codeforces432D Prefixes and Suffixes(kmp+dp)

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D. Prefixes and Suffixes

You have a string s = s₁s₂...s_|s|, where |s| is the length of string s, and s_i its i-th character.

Let's introduce several definitions:

• A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of string s is string s_is_i + 1...s_j.
• The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
• The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].

Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring.

Input

The single line contains a sequence of characters s₁s₂...s_|s| (1 ≤ |s| ≤ 10⁵) — string s. The string only consists of uppercase English letters.

Output

In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers l_i c_i. Numbers l_i c_i mean that the prefix of the length l_i matches the suffix of length l_i and occurs in string s as a substring c_i times. Print pairs l_i c_i in the order of increasing l_i.

Sample test(s)

Input

ABACABA

Output

3
1 4
3 2
7 1

Input

AAA

Output

3
1 3
2 2
3 1

题意：

　　给出一个字符串，问每种既是前缀，又是后缀的字符串出现的次数kmp+dp,利用kmp求出所有的前后缀，考虑到若第i位既是前缀又是后缀，则kmp中的p[i]位也一定是，由此可以得到一个状态转移的方程dp[p[i]]+=dp[i],初始化所有的dp[i]=1，因为还包括它本身。

我队友curs0r是用后缀数组做的。。。。。。

 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>
 using namespace std;
 #define XINF INT_MAX
 #define INF 0x3FFFFFFF
 #define MP(X,Y) make_pair(X,Y)
 #define PB(X) push_back(X)
 #define REP(X,N) for(int X=0;X<N;X++)
 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 #define CLR(A,X) memset(A,X,sizeof(A))
 #define IT iterator
 typedef long long ll;
 typedef pair<int,int> PII;
 typedef vector<PII> VII;
 typedef vector<int> VI;
 #define MAXN 100010
 char s[MAXN];
 int p[MAXN];
 int dp[MAXN];
 int vis[MAXN];
 void getp()
 {
     p[]=;
     int len=strlen(s+);
     for(int i=,j=;i<=len;i++)
     {
         while(j>&&s[j+]!=s[i])j=p[j];
         if(s[j+]==s[i])j+=;
         p[i]=j;
     }
 }

 int main()
 {
     ios::sync_with_stdio(false);
     while(cin>>s+)
     {
         int len=strlen(s+);
         getp();
         for(int i=;i<=len;i++)
         {
             dp[i]=;
         }
         int i=len;
         int k=;
         while(i)
         {
             vis[k]=i;
             k++;
             i=p[i];
         }
         for(i=len;i>;i--)
         {
             dp[p[i]]+=dp[i];
         }
         cout<<k<<endl;
         for(i=k-;i>=;i--)
         {
             cout<<vis[i]<<" "<<dp[vis[i]]<<endl;
         }
     }
     return ;
 }

代码君