PAT甲1077 Kuchiguse【字符串】【暴力】【Hash】【二分】

1077 Kuchiguse （20 分）

The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker's personality. Such a preference is called "Kuchiguse" and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle "nyan~" is often used as a stereotype for characters with a cat-like personality:

• Itai nyan~ (It hurts, nyan~)

• Ninjin wa iyada nyan~ (I hate carrots, nyan~)

Now given a few lines spoken by the same character, can you find her Kuchiguse?

Input Specification:

Each input file contains one test case. For each case, the first line is an integer N (2≤N≤100). Following are N file lines of 0~256 (inclusive) characters in length, each representing a character's spoken line. The spoken lines are case sensitive.

Output Specification:

For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all N lines. If there is no such suffix, write nai.

Sample Input 1:

3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~

Sample Output 1:

nyan~

Sample Input 2:

3
Itai!
Ninjinnwaiyada T_T
T_T

Sample Output 2:

nai

题意：

给n个串求所有串的最长公共后缀。

思路：

觉得自己就是一个傻逼....

看到题目想到的直接是Hash+二分，敲完了WA了去看别人写的发现直接暴力两两找就行了，因为LCS长度是非递增的。

交完了暴力突然明白HashWA了是因为题目的意思是区分大小写，我以为那句话的意思是不区分大小写，还把大写转成了小写。

想太多。

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <map>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 const int maxn = ;
 int n;
 char s1[], s2[];

 int main()
 {
     scanf("%d", &n);
     getchar();
     scanf("%[^\n]", s1);
     int len = inf;
     for(int i = ; i < n; i++){
         getchar();
         scanf("%[^\n]", s2);
         int k1 = strlen(s1) - , k2 = strlen(s2) - ;
         if(k1 > k2){
             swap(s1, s2);
             swap(k1, k2);
         }
         while(k1 >=  && s1[k1] == s2[k2]){
             k1--;k2--;
         }
         //cout<<k1<<endl;
         if(strlen(s1) - k1 -  < len){
             len = strlen(s1) - k1 - ;
         }
         //len = min(len, strlen(s1) - k1);
     }
     //cout<<len<<endl;
     if(!len){
         printf("nai\n");
     }
     else{
         int l = strlen(s1) - ;
         for(int i = l - len + ; i <= l; i++){
             printf("%c", s1[i]);
         }
         printf("\n");
     }
     return ;
 }

Hash + 二分

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <map>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 const int maxn = ;
 int n, len[maxn];
 char s[maxn][];
 unsigned long long h[maxn][], p[];

 unsigned long long get_hash(int i, int j, int id)
 {
     return h[id][j] - h[id][i - ] * p[j - i + ];
 }

 bool check(int mid)
 {
     unsigned long long x = get_hash(len[] - mid + , len[], );
     //cout<<x<<endl;
     for(int i = ; i <= n; i++){
         //cout<<get_hash(len[i] - mid + 1, len[i], i)<<endl;
         if(x != get_hash(len[i] - mid + , len[i], i)){
             return false;
         }
     }
     return true;
 }

 int main()
 {
     scanf("%d", &n);
     p[] = ;
     for(int i = ; i < ; i++){
         p[i] = p[i - ] * ;
     }
     int minlen = inf;
     for(int i = ; i <= n; i++){
         getchar();
         scanf("%[^\n]", s[i] + );
         //printf("%s", s[i] + 1);
         len[i] = strlen(s[i] + );
         minlen = min(minlen, len[i]);
         h[i][] = ;
         for(int j = ; j <= len[i]; j++){
             /*if(s[i][j] >= 'A' && s[i][j] <= 'Z'){
                 s[i][j] = s[i][j] - 'A' + 'a';
             }*/
             h[i][j] = h[i][j - ] *  + (int)s[i][j];
         }
     }

     //cout<<minlen<<endl;
     /*for(int i = 1; i <= n; i++){
         cout<<len[i]<<endl;
     }*/
     int st = , ed = minlen, ans = -;
     while(st <= ed){
         int mid = (st + ed) / ;
         //cout<<mid<<endl;
         if(check(mid)){
             st = mid + ;
             ans = mid;
         }
         else{
             ed = mid - ;
         }
     }

     //cout<<ans<<endl;
     if(ans < ){
         printf("nai\n");
     }
     else{
         bool flag = true;
         for(int j = len[] - ans + ; j <= len[]; j++){
             if(flag && s[][j] == ' '){
                 continue;
             }
             if(flag && s[][j] != ' '){
                 flag = false;
             }
             printf("%c", s[][j]);
         }
         printf("\n");
     }

     return ;
 }