题目不难懂.式子是一个递推式,并且不难发现f[n]都是a的整数次幂.(f[1]=a0;f[2]=ab;f[3]=ab*f[2]c*f[1]...) 我们先只看指数部分,设h[n]. 则 h[1]=0; h[2]=b; h[3]=b+h[2]*c+h[1]; h[n]=b+h[n-1]*c+h[n-1]. h[n]式三个数之和的递推式,所以就可以转化为3x3的矩阵与3x1的矩阵相乘.于是 h[n] c  1  b h[n-1] h[n-1] = 1  0  0 * h[n-2] 1       0…

官方题解: 观察递推式我们可以发现,所有的fi​​都是a的幂次,所以我们可以对f​i​​取一个以a为底的log,g​i​​=log​a​​ f​i​​ 那么递推式变g​i​​=b+c∗g​i−1​​+g​i−2​​,这个式子可以矩阵乘法 这题有一个小trick,注意a mod p=0的情况. 分析:排除了a mod p=0的情况,幂次可以对(p-1)取模,这是由于离散对数定理 相关定理请查阅 算导 吐槽:比赛的时候就是被a mod p=0这种情况给hack掉了,我太弱了 #include <st…

HDU 5667 构造矩阵快速幂 题目描述 解析 我们根据递推公式 设 则可得到Q的指数关系式 求Q构造矩阵 同时有公式 其中φ为欧拉函数,且当p为质数时有 代码 #include <cstdio> using namespace std; long long pow_mod(long long a, long long p, long long mod) { if (p == 0) return 1; long long ans = pow_mod(a, p / 2, mod); ans =…

HDU.2640 Queuing (矩阵快速幂) 题意分析 不妨令f为1,m为0,那么题目的意思为,求长度为n的01序列,求其中不含111或者101这样串的个数对M取模的值. 用F(n)表示串长为n的合法串的个数. 首先不难通过枚举发现F(0) = 0, F(1) =2, F(3) = 6, F(4) = 9, F(5) = 15.然后引用网上如何求解递推公式的详细解释: 用f(n)表示n个人满足条件的结果,那么如果最后一个人是m的话,那么前n-1个满足条件即可,就是f(n-1): 如果最后一个…

HDU - 1005 Number Sequence Problem Description A number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5950 Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers…

Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multiple test cases. Each test case…

A number sequence is defined as follows:  f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.  Given A, B, and n, you are to calculate the value of f(n). Input The input consists of multiple test cases. Each test case contains 3 integers…

题目链接:https://vjudge.net/problem/HDU-5950 思路: 构造矩阵,然后利用矩阵快速幂. 1 #include <bits/stdc++.h> 2 #include <time.h> 3 #include <set> 4 #include <map> 5 #include <stack> 6 #include <cmath> 7 #include <queue> 8 #include <…

题目链接: Sequence 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 not want to eat all of them at once,so he fi…