HDU 5642：King's Order【DP】

King's Order

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 363    Accepted Submission(s): 219


Problem 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 1000000007

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)——The
number of the testcases.

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

 

Output

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

 

Sample Input

2
2
4

 

Sample Output

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

 
题意：给你字符串的长度~让你求出总共有多少种使得最多有三个连续相同的由小写字母组成的字符串。
思路：dp,dp[i][j]代表长度为i,最后的边续相同的数量为j的字符串种数。
AC-code：

#include<cstdio>
#define mod 1000000007
long long  dp[2005][4];
void pre()
{
int i,j;
dp[1][1]=26;
for(i=1;i<=2000;i++)
for(j=1;j<=3;j++)
if(dp[i][j])
{
if(j!=3)
dp[i+1][j+1]=(dp[i][j]+dp[i+1][j+1])%mod;
dp[i+1][1]=(dp[i+1][1]+dp[i][j]*25)%mod;
}
}
int main()
{
int T,n,i;
long long ans;
scanf("%d",&T);
pre();
while(T--)
{
scanf("%d",&n);
ans=0;
for(i=1;i<=3;i++)
ans=(ans+dp
[i])%mod;
printf("%lld\n",ans);
}
return 0;
}