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

題意：t 組數據，每組給出一個字符串，對于字符串的所有前綴，若存在循環節，輸出這個字符串的符合條件的個數與這個字符串的長度

思路：思路與?Seek the Name, Seek the Fame（POJ-2752）類似，但輸出的不是循環節的長度，而是前綴子串的長度，此外需要統計符合條件的子串個數，可利用隊列來統計輸出

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define N 1000001 #define LL long long const int MOD=20091226; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std;int Next[N]; char p[N]; int n; void getNext(){Next[0]=-1;int len=strlen(p);int j=0;int k=-1;while(j<len){if(k==-1||p[j]==p[k]) {k++;j++;Next[j]=k;}else{k=Next[k];}} } int main(){int t;scanf("%d",&t);int Case=1;while(t--){scanf("%s",p);int len=strlen(p);getNext();int i=len;queue<int> Q;while(i){i=Next[i];Q.push(i);}printf("Case #%d: %d\n",Case++,Q.size());printf("%d",len-Q.front());Q.pop();while(!Q.empty()){printf(" %d",len-Q.front());Q.pop();}printf("\n");}return 0; }

?

總結

以上是生活随笔為你收集整理的Period II（FZU-1901）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。