Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8845 Accepted Submission(s): 4104
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
题目链接：HDU 3336
比较有意思的一题，因为Next[i]代表了字符串S[1~i]（是1~i而不是i~len）中最长公共前后缀的长度同时也是最长公共前缀的结束位置，假设当前循环的到了第K个，那可以发现S[1~k]的前缀S[1~Next[k] ]与S[ (len-Next[k]+1)~K]是一样的。
用cnt[i]表示以i结尾的前缀中所出现的所有重复数就有cnt[i]=cnt[Next[k] ]+1即前面重复过的次数加上此时的又出现一次重复
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=200010;
const int mod=10007;
char s[N];
int nxt[N];
int len;
int dp[N];
void getnext()
{
CLR(nxt,0);
int j=0,k=nxt[0]=-1;
len=strlen(s);
while (j<len)
{
if(k==-1||s[j]==s[k])
nxt[++j]=++k;
else
k=nxt[k];
}
}
int main(void)
{
int tcase,i,j,n,ans;
scanf("%d",&tcase);
while (tcase--)
{
ans=0;
CLR(dp,0);
scanf("%d%s",&n,s);
getnext();
for (i=1; i<=n; ++i)
{
dp[i]=dp[nxt[i]]+1;
ans=(ans+dp[i])%mod;
}
printf("%d
",ans);
}
return 0;
}