前四位我们可以算出d[1]=2,d[2]=4,d[3]=6,d[4]=9. 
 
我们可以这样想：一个合法串可以由两个较短的合法串组成 
 
就以d[n]为例：(注意不能重复) 
 
1、n-1个字符的时候: +m 
 
2、n-2: 只能+mm，会和n-1重复,所以不考虑n-2 
 
3、n-3: +mmf 
 
4、n-4: +mmff 
 
5、n-5: 如果是+mmffm，会与n-1重复,+mmmff会与n-4重复,+mmmmf会与n-3重复(不考虑) 
 
所以就考虑n-1,n-3,n-4，DP等式就出来了:dp[n]=dp[n-1]+dp[n-3]+dp[n-4] 
 
矩阵快速幂

#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <time.h>
#include <cstdio>
#include <math.h>
#include <iomanip>
#include <cstdlib>
#include <limits.h>
#include <string.h>
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;#define LL long long
#define MIN INT_MIN
#define MAX INT_MAX
#define PI acos(-1.0)
#define FRE freopen("input.txt","r",stdin)
#define FF freopen("output.txt","w",stdout);
int MOD ;
#define n 4
struct Mat{int mat[4][4];
};
//初始化单位矩阵
Mat init(){Mat E;for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){if(i == j)E.mat[i][i] = 1;elseE.mat[i][j] = 0;}}return E;
}
//重载乘法
Mat operator *(Mat a,Mat b){Mat c;memset(c.mat,0,sizeof(Mat));for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){for(int k = 0; k < n; k++){if(a.mat[i][k] && b.mat[k][j]){c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % MOD;}}}}return c;
}
//重载加法
Mat operator +(Mat a,Mat b){Mat c;memset(c.mat,0,sizeof(Mat));for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){c.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % MOD;}}return c;
}
//重载幂次方
Mat operator ^(Mat A,LL x){if(x == 1)return A;Mat c;c = init();for(; x ; x >>= 1){if(x&1){c = c*A;}A = A*A;}return c;
}
int main(){int l,m;int f[4] = {2,4,6,9};Mat gao = {1,0,1,1,1,0,0,0,0,1,0,0,0,0,1,0};while(scanf("%d%d",&l,&m)!=EOF){if(!l){printf("0\n");}elseif(l<=4)printf("%d\n",f[l-1]);else{MOD = m;Mat ans = gao^(l-4);int res = ans.mat[0][0]*9  + ans.mat[0][1]*6 + ans.mat[0][2]*4 + ans.mat[0][3]*2;printf("%d\n",res%m);}}return 0;
}