2020杭电多校1

Distinct Sub-palindromes
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6754
类型：思维，签到

Problem Description
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≤$10^5$), denoting the number of test cases.

The only line of each test case contains an integer n (1≤n≤$10^9$).

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

翻译： 构造一个长度为n的字符串，使得字符串的子串为回文串的数量最少，求这样的字符串的数量。结果对 998244353取模。
分析：

1. 当n=1时，对于26个任意一个字母构成的字符串，子串都是回文串，回文串的数量为1
2. 当n=2时，构成的字符串有“aa”和“ab”两种情况(排列组合)。这两种字符串子串为回文串的数量都为2.
3. 当n=3时，构成的字符串有“aaa”和“abc”和“aba”三种情况。这三种字符串子串为回文串的数量都为3.
4. 当n>3时，要使子串为回文串的数量最少，构成“abcabcabcabc···”的形势能保证回文串的数量最少，为3

完整代码：

#include
#include
#include
#include
#include
#include
using namespace std;
int quick_mi(int a,int b)
{
	int sum=1;
	while(b)
	{
		if(b&1)
			sum*=a;
		a=a*a;
		b>>=1;
	}
	return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,res;
		scanf("%d",&n);
		if(n<=3)
			res=quick_mi(26,n);
		else
			res=26*25*24;
		cout<<res<<endl;
	}
	return 0;
}