[codeforces1200E]Compress Words

time limit per test : 1 second
memory limit per test : 256 megabytes

分数：2000，但是我觉得这个题挺神的。

Amugae has a sentence consisting of $n$ 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 $n(1≤n≤10^5)$, the number of the words in Amugae’s sentence.

The second line contains $n$ 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 $10^6$.

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

题意：
给定 $n$个字符串，相邻的字符串可以让前一个字符串的一个后缀作为后一个字符串的前缀这样按顺序从头到尾拼接。
询问最终串的最短长度。

题解：
哈希，每次暴力枚举后缀长度即可。
复杂度是 $\sum Len_i$的

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll MOD=200009300333LL;
char s[1000004],a[1000004];
int n;
int main(){
    scanf("%d",&n);
    int lenS=0,lenA=0;
    ll hsa,hss,Gec;
    while(n--){
        scanf("%s",a+1);
        lenA=strlen(a+1);
        int rL=min(lenA,lenS);
        hsa=0;hss=0;
        Gec=1;
        int st=0;
        for(int i=1;i<=rL;i++){
            hsa=(((hsa*256)%MOD)+a[i])%MOD;
            hss=(hss+((Gec*s[lenS-i+1])%MOD))%MOD;
            Gec*=256;
            Gec%=MOD;
            if(hsa==hss)st=i;
        }

        for(int i=st+1;i<=lenA;i++){
            s[++lenS]=a[i];
        }
    }
    s[++lenS]='\0';
    puts(s+1);
    return 0;
}