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 = 105;
int n;
char s1[300], s2[300];

int main()
{
    scanf("%d", &n);
    getchar();
    scanf("%[^\n]", s1);
    int len = inf;
    for(int i = 1; i < n; i++){
        getchar();
        scanf("%[^\n]", s2);
        int k1 = strlen(s1) - 1, k2 = strlen(s2) - 1;
        if(k1 > k2){
            swap(s1, s2);
            swap(k1, k2);
        }
        while(k1 >= 0 && s1[k1] == s2[k2]){
            k1--;k2--;
        }
        //cout<<k1<<endl;
        if(strlen(s1) - k1 - 1 < len){
            len = strlen(s1) - k1 - 1;
        }
        //len = min(len, strlen(s1) - k1);
    }
    //cout<<len<<endl;
    if(!len){
        printf("nai\n");
    }
    else{
        int l = strlen(s1) - 1;
        for(int i = l - len + 1; i <= l; i++){
            printf("%c", s1[i]);
        }
        printf("\n");
    }
    return 0;
}

 

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 = 105;
int n, len[maxn];
char s[maxn][300];
unsigned long long h[maxn][300], p[300];

unsigned long long get_hash(int i, int j, int id)
{
    return h[id][j] - h[id][i - 1] * p[j - i + 1];
}

bool check(int mid)
{
    unsigned long long x = get_hash(len[1] - mid + 1, len[1], 1);
    //cout<<x<<endl;
    for(int i = 2; i <= n; i++){
        //cout<<get_hash(len[i] - mid + 1, len[i], i)<<endl;
        if(x != get_hash(len[i] - mid + 1, len[i], i)){
            return false;
        }
    }
    return true;
}

int main()
{
    scanf("%d", &n);
    p[0] = 1;
    for(int i = 1; i < 300; i++){
        p[i] = p[i - 1] * 13331;
    }
    int minlen = inf;
    for(int i = 1; i <= n; i++){
        getchar();
        scanf("%[^\n]", s[i] + 1);
        //printf("%s", s[i] + 1);
        len[i] = strlen(s[i] + 1);
        minlen = min(minlen, len[i]);
        h[i][0] = 0;
        for(int j = 1; 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 - 1] * 13331 + (int)s[i][j];
        }
    }

    //cout<<minlen<<endl;
    /*for(int i = 1; i <= n; i++){
        cout<<len[i]<<endl;
    }*/
    int st = 0, ed = minlen, ans = -1;
    while(st <= ed){
        int mid = (st + ed) / 2;
        //cout<<mid<<endl;
        if(check(mid)){
            st = mid + 1;
            ans = mid;
        }
        else{
            ed = mid - 1;
        }
    }

    //cout<<ans<<endl;
    if(ans < 1){
        printf("nai\n");
    }
    else{
        bool flag = true;
        for(int j = len[1] - ans + 1; j <= len[1]; j++){
            if(flag && s[1][j] == ' '){
                continue;
            }
            if(flag && s[1][j] != ' '){
                flag = false;
            }
            printf("%c", s[1][j]);
        }
        printf("\n");
    }

    return 0;
}