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…

Mathematician QSC Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 229    Accepted Submission(s): 114 Problem Description QSC dream of becoming a mathematician, he believes that everything in t…

题目链接: Hdu  5451  Best Solver 题目描述: 对于,给出x和mod,求y向下取整后取余mod的值为多少? 解题思路: x的取值为[1, 232],看到这个指数,我的心情是异常崩溃的.(吐血......) 可是仔细观察,它指数大,可是mod小啊,它吓人,可是可以暴力搞啊!! 这个题目一个难点就是要向下取整求余,详解见传送门,本题是向下取整,也就是向上取整加一. 还有就是指数太大,要找到循环节,其实由于mod小,循环节并没有太大,暴力跑就ok啦!  此刻内心是崩溃的 #inc…

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,并且矩阵的更奇妙: g(g(g(n))) mod 109 + 7  最外层MOD=1e9+7  能够算出g(g(n))的循环节222222224.进而算出g(n)的循环节183120LL.然后由内而外计算就可以 凝视掉的是求循环节的代码 //#pragma comment(linker, "/STACK:102400000,102400000")…

题意:求 [(5 + 2*sqrt(6))^(1 + 2^x)]  % M 基于hdu2256可以求(5 + 2*sqrt(6))^ n 但是n特别大,我们可以找矩阵的循环节 两种可能 1.mod-1      2. (mod+1) * (mod-1)    /*(具体ACdreamers的广义裴波那切找循环节) 在知道2256时,自己做了一遍,但是到时想到的是费马小定理(gg) p.s  广义Fibonacci数和循环节方面还是不明白,找机会看看 #include <iostream> #i…

http://acm.hdu.edu.cn/showproblem.php?pid=5451 题意:给定x    求解 思路: 由斐波那契数列的两种表示方法, 之后可以转化为 线性表示 F[n] = F[n-1] + F[n-2] ; 同时可以看出   和 是 一元二次方程的两根, a  = 1, b = -1 又是之后递推式的系数: 同理这里需要构造出两根为 和 ,这时 a = 1, b = –10 得 F[n] = 10F[n-1] – F[n-2]; (当然可以直接打表递推出关系式) 如果…

Fibonacci Check-up Problem Description Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted? Of course, we can number them one by…

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…

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…

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…