题目链接

题面：

题意：
问长度为 n 的字符串中，包含本质不同回文串最少的有多少个（长度为n的串有多少个，其中他们都是包含本质不同回文串最少的串）。

题解：
n = = 1 , a n s = 26 n = = 2 , a n s = 26 ∗ 26 n = = 3 , a n s = 26 ∗ 26 ∗ 26 n > 3 , a n s = 26 ∗ 25 ∗ 24 − − > a b c a b c a b c n==1,ans=26\\n==2,ans=26*26\\n==3,ans=26*26*26\\n>3,ans=26*25*24-->abcabcabc n==1,ans=26n==2,ans=26∗26n==3,ans=26∗26∗26n>3,ans=26∗25∗24−−>abcabcabc

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<string>
#include<queue>
#include<bitset>
#include<map>
#include<unordered_map>
#include<set>
#define ui unsigned int
#define ll long long
#define llu unsigned ll
#define ld long double
#define pr make_pair
#define pb push_back
#define lc (cnt<<1)
#define rc (cnt<<1|1)
#define len(x) (t[(x)].r-t[(x)].l+1)
#define tmid ((l+r)>>1)
using namespace std;

const int inf=0x3f3f3f3f;
const ll lnf=0x3f3f3f3f3f3f3f3f;
const double dnf=1e18;
const int mod=998244353;
const double eps=1e-8;
const double pi=acos(-1.0);
const int hp=13331;
const int maxn=4100;
const int maxm=100100;
const int up=100100;



int main(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        if(n==1) printf("%d\n",26);
        else if(n==2) printf("%d\n",26*26);
        else if(n==3) printf("%d\n",26*26*26);
        else printf("%d\n",26*25*24);
    }
    return 0;
}