根据题意:最后一步是寻求f(b) + f(k + b) + f(2 * k + b) + -+ f((n-1) * k + b) 清除f(b) = A^b 间A = 1 1 1 0 所以sum(n - 1) = A^b(E + A^ k + A ^(2 * k) + - + A ^((n - 1) * k) 设D = A^k sum(n-1) = A^b(E + D + D ^ 2 + - + D ^(n - 1)) 括号中的部分就能够二分递归求出来了 而单个矩阵就能够用矩阵高速幂求出来 /**…

Gauss Fibonacci Time Limit: 3000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 27    Accepted Submission(s): 5 Problem Description Without expecting, Angel replied quickly.She says: "I'v heard that you'r a ve…

题目:Matrix Power Series 传送门:http://poj.org/problem?id=3233 分析: 方法一:引用Matrix67大佬的矩阵十题:这道题两次二分,相当经典.首先我们知道,A^i可以二分求出.然后我们需要对整个题目的数据规模k进行二分.比如,当k=6时,有:$ S(6)= A + A^2 + A^3 + A^4 + A^5 + A^6 =\underline{(A + A^2 + A^3)} + A^3*\underline{(A + A^2 + A^3)}.…

Description Without expecting, Angel replied quickly.She says: "I'v heard that you'r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. " How good an opportunity that Gardon can not give up! The "Prob…

http://acm.hdu.edu.cn/showproblem.php?pid=1588 Problem Description Without expecting, Angel replied quickly.She says: "I'v heard that you'r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. " How good an…

HDU 1588 Gauss Fibonacci(矩阵高速幂+二分等比序列求和) ACM 题目地址:HDU 1588 Gauss Fibonacci 题意:  g(i)=k*i+b;i为变量.  给出k,b,n,M,问( f(g(0)) + f(g(1)) + ... + f(g(n)) ) % M的值. 分析:  把斐波那契的矩阵带进去,会发现这个是个等比序列. 推倒: S(g(i)) = F(b) + F(b+k) + F(b+2k) + .... + F(b+nk) // 设 A = {1…

Gauss Fibonacci Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1706    Accepted Submission(s): 741 Problem Description Without expecting, Angel replied quickly.She says: "I'v heard that you'r…

题目的大意就是求等差数列对应的Fibonacci数值的和,容易知道Fibonacci对应的矩阵为[1,1,1,0],因为题目中f[0]=0,f[1]=1,所以推出最后结果f[n]=(A^n-1).a,所以 f(g(i))= f(k*i+b)= (A^(k*i+b-1)).a,i从 0取到 n-1,取出公因式 A^(b-1)(因为矩阵满足分配率),然后所求结果可化为 A^(b-1) * (A^0 + A^k + A^2k +....+ A^(n-1)k),化到这里后难点就是求和了,一开始我尝试暴力…

欢迎访问~原文出处——博客园-zhouzhendong 去博客园看该题解 题目传送门 - BZOJ3286 题意概括 n,m,a,b,c,d,e,f<=10^1000000 题解 神奇的卡常题目. 在此感谢"zhouzixuan"——bzoj 3286: Fibonacci矩阵 学习他,才15秒卡过此题. 这题的做法应该很明显的,学过矩阵快速幂的大概几眼就看出来了. 对于每一行的转移,是相同的,所以矩阵快速幂可以搞定行与行之间的转移. 然后对于某一行,其实大部分的转移是和abc有…

对每个0<=i<n求f(g(i))的和,其中f(x)为斐波那契数列第x项,g(i)=k*i+b,k,b,n给定,模数给定. 斐波那契数有一种用矩阵乘法求的方法,这个矩阵A自己写,令F[i]为i和i+1的那个矩阵,F[i]=A^b*F[0],然后答案要求F[b]+F[k+b]+F[k*2+b]+……=(A^b+A^(k+b)+A^(2k+b)+……)*F[0]=(E+A^k+……+A^k^(n-1))*A^b*F[0]的[2,1]项.上面括号里就令B=A^k求E+B+B^2+……+B^(n-1)…