E. Compress Words（Hash，KMP）

E. Compress Words

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Amugae has a sentence consisting of nn words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".

Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.

Input

The first line contains an integer nn (1≤n≤1051≤n≤105), the number of the words in Amugae's sentence.

The second line contains nn words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits ('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9'). The total length of the words does not exceed 106106.

Output

In the only line output the compressed word after the merging process ends as described in the problem.

Examples

input

5
I want to order pizza

output

Iwantorderpizza

input

5
sample please ease in out

output

sampleaseinout

算法：Hash 或者 KMP

题解：首先分析题目，题目要求我们将每个连续单词相同的前后缀合并成一个，在我们学过的知识里面，字符串前后缀相同的匹配肯定第一想到的就是KMP，对，这题可以用KMP直接做，但是还有一种使代码更加简洁，思路更加简单的方法，那就是用Hash。只要每次查询的时候直接暴力求前后缀相同的最大字符数就行。

Hash：

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

const int maxn = 1e5+;
const int mod = ;
const int base = ;

string ans, s;

void solve() {
    ll hl = , hr = , tmp = , index = ;
    for(int i = ; i < (int)min(ans.size(), s.size()); i++) {
        hl = (hl * base + ans[ans.size() - i - ]) % mod;   //获取左边数组的后缀Hash值
        hr = (s[i] * tmp + hr) % mod;       //获取右边数组的前缀Hash值
        tmp = (tmp * base) % mod;
        if(hl == hr) {
            index = i + ;      //获取最大的字符匹配位数
        }
    }
    for(int i = index; i < (int)s.size(); i++) {
        ans += s[i];
    }
}  

int main() {
    int n;
    scanf("%d", &n);
    cin >> ans;
    for(int i = ; i < n; i++) {
        cin >> s;
        solve();
    }
    cout << ans << endl;
    return ;
}