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

Problem E	Extend to Palindromes	Time Limit : 3 seconds
	Your task is, given an integer N, to make a palidrome (word that reads the same when you reverse it) of length at least N. Any palindrome will do. Easy, isn't it? That's what you thought before you passed it on to your inexperienced team-mate. When the contest is almost over, you find out that that problem still isn't solved. The problem with the code is that the strings generated are often not palindromic. There's not enough time to start again from scratch or to debug his messy code. Seeing that the situation is desperate, you decide to simply write some additional code that takes the output and adds just enough extra characters to it to make it a palindrome and hope for the best. Your solution should take as its input a string and produce the smallest palindrome that can be formed by adding zero or more characters at its end.
	Input
	Input will consist of several lines ending in EOF. Each line will contain a non-empty string made up of upper case and lower case English letters ('A'-'Z' and 'a'-'z'). The length of the string will be less than or equal to 100,000.
	Output
	For each line of input, output will consist of exactly one line. It should contain the palindrome formed by adding the fewest number of extra letters to the end of the corresponding input string.
	Sample Input	Sample Output
	aaaa abba amanaplanacanal xyz	aaaa abba amanaplanacanalpanama xyzyx

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做法 : KMP

做出另外一個反轉字串的字串, 與原先的字串做KMP匹配

#include<stdio.h>
#include<string.h>
#define MaxL 100005
char A[MaxL], B[MaxL];
int P[MaxL];
void KMPTable(char *A) {
    int i, j;
    P[0] = -1, i = 1, j = -1;
    while(A[i]) {
        while(j >= 0 && A[j+1] != A[i])
            j = P[j];
        if(A[j+1] == A[i])
            j++;
        P[i++] = j;
    }
}
int KMPMatching(char A[], char B[]) {
    KMPTable(B);
    int i, j, Alen, Blen;
    Alen = strlen(A), Blen = strlen(B);
    for(i = 0, j = -1; i < Alen; i++) {
        while(j >= 0 && B[j+1] != A[i])
            j = P[j];
        if(B[j+1] == A[i])    j++;
/*        if(j == Blen-1)
            return true;
*/
    }
    return j;
}
void Solve(char A[]) {
   
    int Alen = strlen(A), Blen = Alen;
    int i, j;
    for(i = 0, j = Blen-1; i < Alen; i++, j--)
        B[j] = A[i];
    B[Blen] = '\0';
    int tail = KMPMatching(A, B);
    for(i = Blen-1; i > tail; i--)
        printf("%c", B[i]);
    printf("%s\n", B);
}
int main() {
    while(scanf("%s", A) == 1)
        Solve(A);
    return 0;
}