z.size(); ++i) {
        cout << z[i];
        if (i + 1 == z.size()) cout << endl;
        else cout << " ";
      }
    }
    
   int main() {
      int t;
      cin >> t;
      while (t--) solve();
      return 0;
    }

 3.用vector 来存储偏移量(向量)Problem - A - Codeforces

#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl

const int N = 1e5 + 10,inf = 1e9;
const int mode = 1e9 + 7;
int _;
int n,m;

int main(){
    cin >> _;
    while(_ -- ){
    int a, b;
    cin >> a >> b;
    int xK, yK;
    cin >> xK >> yK;
    int xQ, yQ;
    cin >> xQ >> yQ;
    vector<int> dx = {a, a, -a, -a, b, b, -b, -b};
    vector<int> dy = {b, -b, b, -b, a, -a, a, -a};
    int ans = 0;
    for (int i = 0; i < 8; i ++){
      for (int j = 0; j < 8; j ++){
        if (xK + dx[i] == xQ + dx[j] && yK + dy[i] == yQ + dy[j]){
          ans++;
        }
      }
    }
    if (a == b){
      ans /= 4;
    }
    cout << ans << endl;
  }
}

 4.string 用法 Problem - F - Codeforces

首先用n个字符表示n1对 01字符串，然后在第一个0前面加n1个零构成n1对00字符串，在第一个1前面加n2个1构成n2个11字符串

然后特殊情况特判一下就好了，这里用string(n,'0')的用法

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
int _,n,m;
const int N = 1e3 + 10,inf = 1e9;
const int mod = 1e9 + 7;
int a[N],s[N],t[N];

int main()
{
    cin >> _;
    while(_ -- )
    {
        int n0,n1,n2;
        cin >> n0 >> n1 >> n2;
        if(n1 == 0){
            if(n0 != 0) cout << string(n0 + 1,'0') << endl;
            else cout << string(n2 + 1,'1') << endl;
            continue;
        }

        string ans;
        for (int i = 0; i < n1 + 1; i ++ ){
            if(i & 1) ans += '0';
            else ans += '1';
        }
        ans.insert(1,string(n0,'0'));
        ans.insert(0,string(n2,'1'));
        
        cout << ans << endl;
    }
    return 0;
}