当要求递推数列的第n项且n很大时,怎么快速求得第n项呢?可以用矩阵快速幂来加速计算.我们可以用矩阵来表示数列递推公式比如fibonacci数列 可以表示为 [f(n)   f(n-1)] = [f(n-1)    f(n-2)] [ 1 1 ]     [ 1 0 ] 设A = [ 1 1 ]  [ 1 0 ] [f(n)   f(n-1)] = [f(n-2)   f(n-3)]*A*A[f(n)   f(n-1)] = [f(2)   f(1)]*A^(n-2)矩阵满足结合律,所以先计算A^…

http://codeforces.com/contest/719/problem/E 题目大意:给你一串数组a,a[i]表示第i个斐波那契数列,有如下操作 ①对[l,r]区间+一个val ②求出[l,r]区间的和. 定义区间的和为该区间内每个a[i]所对应的斐波那契数列的和. 思路:线段树保存区间val,和区间更新,用矩阵快速幂求解复杂度是m*logn*logk //看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h>…

Sequence  Accepts: 59  Submissions: 650  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) Problem Description \ \ \ \    Holion August will eat every thing he has found. \ \ \ \    Now there are many foods,but he does…

graph  Accepts: 9 Submissions: 61  Time Limit: 8000/4000 MS (Java/Others)  Memory Limit: 65536/65536 K (Java/Others) 问题描述 在一个NN个点(标号11~nn),MM条边的有向图上,一开始我在点uu,每一步我会在当前点的出边中等概率的选一条走过去,求走了恰好KK步后走到每个点的概率. 输入描述 第一行两个正整数N,MN,M,表示点数和边数. 接下来MM行,每行两个正整数X,YX,Y…

GTY's math problem Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description GTY is a GodBull who will get an Au in NOI . To have more time to learn alg…

#include<cstdio> #include<string> #include<iostream> #include<vector> #include<set> #include<map> #include<math.h> #include<queue> #include<stdlib.h> #include<cstring> #include<algorithm> u…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec Problem Description Input The input contains a single line consisting of 2 integers N and M (1≤N≤10^18, 2≤M≤100). Output Print one integer, the total n…

题目链接:http://codeforces.com/problemset/problem/678/D 简单的矩阵快速幂模版题 矩阵是这样的: #include <bits/stdc++.h> using namespace std; typedef __int64 LL; struct data { LL mat[][]; }; LL mod = 1e9 + ; data operator *(data a , data b) { data res; ; i <= ; ++i) { ;…

题目链接:http://codeforces.com/problemset/problem/450/B 题意很好懂,矩阵快速幂模版题. /* | 1, -1 | | fn | | 1, 0 | | fn-1 | */ #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef __int64 LL; LL mod = 1e9 + ; struct data {…

https://codeforces.com/contest/1106/problem/F 题意 数列公式为\(f_i=(f^{b_1}_{i-1}*f^{b_2}_{i-2}*...*f^{b_k}_{i-k})\)mod\(P\),给出\(f_{1}...f_{k-1}\)和\(f_{n}\),求\(f_{k}\),其中\(P\)等于998244353 题解 3是998244353的离散对数,所以\(f^{b_1}_{i-1} \equiv 3^{h_i*b_1}(modP)\),怎么求离散…