hdu3336 Count the string(KMP应用)

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5460    Accepted Submission(s): 2575

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 

/*
因为next数组next[k]表示的是以k结尾的字符串的前缀和后缀公共长度。挺经典的。
dp[k]就可以用他前缀的总个数加它本身 dp[next[k]]+1，
题目每个前缀在总串中出现的次数实际为1--n每个前缀个数之和，即为每个串的前缀之和
Time:2015-4-13 17:56
*/
#include
#include
#include
using namespace std;
const int MAX=200000+10;
const int mod=10007;
int nxt[MAX];
int dp[MAX];
char str[MAX];
void Get_next(){
    int i,j;
    memset(nxt,0,sizeof(nxt));
    i=0;j=-1;
    nxt[0]=-1;
    while(str[i]){
        if(j==-1||str[i]==str[j]){
            i++;j++;
            nxt[i]=j;
        }else
            j=nxt[j];
    }
}
int main(){
    int T,len;
    scanf("%d",&T);
    while(T--){
        scanf("%d %s",&len,str);
        Get_next();
        memset(dp,0,sizeof(dp));
        int sum=0;
        for(int i=1;i<=len;i++){
            //printf("next=%d ",nxt[i]);
            dp[i]=(dp[nxt[i]]+1)%mod;
            sum=(sum+dp[i])%mod;
            //printf("dp=%d\n",dp[i]);
        }//puts("");
        printf("%d\n",sum);
    }
return 0;
}