Number Sequence  HDU-1005 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 162730    Accepted Submission(s): 40016 Problem Description A number sequence is defined as follows:f(1) = 1, f(2) = 1,…

//Accepted hdu1005 0MS 248K #include <cstdio> #include <cstring> #include <iostream> #include <queue> #include <cmath> #include <algorithm> using namespace std; /** * This is a documentation comment block * 如果有一天你坚持不下去了…

矩阵快速幂是基于普通的快速幂的一种扩展,如果不知道的快速幂的请参见http://www.cnblogs.com/Howe-Young/p/4097277.html.二进制这个东西太神奇了,好多优秀的算法都跟他有关系,这里所说的矩阵快速幂就是把原来普通快速幂的数换成了矩阵而已,只不过重载了一下运算符*就可以了,也就是矩阵的乘法,  当然也可以写成函数,标题中的这三个题都是关于矩阵快速幂的基础题.拿来练习练习熟悉矩阵快速幂,然后再做比较难点的,其实矩阵快速幂比较难的是构造矩阵.下面还是那题目直接说话…

http://acm.hdu.edu.cn/showproblem.php?pid=1005 1.一开始就注意到了n的数据范围 <=100 000 000,但是还是用普通的循环做的,自然TLE了,然后朴素打表,= =运行不了,(怎么可能能把数组开到那么大).再然后就想到了寻找下一个1 1 连续在一起的,那就能开始下一轮循环了. 但是,各种WA--(将数组开大一点,寻找到a[ i ] = a[ i -1 ] ==1 即跳出),这个AC代码将102改成100,150,200都可以,但是108,49…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1005 #include<cstdio> using namespace std; int n,a,b; struct Matrix { int m[2][2]; void init() { m[0][0] = a; m[0][1] = 1; m[1][0] = b; m[1][1] = 0; } void init0() { m[0][0] = a+b; m[0][1] = 1; m[1][0]…

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 147161    Accepted Submission(s): 35755 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A…

Number Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 86102 Accepted Submission(s): 20423 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n…

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…

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 contains…

主题链接: pid=1005">huangjing 题意: 就是给了一个公式,然后求出第n项是多少... 思路: 题目中n的范围实在是太大,所以肯定直接递推肯定会超时,所以想到的是暴力打表,找循环节,可是也不是那么easy发现啊,所以这时候分析一下,由于最后都会mod7,所以总共同拥有7X7总情况,即A 0,1,2,3,4,5,6,7.B也是如此,所以循环节为49,这么这个问题就攻克了. .. 题目: Number Sequence Time Limit: 2000/1000 MS (Ja…