Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14288 Accepted Submission(s): 6544
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
/// 找字符串的每个前缀在该字符串中出现过的次数总和
#include <bits/stdc++.h>
using namespace std;
const int mn = 200010, mod = 10007;
char ch[mn];
int nx[mn], dp[mn];
void cal_next(char ch[], int len)
{
nx[0] = -1;
int k = -1;
for (int i = 1; i < len; i++)
{
while (k > -1 && ch[k + 1] != ch[i])
k = nx[k];
if (ch[k + 1] == ch[i])
k++;
nx[i] = k;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:\\in.txt", "r", stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d", &T);
while (T--)
{
int len;
scanf("%d %s", &len, ch);
cal_next(ch, len);
int ans = 0;
/// DP 数组代表 加入 i 字符后新增的满足条件的个数
for (int i = 0; i < len; i++)
{
if (nx[i] == -1)
dp[i] = 1;
else // 本身 + nx[i]时新增
dp[i] = 1 + dp[nx[i]];
ans = (ans + dp[i]) % mod;
}
printf("%d\n", ans);
}
return 0;
}