【KMP nx数组 + DP】K - Count the string HDU - 3336

博客围绕字符串前缀匹配次数求和问题展开。给定字符串,需写出其所有非空前缀并统计各前缀在原字符串中的匹配次数,最后求所有前缀匹配次数之和并对10007取模。还给出了输入输出格式及示例。

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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;
}

 

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