Problem Description
For each prefix with length P of a given string S,if
S[i]=S[i+P] for i in [0..SIZE(S)-p-1],
then the prefix is a “period” of S. We want to all the periodic prefixs.
Input
Input contains multiple cases.
The first line contains an integer T representing the number of cases. Then following T cases.
Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.
Output
For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.
Sample Input
4
ooo
acmacmacmacmacma
fzufzufzuf
stostootssto
Sample Output
Case #1: 3
1 2 3
Case #2: 6
3 6 9 12 15 16
Case #3: 4
3 6 9 10
Case #4: 2
9 12
题意:求出一个字符串的所有可能循环节的长度
思路:kmp的运用,题目的意思不是很清楚 j=next[j] 就是次大的匹配个数
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1}; int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1}; const int inf=0x3f3f3f3f; const ll mod=1e9+7; int nextt[1000007]; void getnext(string s){ nextt[1]=0; int len=s.length(); for(int i=2,j=0;i<=len;i++){ while(j>0&&s[i-1]!=s[j]) j=nextt[j]; if(s[i-1]==s[j]) j++; nextt[i]=j; } } int main(){ ios::sync_with_stdio(false); int t; cin>>t; int w=0; while(t--){ string s; cin>>s; getnext(s); int len=s.length(); queue<int> q; int j=nextt[len]; while(j>0){ q.push(j); j=nextt[j]; } q.push(0); bool f=1; cout<<"Case #"<<++w<<": "<<q.size()<<endl; while(!q.empty()){ int temp=q.front(); q.pop(); if(f){ cout<<len-temp; f=0; }else{ cout<<" "<<len-temp; } } cout<<endl; } return 0; }