传送门
文章目录
- 题意：
- 思路：
题意：
思路：
通过完美子串的定义，我们不难发现满足条件的子串就是kmpkmpkmp中ne[n]ne[n]ne[n]不断向前跳得到的串，现在问题就是如何求这些前缀串在串中出现的次数了。
考虑一个前缀iii，那么一直暴跳ne[i]ne[i]ne[i]，能跳到的前缀都需要加上前缀iii的出现次数。显然不能直接跳，所以考虑dpdpdp，dp[i]dp[i]dp[i]表示前缀iii的出现次数，那么转移方程就是dp[ne[i]]+=dp[i]dp[ne[i]]+=dp[i]dp[ne[i]]+=dp[i]
这样就可以避免暴跳了。
复杂度O(N)O(N)O(N)
// Problem: D. Prefixes and Suffixes
// Contest: Codeforces - Codeforces Round #246 (Div. 2)
// URL: https://codeforces.com/contest/432/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
char s[N];
int ne[N],f[N];
vector<PII>ans;int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);cin>>(s+1);n=strlen(s+1);for(int i=2,j=0;i<=n;i++) {while(j&&s[i]!=s[j+1]) j=ne[j];if(s[i]==s[j+1]) j++;ne[i]=j;}for(int i=1;i<=n;i++) f[i]=1;for(int i=n;i>=1;i--) f[ne[i]]+=f[i];while(n) {ans.pb({n,f[n]});n=ne[n];}sort(ans.begin(),ans.end());printf("%d\n",ans.size());for(auto x:ans) printf("%d %d\n",x.X,x.Y);return 0;
}
/**/
