max str 時間復雜度 結果 con lib one == ont

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

InputThe first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.
OutputFor each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.Sample Input

1
4
abab

Sample Output

6
題意：找每個前綴在字符串中的出現次數
題解：kmp的next數組處理，剛開始從後往前遍歷，每個next數值往前遞歸加上次數，最後就是結果，交了之後發現tle（沒有發現最差的時間復雜度居然有O（n*n）），於是乎只能改進一下，用num數組來存次數，每次遞歸的時候就把當前num++，這樣只需要遍歷一遍就行了
時間復雜度變成了O（n）果然不會Tle了

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include
 
<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 10007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const
 
 double g=10.0,eps=1e-9;
const int N=200000+5,maxn=60+5,inf=0x3f3f3f3f;

ll num[N];
int Next[N],slen;
string str;

void getnext()
{
    int k=-1;
    Next[0]=-1;
    for(int i=1;i<slen;i++)
    {
        while(k>-1&&str[k+1]!=str[i])k=Next[k];
        if(str[k+1]==str[i])k++;
        Next[i]=k;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t,n;
    cin>>t;
    while(t--){
        cin>>n>>str;
        slen=str.size();
        getnext();
        ll ans=0;
        memset(num,0,sizeof num);
        for(int i=slen-1;i>=0;i--)
        {
            ll j=i;
            while(j!=-1)num[j]++,j=Next[j];
        }
        for(int i=0;i<slen;i++)ans=ans%mod+num[i]%mod;
        cout<<ans<<endl;
    }
    return 0;
}

View Code

hdu3336 kmp