Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

You have been tasked to infiltrate a tight-lipped society for fun and profit: the ACM ICPC regional judges. Through the PC² ``submission" software, you know that classified information is accessible through the log-ins of the judges tasked to a particular ``regional site". However, you are not certain that any particular judge has access to all the relevant information, so several log-ins will be required. You have been handed down a list of usernames, and the passwords used can be derived from these usernames, as follows:

Input 

The input will only have capital letters (denoting the usernames) and carriage returns. Each line (thus each username) will not be longer than twenty characters, and there will not be more than 12 ``judges" whose log-ins you will need to infiltrate. Strangely, no username uses any letter more than once.

Output 

For each username, you must produce a line containing the password which that username uses. The password for a given username is determined from the twenty-one lexicographically consecutive permutations of the username, the eleventh (middle) of which is the username itself. For example, if the username is WORDFISH, the lexicographic permutations of WORDFISH contain, in order:

..., WOISHRFD, WOISRDFH, WOISRDHF, WOISRFDH, WOISRFHD, WOISRHDF, WOISRHFD, WORDFHIS, WORDFHSI, WORDFIHS, WORDFISH, WORDFSHI, WORDFSIH, WORDHFIS, WORDHFSI, WORDHIFS, WORDHISF, WORDHSFI, WORDHSIF, WORDIFHS, WORDIFSH, ...

The password is then the permutation among the twenty-one lexicographically consecutive permutations of the username which has the largest minimum absolute distance between consecutive letters (and the first amongst the lexicographically ordered, if several permutations have the largest minimum absolute distance), followed by that minimum absolute distance. For the username WORDFISH, the password is WORDHSFI3¹ .

¹ Disclaimer: The above story is completely fictional, and in no way represents any fact, regarding ACM regional judges, their passwords or world domination.

Sample Input 

WORDFISH

Sample Output 

WORDHSFI3

---

翻譯已經放上不幸狗
http://unfortunatedog.blogspot.tw/2013/07/1209-wordfish.html

這裡提供一個不用內建找前後的字典順序的寫法。

接著就是模擬找一個字典順序最小的差。


#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
using namespace std;
char s[128], cnt[128], pwd[128], path[128];
int pcnt, first, ret, n;
void next(int nd) {
    if(pcnt == 10) return;
    int i;
    if(nd == n) {
        if(first == 0)  {
            first = 1;
            return;
        }
        pcnt++;
        path[nd] = '\0';
        int mn = 0xffffff;
        for(i = 1; i < n; i++)
            mn = min(mn, abs(path[i]-path[i-1]));
        if(mn > ret) {
            strcpy(pwd, path);
            ret = mn;
        }
        return;
    }
    for(i = 'A'; i <= 'Z'; i++) {
        if(first == 0)
            i = s[nd];
        if(cnt[i]) {
            cnt[i]--;
            path[nd] = i;
            next(nd+1);
            cnt[i]++;
        }
    }
}
void prev(int nd) {
    if(pcnt == 10) return;
    int i;
    if(nd == n) {
        if(first == 0)  {
            first = 1;
            return;
        }
        pcnt++;
        path[nd] = '\0';
        int mn = 0xffffff;
        for(i = 1; i < n; i++)
            mn = min(mn, abs(path[i]-path[i-1]));
        if(mn >= ret) {
            strcpy(pwd, path);
            ret = mn;
        }
        return;
    }
    for(i = 'Z'; i >= 'A'; i--) {
        if(first == 0)
            i = s[nd];
        if(cnt[i]) {
            cnt[i]--;
            path[nd] = i;
            prev(nd+1);
            cnt[i]++;
        }
    }
}
int main() {
    int i;
    while(scanf("%s", s) == 1) {
        memset(cnt, 0, sizeof(cnt));
        n = strlen(s);
        for(i = 0; s[i]; i++)
            cnt[s[i]]++;
        ret = 0xffffff;
        for(i = 1; i < n; i++)
            ret = min(ret, abs(s[i]-s[i-1]));
        strcpy(pwd, s);
        first = 0, pcnt = 0;
        next(0);
        first = 0, pcnt = 0;
        prev(0);
        printf("%s%d\n", pwd, ret);
    }
    return 0;
}