Rikka with Nickname

鏈接：https://www.nowcoder.com/acm/contest/148/J
來源：牛客網
 

時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述

Sometimes you may want to write a sentence into your nickname like "lubenwei niubi". But how to change it into a single word? Connect them one by one like "lubenweiniubi" looks stupid.

To generate a better nickname, Rikka designs a non-trivial algorithm to merge a string sequence s1...sn into a single string. The algorithm starts with s=s1 and merges s2...sn into s one by one. The result of merging t into s is the shortest string r which satisfies s is a prefix of r and t is a subsequence of r.(If there are still multiple candidates, take the lexicographic order smallest one.)

String s is a prefix of r if and only if |s| ≤ |r| and for all index i ∈ [1, |s|], si = ri.

String s is a subsequence of r if and only if there is an index sequence which satisfies .

For example, if we want to generate a nickname from "lubenwei niubi", we will merge "niubi" into "lubenwei", and the result is "lubenweiubi".

Now, given a sentence s1...sn with n words, Rikka wants you to calculate the resulting nickname generated by this algorithm.

輸入描述:

The first line contains a single number t(1 ≤ t ≤ 3), the number of testcases.

For each testcase, the first line contains one single integer n(1 ≤ n ≤ 106).

Then n lines follow, each line contains a lowercase string .

輸出描述:

For each testcase, output a single line with a single string, the result nickname.

 

示例1

輸入

2
2
lubenwei
niubi
3
aa
ab
abb

輸出

lubenweiubi
aabb

題解：從第一個字符串開始，第二個字符串以順序的方式融入前一個字符串，融不進的就把剩下的字符串添加在總字符串尾；

           很簡單，想複雜了怎麼也搞不出來；

tips：string定義的字符串變量可以直接加；

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t,n;
    string s,s1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cin>>s;
        for(int i=1;i<n;i++)
        {
            cin>>s1;
            int k=0,j=0;
            while(s[j]&&s1[k])
            {
                if(s[j]==s1[k])
                {
                    k++;
                }
                j++;
            }
            while(s1[k])
            {
                s+=s1[k];//s字符串變量可以用加的方式把字符直接加在末尾
                k++;
            }
            s1="";//定義s1字符串爲空
        }
        cout<<s<<endl;
    }
}