题意:给你一个n,输出Fibonacci (n)%10000的结果 思路:裸矩阵快速幂乘,直接套模板 代码: #include <cstdio> #include <cstring> #include <iostream> using namespace std; typedef long long ll; ,M=,P=; ; struct Matrix { ll m[N][N]; }; Matrix A={,, ,}; Matrix I={,, ,}; Matrix…

Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … An alternative formula for the Fibonacci sequence is…

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18607   Accepted: 12920 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequen…

题意: 求出斐波那契数列的第n项的后四位数字 思路:f[n]=f[n-1]+f[n-2]递推可得二阶行列式,求第n项则是这个矩阵的n次幂,所以有矩阵快速幂模板,二阶行列式相乘, sum[ i ] [ j ]+=v[i][k]*u[k][j] ___k==1->j: 模板: #include<stdio.h> #include<map> using namespace std; //矩阵快速幂 struct node { int m[10][10]; }a,b; node ju…

Gauss Fibonacci Time Limit: 3000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 27    Accepted Submission(s): 5 Problem Description Without expecting, Angel replied quickly.She says: "I'v heard that you'r a ve…

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12329   Accepted: 8748 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequenc…

                             Modular Fibonacci The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:F0 = 0F1 = 1Fi = Fi−1 + Fi−2 for i > 1Write a program which calculates Mn = Fn mod 2m for given pair of n and…

搞懂了什么是矩阵快速幂优化.... 这道题的重点不是DP. /* 题意: 小明要走某条路,按照个人兴致,向前走一步的概率是p,向前跳两步的概率是1-p,但是地上有地雷,给了地雷的x坐标,(一维),求小明安全到达最后的概率. 思路: 把路分成好多段,小明安全走完每一段的概率乘起来就是答案. dp[i]=p*dp[i-1]+(1-p)*dp[i-2]; 参考fib数列构造矩阵进行快速幂. 注意初始化的时候,起点概率看作1,起点减一也就是有地雷的地方概率看作0.//屌丝一开始在这里没搞明白. */ #…

学了线代之后 终于明白了矩阵的乘法.. 于是 第一道矩阵快速幂.. 实在是太水了... 这差不多是个模板了 #include <cstdlib> #include <cstring> #include <cstdio> #include <iostream> using namespace std; int N; struct matrix { int a[3][3]; }origin,res; matrix multiply(matrix x,matrix…

设人数为 $n$,构造 $(n + 1) \times (n + 1)$ 的矩阵 得花生:将改行的最后一列元素 $+ 1$ \begin{gather}\begin{bmatrix}1 & 0 & 0 & 1 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{bmatrix}\times\begin{bmatrix}x \\y \\z \\1 \\\end{…