Codeforces 432 D. Prefixes and Suffixes

全栈程序员-站长 • 2022年1月14日下午2:00 • 未分类 • 阅读 54

大家好，又见面了，我是全栈君，今天给大家准备了Idea注册码。

随着扩展KMP做一个简单的努力…..

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.

         Sample test(s)
       

           input
         
ABACABA

           output
         
3
1 4
3 2
7 1

           input
         
AAA

           output
         
3
1 3
2 2
3 1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=100100;

char T[maxn],P[maxn];
int next[maxn],ex[maxn];

void pre_exkmp(char P[])
{
    int m=strlen(P);
    next[0]=m;
    int j=0,k=1;
    while(j+1<m&&P[j]==P[j+1]) j++;
    next[1]=j;
    for(int i=2;i<m;i++)
    {
        int p=next[k]+k-1;
        int L=next[i-k];
        if(i+L<p+1) next[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<m&&P[i+j]==P[j]) j++;
            next[i]=j; k=i;
        }
    }
}

void exkmp(char P[],char T[])
{
    int m=strlen(P),n=strlen(T);
    pre_exkmp(P);
    int j=0,k=0;
    while(j<n&&j<m&&P[j]==T[j]) j++;
    ex[0]=j;
    for(int i=1;i<n;i++)
    {
        int p=ex[k]+k-1;
        int L=next[i-k];
        if(i+L<p+1) ex[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<n&&j<m&&T[i+j]==P[j]) j++;
            ex[i]=j; k=i;
        }
    }
}

int pos[maxn],sum[maxn],mx=-1;

struct ANS
{
    int a,b;
}ans[maxn];
int na=0;

bool cmp(ANS x,ANS y)
{
    if(x.a!=y.a)return x.a<y.a;
    return x.b<y.b;
}

int lisan[maxn],nl;

int main()
{
    cin>>P;
    pre_exkmp(P);
    int n=strlen(P);
    for(int i=0;i<n;i++)
    {
        pos[next[i]]++;
        lisan[nl++]=next[i];
        mx=max(mx,next[i]);
    }
    sort(lisan,lisan+nl);
    int t=unique(lisan,lisan+nl)-lisan;
    for(int i=t-1;i>=0;i--)
    {
        sum[lisan[i]]=sum[lisan[i+1]]+pos[lisan[i]];
    }
    for(int i=0;i<n;i++)
    {
        if(next[i]==n-i)
        {
            ans[na++]=(ANS){next[i],sum[next[i]]};
        }
    }
    sort(ans,ans+na,cmp);
    printf("%d\n",na);
    for(int i=0;i<na;i++)
    {
        printf("%d %d\n",ans[i].a,ans[i].b);
    }
    return 0;
}

发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/116863.html原文链接：https://javaforall.net