官方题解 | Word

B. Word

Subtask 100 pt

根据题意模拟即可。本题难点在于字符串函数的使用，以下是本题需用到的函数：

• size()：得到字符串的长度
• erase()：删除字符串的指定区域
• insert()：在字符串指定区域插入内容
• empty()：判断字符串是否为空
• substr()：截取字符串的指定区域

AC Code

#include<bits/stdc++.h>
#define int long long
using namespace std;
signed main(){
    int n; cin >> n;
    string wd = "",jtb = "";
    int gb = 0,xz_l = -1,xz_r = -1;
    for(int _ = 1; _ <= n;_++){
        string op; cin >> op;
        if(op == "I"){
            string s; cin >> s;
            if(xz_l != -1 && xz_l <= xz_r && xz_r < (int)wd.size()){
                wd.erase(xz_l,xz_r - xz_l + 1);
                wd.insert(xz_l,s);
                gb = xz_l + s.size();
            }else{
                wd.insert(gb,s);
                gb += s.size();
            }
            xz_l = xz_r = -1;
        }else if(op == "A"){
            if(wd.empty()) xz_l = xz_r = -1;
            else{
                xz_l = 0;
                xz_r = (int)wd.size() - 1;
            }
            gb = wd.size();
        }else if(op == "C"){
            if(xz_l != -1 && xz_l <= xz_r && xz_r < (int)wd.size()) jtb = wd.substr(xz_l,xz_r - xz_l + 1);
            xz_l = xz_r = -1;
        }else if(op == "V"){
            if(xz_l != -1 && xz_l <= xz_r && xz_r < (int)wd.size()){
                wd.erase(xz_l,xz_r - xz_l + 1);
                wd.insert(xz_l,jtb);
                gb = xz_l + jtb.size();
            }else{
                wd.insert(gb,jtb);
                gb += jtb.size();
            }
            xz_l = xz_r = -1;
        }else if(op == "P"){
            int i; cin >> i;
            gb = min(i,(int)wd.size());
            xz_l = xz_r = -1;
        }else if(op == "TP"){
            int i; cin >> i;
            if(wd.empty()){
                xz_l = xz_r = -1;
                gb = 0;
            }else{
                i = max(1LL,min(i,(int)wd.size()));
                xz_l = xz_r = i - 1;
                gb = i;
            }
        }
    }
    cout << wd;
    return 0;
}

时间复杂度：$O(n)$​