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BestCoder Round #75 King's Order dp：数位dp
King‘s Order
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.
The first line contains a number T(T≤10)——The number of the test cases.
For each test case, the first line and the only line contains a positive number n(n≤2000).
For each test case, 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
Source
The question is from BC
and hduoj 5642.
My Solution
数位dp
状态： d[i][j][k] 为处理完i 个字符 , 结尾字符为′a′+j , 结尾部分已反复出现了
k 次的方案数;
边界：d[1][j][1] = 1; (1 <= j <= 26 );
状态转移方程：看代码吧;
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 2000+6;
const int HASH = 1000000007;
long long d[maxn][27][4];
int main()
{
int T, n;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
memset(d, 0, sizeof(d));
for(int j = 1; j <= 26; j++){
d[1][j][1] = 1; //d[0][j][2] = 0; d[0][j][3] = 0;//they are not needed.
//d[1][j][2] = 1; these two are wrong and not needed.
//d[2][j][3] = 1; these two are wrong and not needed.
}
for(int i = 1; i <= n; i++){
for(int j = 1; j <= 26; j++){
d[i][j][2] = (d[i][j][2]+d[i-1][j][1])%HASH; // 2
d[i][j][3] = (d[i][j][3]+d[i-1][j][2])%HASH; // 3
for(int k = 1; k <= 26; k++){
if(j != k) d[i][j][1] = ( ( (d[i][j][1]+d[i-1][k][1])%HASH + d[i-1][k][2])%HASH + d[i-1][k][3])%HASH;
}
}
}
long long ans = 0;
for(int j = 1; j <= 26; j++){
for(int k = 1; k <= 3; k++ ){
ans = (ans + d[n][j][k]) %HASH;
}
}
cout<<ans<<endl;
//printf("%lld\n", ans); this website : %I64d
}
return 0;
}
Thank you!
BestCoder Round #75 King's Order dp：数位dp