【遞推&矩陣快速冪】hdu 2604

前四位我們可以算出d[1]=2,d[2]=4,d[3]=6,d[4]=9. 
 
我們可以這樣想：一個合法串可以由兩個較短的合法串組成 
 
就以d[n]為例：(注意不能重複) 
 
1、n-1個字元的時候: +m 
 
2、n-2: 只能+mm，會和n-1重複,所以不考慮n-2 
 
3、n-3: +mmf 
 
4、n-4: +mmff 
 
5、n-5: 如果是+mmffm，會與n-1重複,+mmmff會與n-4重複,+mmmmf會與n-3重複(不考慮) 
 
所以就考慮n-1,n-3,n-4，DP等式就出來了:dp[n]=dp[n-1]+dp[n-3]+dp[n-4] 
 
矩陣快速冪

#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 4struct Mat{    int mat[4][4];};//初始化單位矩陣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;            else            E.mat[i][j] = 0;        }    }    return E;}//重載乘法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;}//重載加法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;}//重載冪次方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");        }        else        if(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;}