HDU 2604_递推+矩阵快速幂

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我是靠谱客的博主甜美柠檬，最近开发中收集的这篇文章主要介绍HDU 2604_递推+矩阵快速幂，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5546 Accepted Submission(s): 2414

Problem Description

Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.

Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2 ^L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.

Input

Input a length L (0 <= L <= 10 ⁶) and M.

Output

Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.

Sample Input

Sample Output

Author

WhereIsHeroFrom

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
using namespace std;  
#include <cmath>  
#include <cstdlib>  
typedef long long ll;  
  
const int N = 0;  
const int SIZE = 4;  
  
int l, MOD;  
  
struct Mat{  
    ll v[SIZE][SIZE];   // value of matrix  
  
    Mat() {  
        memset(v, 0, sizeof(v));  
    }  
  
    void init(ll _v) {  
    	for(int i=0;i<4;i++){
    		v[i][i]=_v;
		}
	}  
};  
  
Mat operator * (Mat a, Mat b) {  
    Mat c;  
    for(int i=0;i<4;i++) {  
        for(int j=0;j<4;j++) {  
           c.v[i][j] = 0;  
            for(int k=0;k<4;k++){  
                c.v[i][j] += (a.v[i][k] * b.v[k][j]) % MOD;  
                c.v[i][j] %= MOD;  
            }  
        }  
    }  
    return c;  
}  
  
Mat operator ^ (Mat a, ll k) {  
    Mat c;  
    c.init(1);  
    while (k) {  
        if (k&1) c = a * c;  
        a = a * a;  
        k >>= 1;  
    }  
    return c;  
}
Mat mul(Mat a,ll k ){
	Mat tmp;
	int i; 
	for(i=0;i<4;i++){
		tmp.v[i][i]=1;
	}
  while(k){
	if(k&1){
		tmp=(tmp*a);
	}
		a=a*a;
		k>>=1;
	}
	return tmp;
}
int main() {  
    Mat a, b, c;  
    // a  
    a.v[0][0] = 9;  
    a.v[1][0] = 6;  
    a.v[2][0] = 4;  
    a.v[3][0] = 2;  
      
    // b  
    b.v[0][0] = b.v[0][2] = b.v[0][3] = b.v[1][0] = b.v[2][1] = b.v[3][2] = 1;  
  //初始化地推矩阵 
    while (scanf("%d%d", &l, &MOD)==2) {  
        if (l == 0) {  
            puts("0");  
        } else if (l <= 4) {  
            printf("%lldn", a.v[4 - l][0] % MOD);  
        } else {  
            c =mul(b,(l-4));  
            c = c * a;  
            printf("%lldn", c.v[0][0] % MOD);  
        }  
    }  
  
    return 0;  
}