http://www.lightoj.com/volume_showproblem.php?problem=1070 思路:\({(a+b)}^n =(a+b){(a+b)}^{n-1} \) \((ab)C_{n}^{r}a^{n-r}b{r} = C_{n+2}^{r}a^{n-r+2}b{r} - a^{n+2} - b^{n+2} \) 综上\( f(n) = (a+b)f(n-1)-(ab)f(n-2) \) /** @Date : 2016-12-19-19.53 * @Author…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1070 题意: 给你a+b和ab的值,给定一个n,让你求a^n + b^n的值(MOD 2^64). 题解: a + b也就是a^1 + b^1,然后要从这儿一直推到a^n + b^n. 矩阵快速幂?o( ̄▽ ̄)d 那么主要解决的就是如何从a^n + b^n推到a^(n+1) + b^(n+1). 下面是推导过程: 由于推a^(n+1) + b^(n+1)要用到a^n + b^n和a^…
题目链接:problem=1070">LightOJ 1070 Algebraic Problem 题意:已知a+b和ab的值求a^n+b^n.结果模2^64. 思路: 1.找递推式 watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt=""> 得到递推式之后就是矩…
题链:http://lightoj.com/volume_showproblem.php?problem=1070 1070 - Algebraic Problem PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Given the value of a+b and ab you will have to find the value of an+bn. a and b not necessar…
题目地址:HDU 1757 最终会构造矩阵了.事实上也不难,仅仅怪自己笨..= =! f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + -- + a9 * f(x-10) 构造的矩阵是:(我代码中构造的矩阵跟这个正好是上下颠倒过来了) |0 1 0 ......... 0|    |f0|   |f1 | |0 0 1 0 ...... 0|    |f1|   |f2 | |...................1| *  |..| = |...…
和这一题构造的矩阵的方法同样. 须要注意的是.题目中a0~a9 与矩阵相乘的顺序. #include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <cstring> #define N 10 using namespace std; int mod; typedef long long LL; struct matrix { LL a[10…
题目大意:求(1^K + 2^K + 3K + - + N^K) % 2^32 解题思路: 借用别人的图 能够先打表,求出Cnm,用杨辉三角能够高速得到 #include<cstdio> typedef unsigned long long ll; const int N = 55; const ll mod = (1LL << 32); struct Matrix{ ll mat[N][N]; }A, B, tmp; ll n, num[N]; ll C[N][N]; int K…
题目地址:HDU 2256 思路: (sqrt(2)+sqrt(3))^2*n=(5+2*sqrt(6))^n; 这时要注意到(5+2*sqrt(6))^n总能够表示成an+bn*sqrt(6); an+bn*(sqrt(6))=(5+2*sqrt(6))*(a(n-1)+b(n-1)*sqrt(6)) =(5*a(n-1)+12*b(n-1))+(2*a(n-1)+5*b(n-1))*sqrt(6); 显然,an=5*a(n-1)+12*b(n-1);bn=2*a(n-1)+5*b(n-1);…
Cellular Automaton Time Limit: 12000MS   Memory Limit: 65536K Total Submissions: 3048   Accepted: 1227 Case Time Limit: 2000MS Description A cellular automaton is a collection of cells on a grid of specified shape that evolves through a number of dis…
矩阵高速幂: 依据关系够建矩阵 , 高速幂解决. Arc of Dream Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 2164    Accepted Submission(s): 680 Problem Description An Arc of Dream is a curve defined by following fun…