King's Order
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 822 Accepted Submission(s): 451
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.
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).
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
原題:點擊打開鏈接
數組中的三列數分別代表開頭(末尾)是一個連續的字母,兩個連續的字母和三個連續的字母的排列方法,
畫個圖,自己推推就發現規律了.
#include <iostream>
#include <stdio.h>
#include <memory.h>
#define mod 1000000007
#define ll long long
using namespace std;
ll data[2005][4];
void dp()
{
memset(data,0,sizeof(data));
data[1][1]=26;
for(int i=1;i<=2000;i++)
{
for(int j=1;j<=3;j++)
{
if(data[i][j])
{
if(j!=3)
data[i+1][j+1]=data[i][j];
data[i+1][1]=(data[i+1][1]+data[i][j]*25)%mod;
}
<span style="white-space:pre"> </span>}
}
//for(int i=1;i<=2000;i++){for(int j=0;j<3;j++)printf("%lld ",data[i][j]);printf("\n");}
}
int main()
{
int T,n;
ll ans;
dp();
cin>>T;
for(int i=0;i<T;i++)
{
cin>>n;
ans=0;
for(int j=1;j<=3;j++)
ans=(ans+data[n][j])%mod;
cout<<ans<<endl;
}
return 0;
}