D. The Next Good String time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output
In problems on strings one often has to find a string with some particular properties. The problem authors were reluctant to waste time on thinking of a name for some string so they called it good. A string is good if it doesn't have palindrome substrings longer than or equal to d.
You are given string s, consisting only of lowercase English letters. Find a good string t with length |s|, consisting of lowercase English letters, which is lexicographically larger than s. Of all such strings string t must be lexicographically minimum.
We will call a non-empty string s[a ... b]?=?sasa?+?1... sb (1?≤?a?≤?b?≤?|s|) a substring of string s?=?s1s2... s|s|.
A non-empty string s?=?s1s2... sn is called a palindrome if for all i from 1 to n the following fulfills: si?=?sn?-?i?+?1. In other words, palindrome read the same in both directions.
String x?=?x1x2... x|x| is lexicographically larger than string y?=?y1y2... y|y|, if either |x|?>?|y| and x1?=?y1,?x2?=?y2,?... ,?x|y|?=?y|y|, or there exists such number r (r?
x|,?r?
y|), that x1?=?y1,?x2?=?y2,?... ,?xr?=?yr and xr?+?1?>?yr?+?1. Characters in such strings are compared like their ASCII codes.
Input
The first line contains integer d (1?≤?d?≤?|s|).
The second line contains a non-empty string s, its length is no more than 4?105 characters. The string consists of lowercase English letters.
Output
Print the good string that lexicographically follows s, has the same length and consists of only lowercase English letters. If such string does not exist, print "Impossible" (without the quotes).
Sample test(s) input
3
aaaaaaa
output
aabbcaa
input
3
zzyzzzz
output
Impossible
input
4
abbabbbabbb
output
abbbcaaabab
搜索+hash，
长度为L的回文串一定包含长度为L－2的回文串。
判断回文的时候只要判断d和d+1不是回文串就可以了。
#include
#include
#include
#include
using namespace std; typedef unsigned long long int ull; const int maxn=500100; char s[maxn],r[maxn]; int n,d; ull p[maxn],hash[maxn],rhash[maxn]; bool ok(int ed,int d) { ed++; int st=ed-d+1; if(st<0) return true; if((rhash[ed]-rhash[st-1]*p[d])*p[st-1]!=hash[ed]-hash[st-1]) return true; return false; } bool dfs(int x,int t) { if(x==n) { puts(r); return true; } for(r[x]=(t?s[x]:'a');r[x]<='z';r[x]++) { hash[x+1]=hash[x]+r[x]*p[x]; rhash[x+1]=rhash[x]*175+r[x]; if(ok(x,d)&&ok(x,d+1)&&dfs(x+1,t&&(r[x]==s[x]))) return true; } return false; } int main() { scanf("%d %s",&d,s); n=strlen(s); int i=n-1; for(;i>=0&&s[i]=='z';i--) s[i]='a'; if(i<0) { puts("Impossible"); return 0; } s[i]++;p[0]=1; for(int i=1;i