HDU5015 233 Matrix(矩阵高速幂) 题目链接 题目大意: 给出n∗m矩阵,给出第一行a01, a02, a03 ...a0m (各自是233, 2333, 23333...), 再给定第一列a10, a10, a10, a10,...an0.矩阵中的每一个元素等于左边的加上上面的,求出anm. 解题思路: 先要依据矩阵元素的特征得出相乘的矩阵T, 然后就是求这个矩阵T的m次幂(这里就能够用矩阵高速幂),最后再和给定的第一列所形成的矩阵相乘,就能得到anm. 求矩阵T请參考 代码:…

题目链接:https://vjudge.net/problem/HDU-5015 233 Matrix Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2805    Accepted Submission(s): 1611 Problem Description In our daily life we often use 233 t…

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333...…

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333...…

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333...…

题意:给出矩阵的第0行(233,2333,23333,...)和第0列a1,a2,...an(n<=10,m<=10^9),给出式子: A[i][j] = A[i-1][j] + A[i][j-1],要求A[n][m]. 解法:看到n<=10和m<=10^9 应该对矩阵有些想法,现在我们假设要求A[a][b],则A[a][b] = A[a][b-1] + A[a-1][b] = A[a][b-1] + A[a-1][b-1] + A[a-2][b] = ... 这样相当于右图:,红…

HDU 4965 Fast Matrix Calculation 题目链接 矩阵相乘为AxBxAxB...乘nn次.能够变成Ax(BxAxBxA...)xB,中间乘n n - 1次,这样中间的矩阵一个仅仅有6x6.就能够用矩阵高速幂搞了 代码: #include <cstdio> #include <cstring> const int N = 1005; const int M = 10; int n, m; int A[N][M], B[M][N], C[M][M], CC[N…

[HDU5015]233 Matrix 试题描述 In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it woul…

http://acm.hdu.edu.cn/showproblem.php?pid=3221 一晚上搞出来这么一道题..Mark. 给出这么一个程序.问funny函数调用了多少次. 我们定义数组为所求:f[1] = a,f[2] = b, f[3] = f[2]*f[3]......f[n] = f[n-1]*f[n-2].相应的值表示也可为a^1*b^0%p.a^0*b^1%p,a^1*b^1%p,.....a^fib[n-3]*b^fib[n-2]%p.即a,b的指数从n=3以后与fib数列…

题目地址:HDU 1575 矩阵高速幂裸题. 初学矩阵高速幂.曾经学过高速幂.今天一看矩阵高速幂,原来其原理是一样的,这就好办多了.都是利用二分的思想不断的乘.仅仅只是把数字变成了矩阵而已. 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #i…