2020 HDU多校 第一场 04-Distinct Sub-palindromes

题目链接: 04-Distinct Sub-palindromes

Description

题意：给定n，求长度为n的字符串中包含的不同次回文串的数目的最大值

S is a string of length n. S consists of lowercase English alphabets.

Your task is to count the number of different S with the minimum number of distinct sub-palindromes. Sub-palindrome is a palindromic substring.

Two sub-palindromes u and v are distinct if their lengths are different or for some i (0 ≤ i ≤ length), ui≠vi. For example, string “aaaa” contains only 4 distinct sub-palindromes which are “a”, “aa”, “aaa” and “aaaa”.

Input

The first line contains an integer T (1 ≤ T ≤ 105), denoting the number of test cases.

The only line of each test case contains an integer n (1 ≤ n ≤ 109).

Output

For each test case, output a single line containing the number of different strings with minimum number of distinct sub-palindromes.

Since the answer can be huge, output it modulo 998244353.

Sample Input

2
1
2

Sample Output

26
676

Method

• n == 1 时， 显然 不同次回文子串数 最大值为 26；
• n == 2 时，aa, ab 均包含两个 次回文子串，最大值为 26²;
• n == 3 时，有 aaa, aab, aba, abc 这四种次回文串，且均为 3 个， 最大值为 26³;
• n >= 4 时，考虑在 n == 3 的情况上加上一个字符，只需考虑abcd这种情况，回文子串数最少为3，所以最大值 为 26×25×24;

Code

详见注释

#include 
#include 

using namespace std;
#define Max 1e5+5
#define ll long long
#define mod 998244353
template<typename T> T get(T a, T b) { 		//快速幂(可以不用快速幂,连乘即可)
	ll s = 1%mod;
	while (b) {
		if (b%2 == 1) {
			s = (s*a)%mod;
		}
		b /= 2;
		a = (a%mod)*(a%mod)%mod;
	}
	return s;
}
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		ll sum, n, k=26;
		scanf("%lld", &n);
		if(n <= 3) sum = get(k, n)%mod;		//如果n <= 3, 次回文串的个数为 26^n 
		else sum = 26*25*24%mod;			//如果n > 3，次回文串的个数为 26*25*24 
		printf("%lld\n", sum);
	}
	return 0;
}

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