hdu 1757 矩阵

HDU 1757 题意: If x < 10, f(x) = x; If x >= 10, f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10);  给出k和mod,求f(k). 总结: 1.特别注意,矩阵相乘不满足交换律,即a*b != b*a.  2.感觉推方程有点困难. 3.矩阵初始化注意. f(x-10)   0 0 0 0 0 0 0 0 0        ( first矩阵 )       f(x-9)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1757 题意不难理解,当x小于10的时候,数列f(x)=x,当x大于等于10的时候f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10); 所求的是f(x)取m的模,而x,m,a[0]至a[9]都是输入项 初拿到这道题,最开始想的一般是暴力枚举,通过for循环求出f(x)然后再取模,但是有两个问题,首先f(x)可能特别大,其

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4307    Accepted Submission(s): 2586 Problem Description Lele now is thinking about a simple function f(x). If x < 10 f(x) =

一看正确率这么高,以为是水题可以爽一发,结果是没怎么用过的矩阵快速幂,233 题解链接:点我 1 #include<iostream> 2 #include<cstring> 3 const int N=10; 4 using namespace std; 5 int k,m; 6 struct Matrix{ 7 int map[N][N]; 8 }; 9 10 Matrix matrix; 11 12 void Initiate(){ 13 for(int i=0;i<N

题目地址:HDU 1757 终于会构造矩阵了.其实也不难,只怪自己笨..= =! f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10) 构造的矩阵是:(我代码中构造的矩阵跟这个正好是上下颠倒过来了) |0 1 0 ......... 0|    |f0|   |f1 | |0 0 1 0 ...... 0|    |f1|   |f2 | |...................1| *  |..| = |...|

A Simple Math Problem Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2441    Accepted Submission(s): 1415 Problem Description Lele now is thinking about a simple function f(x).If x < 10 f(x) =

HDU 1575 Tr A Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2912    Accepted Submission(s): 2167 Problem Description A为一个方阵,则Tr A表示A的迹(就是主对角线上各项的和),现要求Tr(A^k)%9973. Input 数据的第一行是一个T,表示有T组数据.每组

Description Lele now is thinking about a simple function f(x). If x < 10 f(x) = x. If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10); And ai(0<=i<=9) can only be 0 or 1 . Now, I will give a0 ~ a9 and two positive in

http://acm.hdu.edu.cn/showproblem.php?pid=4291 凡是取模的都有循环节-----常数有,矩阵也有,而且矩阵的更神奇: g(g(g(n))) mod 109 + 7  最外层MOD=1e9+7  可以算出g(g(n))的循环节222222224,进而算出g(n)的循环节183120LL,然后由内而外计算即可 注释掉的是求循环节的代码 //#pragma comment(linker, "/STACK:102400000,102400000")