Recursive sequence 矩陣快速冪 + 組合數非線性變線性，利用到了組合數（楊輝三角求解快）

阿新 • • 發佈：2019-01-17

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and

. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

Recursive sequence 矩陣快速冪 + 組合數非線性變線性，利用到了組合數（楊輝三角求解快）

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a

【２０１６ICPC 瀋陽onsite C】Recursive sequence(矩陣快速冪)

題面給你一個遞推式 F [ n ]

HDU 5950 - Recursive sequence - [矩陣快速冪加速遞推][2016ACM/ICPC亞洲區瀋陽站 Problem C]

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=5950 Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive

HDU5950-Recursive sequence(矩陣快速冪)

題意：給出n頭母牛，第一頭報a，第二頭報b，第i頭報f[i-2]*2+f[i-1]+i^4,問第n頭母牛報數多少。思路：就是推公式啊，可惜沒經驗，白搭。自己推得時候一直不知道i^4怎麼消去，原來是可以加上i^3,i^2,i，再算的。總之相乘的矩陣必須都是常數，

HDU5950-Recursive sequence(矩陣快速冪）

題意：給出n頭母牛，第一頭報a，第二頭報b，第i頭報f[i-2]*2+f[i-1]+i^4,問第n頭母牛報數多少題目大意：　　Fi=Fi-1+2Fi-2+i4。給定F1和F2求Fn。題目思路：　　【遞推+矩陣快速冪】　　現場用算了1個多小時的公式過了。

Recursive sequence 矩陣快速冪解遞推公式

1. 通常首先能用矩陣快速冪優化的遞推型別是f[n]=5f[n-3]+6f[n-2]+2f[n-1]+n^2+n+8之類的也就是說遞推是線性遞推且f[n-i]前面的係數是常數，可以含有與n有關的多項式，也可以含有常數的這種遞推2.比如以下fn=2fn−2+fn−1+n4通常左

UVA 10689 Yet another Number Sequence 矩陣快速冪水呀水

技術分享 ont truct string esp while .com tdi 快速冪 #include <iostream> #include <cstdio> #include <cstring> #include &

Sequence 矩陣快速冪 + 費馬小定理

f[1] = 1 f[2] = a ^ b 其實不是很好的去想到取log的兩邊同時取log 然後 F[2] = b F[1] = 0 則 f[n] = a^(F[n]) % p 費馬小定理： ① 判斷素數，對於大素數的判定，Miller-R

HDU 6395 Sequence 矩陣快速冪+分塊

Let us define a sequence as below Your job is simple, for each task, you should output Fn module 109+7. Input The first line has only

【HDU1005】Number Sequence(矩陣快速冪)

記錄一個菜逼的成長。。題目連結題目大意： f[1] = 1,f[2] = 1,f[n] = (a*f[n-1]+b*f[n-2])%7(n > 2) 給你a,b。求f[n]。

hdu 5667 Sequence(矩陣快速冪+費馬小定理+快速冪)

#include <cstdio> #include <iostream> #include <cstring> #include <string> #include <cstdlib> #include <algorithm> #inc

HDU 5667 Sequence 矩陣快速冪 + 費馬小定理

olion August will eat every thing he has found. Now there are many foods,but he does not want to eat all of them at once,so he find a

HDU 5667 Sequence (矩陣快速冪+費馬小定理)

Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2687 Acce

HDU 5667 Sequence(矩陣快速冪+費馬小定理)

大意： He gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But there are only p foods,so you should tell him fn mod p. Inpu

hdu 5667 Sequence(矩陣快速冪)

Holion August will eat every thing he has found.Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.fn=⎧⎩⎨⎪⎪1

洛谷2822 組合數問題（楊輝三角）

傳送門【題目分析】不得不說NOIP DAY2還是有點東西啊。。。。。。考慮組合數的計算：，又有，那麼就根據這個打印出前5行組合數：眼熟的感覺。。。。沒錯這就是楊輝三角。。。。。所以在預處理2000*2000的楊輝三角的時候記錄一下當前列有多少為k的倍數，然後直接O

poj 3252 Round Numbers （楊輝三角求組合數）

題目連結：poj 3252 題意：給出範圍為 [a , b] 的區間，問在這區間內的每個數字，假如它的二進位制位0的個數大於1的個數，就說明它是Round Numbers，問你有多少個Round Numbers數？題解：首先楊輝三角求組合數學，見程式碼。 ///此

求組合數取模（楊輝三角打表 & 求逆元（擴充套件歐幾里得、費馬小定理、尤拉定理、線性求法） & Lucas）

在acm競賽中，組合數取模的題目還是經常會見到的，所以這是有必要掌握的一個演算法。我本人就因為這個東西而被坑了很多次了= =之前的部落格也都扯過了，就不多說了，下面進入正題。（1）楊輝三角求組合數楊輝三角這個東西應該都不陌生，三角的兩邊始終為一，之後向

組合數取模（楊輝三角+Lucas定理+模合數）

/* (1) 1 <= m <= n <= 1000 和 1 <= p <= 10^9 ( p可以是任何數 ) 這個問題比較簡單，組合數的計算可以靠楊輝

組合數（楊輝三角）

原來組合數和楊輝三角是有關係的：楊輝三角上的每一個數字都等於它的左上方和右上方的和（除了邊界）第n行，第m個就是，就是C(n, m) （從0開始）所以以後求楊輝三角或者組合數都可以用到下面的遞推公式： #include<cstdio> c