Codeforces Round #246 (Div. 2)-D. Prefixes and Suffixes(KMP+DP)

原题链接

D. Prefixes and Suffixes

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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

Let's introduce several definitions:
A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of
string s is string sisi + 1...sj.
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 li ci.
Numbers li ci mean
that the prefix of the length li matches
the suffix of length li and
occurs in string s as a substringci times.
Print pairs li ci in
the order of increasing li.

Examples

input

ABACABA

output

3
1 4
3 2
7 1

input

AAA

output

3
1 3
2 2
3 1

某大牛思路:

题目分析 :首先得到母串的next数组，next数组的含义是next[j]的值表示str[0...j-1]（我的next[0]是-1）这个子串的前后缀匹配的最长长度，如样例1

index  0  1  2  3  4  5  6  7

str    A  B  A  C  A  B  A

next   -1 0  0  1  0  1  2  3

next[6] = 2即ABACAB这个子串的前后缀最长匹配是2(AB)

由此性质我们可以发现满足条件的子串即是next[next[len。。。]]不断往前递归直到为0，因为长的可能会包含短的，我们可以递归得到dp数组(dp记录的就是子串出现的次数)re数组的递归式为dp[next[i]] += dp[i];

#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;

char str[maxn];
int next[maxn], dp[maxn];
int ans1[maxn], ans2[maxn], cnt;
void KMP(){

next[0] = -1;
int i = 0, j = -1;
while(str[i]){
if(j == -1 || str[i] == str[j]){
i++;
j++;
next[i] = j;
}
else
j = next[j];
}
}
void solve(){

int len = strlen(str);
for(int i = 0; i <= len; i++){
dp[i] = 1;
}
for(int i = len; i >= 1; i--){
if(next[i] != -1){
dp[next[i]] += dp[i];
}
}
int pos = len;
while(pos){
ans1[cnt] = pos;
ans2[cnt++] = dp[pos];
pos = next[pos];
}
}
int main(){
//	freopen("in.txt", "r", stdin);
scanf("%s", str);
KMP();
solve();
printf("%d\n", cnt);
while(cnt--){
printf("%d %d\n", ans1[cnt], ans2[cnt]);
}
return 0;
}