链接:http://acm.hdu.edu.cn/showproblem.php?pid=5318 The Goddess Of The Moon Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 438    Accepted Submission(s): 150 Problem Description Chang'e (嫦娥) is…

The Goddess Of The Moon Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 589    Accepted Submission(s): 251 Problem Description Chang’e (嫦娥) is a well-known character in Chinese ancient mythology…

The Goddess Of The Moon Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 943    Accepted Submission(s): 426 Problem Description Chang’e (嫦娥) is a well-known character in Chinese ancient mythology…

题目链接 题意 :给你m和k, 让你求f(k)%m.如果k<10,f(k) = k,否则 f(k) = a0 * f(k-1) + a1 * f(k-2) + a2 * f(k-3) + …… + a9 * f(k-10);思路 :先具体介绍一下矩阵快速幂吧,刚好刚刚整理了网上的资料.可以先了解一下这个是干嘛的,怎么用. 这个怎么弄出来的我就不说了,直接看链接吧,这实在不是我强项,点这儿,这儿也行 //HDU 1757 #include <iostream> #include <s…

Recursive sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 249    Accepted Submission(s): 140 Problem Description Farmer John likes to play mathematics games with his N cows. Recently, t…

题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=6030 Problem Description Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads. Little Q desperately wants to impress his girlfriend,…

http://acm.hdu.edu.cn/showproblem.php?pid=2604 这题居然O(9 * L)的dp过不了,TLE,  更重要的是找出规律后,O(n)递推也过不了,TLE,一定要矩阵快速幂.然后立马GG. 用2代表m,1代表f.设dp[i][j][k]表示,在第i位,上一位站了的人是j,这一位站的人是k,的合法情况. 递推过去就是,如果j是1,k是2,那么这一位就只能放一个2,这个时猴dp[i][k][2] += dp[i - 1][j][k]; 其他情况分类下就好,然后…

分析:假设g(g(g(n)))=g(x),x可能非常大,但是由于mod 10^9+7,所以可以求出x的循环节 求出x的循环节后,假设g(g(g(n)))=g(x)=g(g(y)),即x=g(y),y也可能非常大,但是由x的循环节可以求出y的循环节 所以最终结果只要进行矩阵快速幂即可求出 循环节 #include<stdio.h> ;//第一次是MOD=1000000007 找出循环节是222222224 //第二次是MOD=222222224,找出循环节183120 int main() {…

题目 也是和LightOJ 1096 和LightOJ 1065 差不多的简单题目. #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int num,mod; struct matrix { ][]; }; matrix multiply(matrix x,matrix y)//矩阵乘法 { matrix temp; ;i<num;i++) { ;j<…

题目链接 Problem Description Function Fx,ysatisfies: For given integers N and M,calculate Fm,1 modulo 1e9+7. Input There is one integer T in the first line. The next T lines,each line includes two integers N and M . 1<=T<=10000,1<=N,M<2^63. Output…