Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3797 Accepted Submission(s): 1776
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.
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.
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
Recommend
数据结构：串：KMP算法。
没怎么看明白KMP算法的实现，看别人的代码也似懂非懂，总之最后参考着改还是AC了。
KMP算法分两步：1、产生next[]数组。2、KMP处理。
这道题只用到第一步，在生成next[]的函数中做些改动，即可通过它获得结果。
没搞透，有时间再看看。
1 #include <iostream>
2 #include <string.h>
3 using namespace std;
4 char s[200001];
5 int next[200001];
6 int c[200001];
7 int ans;
8 void GetNext(char t[],int next[])
9 {
10 memset(c,0,sizeof(c));
11 int j,k;
12 j=0;k=-1;next[0] = -1;
13 int length;
14 for(length = 0;t[length]!=' ';length++);
15 while(j<length){
16 if(k==-1 || t[j]==t[k]){
17 if(k!=-1){
18 c[j] = c[k] + 1;
19 ans+=c[j];
20 }
21 j++;k++;
22 next[j] = k;
23 }
24 else
25 k = next[k];
26 }
27 }
28 int main()
29 {
30 int T,n;
31 cin>>T;
32 while(T--){
33 cin>>n;
34 cin>>s;
35 ans = 0;
36 GetNext(s,next);
37 cout<<(ans+n)%10007<<endl;
38 }
39 return 0;
40 }
重又做了一遍这道题，这次大体明白了KMP算法是怎么回事，但是这道题的DP部分还是不太懂。
DP公式为 dp[j]=dp[next[j]]+1; 代表以前j个字母为前缀的字符串在总字符串中出现的次数-1。
代码：
1 #include <iostream>
2
3 using namespace std;
4 char a[200001];
5 int next[200001];
6 int dp[200001];
7 int n,ans;
8 void GetNext(void) //KMP算法自己写的
9 {
10 int j=0,k=-1;
11 next[0] = -1;
12 while(j<n){
13 if(k==-1){
14 j++;k++;
15 next[j] = 0;
16 }
17 else if(a[j]==a[k]){
18 dp[j] = dp[k] + 1; //DP公式
19 ans += dp[j]; //累加次数
20 next[j+1] = k + 1;
21 j++;
22 k = next[j];
23 }
24 else
25 k = next[k];
26 }
27 }
28 int main()
29 {
30 int T;
31 cin>>T;
32 while(T--){
33 cin>>n;
34 cin>>a;
35 ans = 0;
36 GetNext();
37 cout<<(ans+n)%10007<<endl;
38 }
39 return 0;
40 }
Freecode : www.cnblogs.com/yym2013