题目大意：给定一个字符串，输出最少须要加入多少个字符使得字符串变成回文串。

解题思路：以字符串的转置做KMP，然后用原串匹配就可以。最后匹配长度即为反复长度。

#include 
     
      #include 
      
       #include 
       
        using namespace std;const int maxn = 1e5+5;char s[maxn], t[maxn];int n, jump[maxn];void get_jump () {    int p = 0;    for (int i = 2; i <= n; i++) {        while (p && s[p + 1] != s[i])            p = jump[p];        if (s[p + 1] == s[i])            p++;        jump[i] = p;    }}int find () {    int p = 0;    for (int i = 1; i <= n; i++) {        while (p && s[p + 1] != t[i])            p = jump[p];        if (s[p + 1] == t[i])            p++;    }    return p;}int main () {    while (scanf("%s", s + 1) == 1) {        printf("%s", s+1);        n = strlen(s + 1);        for (int i = 1; i <= n + 1; i++)            t[i] = s[i];        reverse(s + 1, s + n + 1);        get_jump();        int k = find();        for (int i = k + 1; i <= n; i++)            printf("%c", s[i]);        printf("\n");    }    return 0;}