题意:已知X,数组arr[n],求一个分式的分子与分母的最大公因数.分子为ΣX^arr[i],分母为X^Σarr[i],数组为不递减序列. 思路:比赛的时候以为想出了正确思路,WA掉了很多发,看了别人写的代码才发现自己漏掉很多细节. 1.容易想到,分子的最低次幂即可能为所需答案 2.由于arr里存在相同的数,因此分子的各个幂存在可以合并同类项的情况,所以应该先彻底完成合并同类项,再进行步骤1. 3.一个小技巧,并不需要完成所有项的合并,当且仅当当前最小项的系数可以被X整除时才需要继续合并,否则当…

Simon has a prime number x and an array of non-negative integers a1, a2, ..., an. Simon loves fractions very much. Today he wrote out number  on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a frac…

题目链接 Eugene and big number 题目转化为 $f(n) = m * f(n - 1) + a$ $f(n + 1) = m * f(n) + a$ 两式相减得 $f(n + 1) = (m + 1) * f(n) - m * f(n - 1)$ 求$f(n)$ 其中$m$为$10^{k}$ ($k$为$a$的位数) 那么利用矩阵快速幂加速就可以了. #include <bits/stdc++.h> using namespace std; #define rep(i, a…

Let’s define another number sequence, given by the following function: f(0) = a f(1) = b f(n) = f(n − 1) + f(n − 2), n > 1 When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and b, you can get many different se…

                  Yet another Number Sequence Let’s define another number sequence, given by the following function:f(0) = af(1) = bf(n) = f(n − 1) + f(n − 2), n > 1When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values…

Description Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation: F1 = 1, F2 = 2, Fi = Fi - 1 + Fi - 2 (i > 2). We'll define a new number sequence Ai(k) by the formula: Ai(k) = Fi × ik (i ≥ 1). In thi…

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…

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…

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…

一开始找矩阵快速幂的题来做时就看到了这题,题意就是让你求出如图所示的第n个三角形中指向向上的小三角形个数.从图中已经很容易看出递推关系了,我们以f[n]表示第n个大三角形中upward的小三角形个数,g[n]表示第n个大三角形中downward的小三角形个数,然后,递推关系就是: f[n]= 3*f[n-1]+1*g[n-1]; (1) g[n]= 3*g[n-1]+1*f[n-1];  (2) 其中f[0]= 1,g[0]= 0(一开始的纯三角形是从n=0开始的),然后……就没有然后了,直觉上…