# 题目

## Description

After the king's speech , everyone is encouraged. But the war is not over. The king needs to give orders from time to time. But sometimes he can not speak things well. So in his order there are some ones like this: "Let the group-p-p three come to me". As you can see letter 'p' repeats for 3 times. > > Poor king!

Now , it is war time , because of the spies from enemies , sometimes it is pretty hard for the general to tell which orders come from the king. But fortunately the general know how the king speaks: the king never repeats a letter for more than 3 times continually .And only this kind of order is > > legal. For example , the order: "Let the group-p-p-p three come to me" can never come from the king. While the order:" Let the group-p three come to me" is a legal statement.

The general wants to know how many legal orders that has the length of n

To make it simple , only lower case English Letters can appear in king's order , and please output the answer modulo 10000000071000000007

We regard two strings are the same if and only if each charactor is the same place of these two strings are the same.

## Input

The first line contains a number T(T≤10)T(T≤10)――The number of the testcases.

For each testcase, the first line and the only line contains a positive number n(n≤2000)n(n≤2000).

## Output

For each testcase, print a single number as the answer.

2
2
4

676
456950

## hint:

All the order that has length 2 are legal. So the answer is 26*26. For the order that has length 4. The illegal order are : "aaaa" , "bbbb"…….."zzzz" 26 orders in total. So the answer for n == 4 is 26^4-26 = 456950

# 题解

n为层数 m为连续数目
`dp[n][m]=dp[n-1][m+1]+dp[n-1][m]*25`

`dp[n-1][m+1]`代表选择和上一位一样的

`dp[n-1][m]`代表选择和上一位不一样的

`(a+b) mod c = ((a mod c) + (b mod c)) mod c`
`(a*b) mod c = ((a mod c) * (b mod c)) mod c`

`(dp[n][m] = (DP(n - 1, m + 1) % mod) + ((25 * (DP(n - 1, 1) % mod)) % mod)) % mod; `

# 代码

```/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！

*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

const int mod = 1000000007;
const int maxn = 2005;
typedef unsigned long long LL;
LL dp[maxn][5];

//n层数 m连续数目
LL DP(int n, int m) {
if (n == 0) {
if (m == 1)
return 0;
else
return 1;
}

if (m == 4)
return 0;

if (dp[n][m] == 0) {
//(a+b) mod c = ((a mod c) + (b mod c)) mod c
//(a*b) mod c = ((a mod c) * (b mod c)) mod c
return (dp[n][m] = (DP(n - 1, m + 1) % mod) + ((25 * (DP(n - 1, 1) % mod)) % mod)) % mod;
}
else {
return dp[n][m];
}

}

void Do() {
int n;
scanf("%d", &n);
LL ans = (DP(n, 1) % mod) * 26;
printf("%llu\n", ans % mod);
return;
}

int main() {
memset(dp, 0, sizeof(dp));
int T;
scanf("%d", &T);
while (T--) {
Do();
}
return 0;
}
```