HDU 3336 Count the string (kmp+dp)

Count the string

Problem Description

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.

 

Input

The 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.

 

Output

For 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

 

Author

foreverlin@HNU

 

Source

HDOJ Monthly Contest – 2010.03.06 

 

题意: 求字符串的所有前缀在字符串中出现的次数和

 

dp+求kmp中的next数组 

状态转移方程：dp[i]+=dp[next[i]]+1,ans+=dp[i]

 

#include <cctype>
#include <cstdio>

const int mod=10007;
const int MAXN=10000010;

int n,ans,T;

int next[MAXN],dp[MAXN];

char s[MAXN];

inline void read(int&x) {
    int f=1;register char c=getchar();
    for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
    for(;isdigit(c);x=x*10+c-48,c=getchar());
    x=x*f;
}

inline void KMP_next() {
    for(int i=1;i<=n;++i) next[i]=0;
    next[1]=0;  
    int j=0;  
    for(int i=2;i<=n;i++){  
        while(j>0&&s[j+1]!=s[i]) j=next[j];  
        if(s[j+1]==s[i]) j+=1;  
        next[i]=j;  
    }  
}

int hh() {
    freopen("ha.in","r",stdin);
    freopen("ha.out","w",stdout);
    read(T);
    while(T--) {
        read(n);
        scanf("%s",s+1);
        KMP_next();
        ans=0;
        for(int i=1;i<=n;++i) {
            dp[i]=(dp[next[i]]+1)%mod;
            ans=(ans+dp[i])%mod;
        }
        printf("%d\n",ans);
    }
    fclose(stdin);
    fclose(stdout);
    return 0;
}

int sb=hh();
int main(int argc,char**argv) {;}