题意：
求最长回文串 长度要大于等于2 且输出起点和终点 输出回文串字符 这个字符还是要以给出的字符为起点a 输出
解析：
分析一下s_new串就好了
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cmath> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 200010, INF = 0x7fffffff; char s[maxn], s_new[maxn<<1]; int p[maxn<<1]; int tmp; int init() { int len = strlen(s); s_new[0] = '$'; s_new[1] = '#'; int j = 2; for(int i=0; i<len; i++) s_new[j++] = s[i], s_new[j++] = '#'; s_new[j++] = '\0'; return j; } int manacher() { int len = init(); int max_len = -1; int id, mx = 0; for(int i=1; i<len; i++) { if(i < mx) p[i] = min(p[2*id - i], mx - i); else p[i] = 1; while(s_new[i-p[i]] == s_new[i+p[i]]) p[i]++; if(mx < i+p[i]) { mx = i+p[i]; id = i; } if(max_len < p[i] - 1) max_len = p[i] - 1, tmp = i; } return max_len; } char c; int main() { while(~scanf("%c%s", &c, s)) { getchar(); int max_len = manacher(); if(max_len < 2) printf("No solution!\n"); else { int a = (tmp-p[tmp])/2, b = (tmp-p[tmp])/2+p[tmp]-2; printf("%d %d\n", a, b); for(int i=a; i<=b; i++) { if(s[i] >= c) printf("%c", s[i] - c + 'a'); else printf("%c", 'z' - c + s[i] + 1); } printf("\n"); } } return 0; }