UVA 11551 Experienced Endeavour】的更多相关文章

UVA 11551 - Experienced Endeavour 题目链接 题意:给定一列数,每一个数相应一个变换.变换为原先数列一些位置相加起来的和,问r次变换后的序列是多少 思路:矩阵高速幂,要加的位置值为1.其余位置为0构造出矩阵,进行高速幂就可以 代码: #include <cstdio> #include <cstring> const int N = 55; int t, n, r, a[N]; struct mat { int v[N][N]; mat() {mem…
矩阵快速幂. 题意事实上已经告诉我们这是一个矩阵乘法的运算过程. 构造矩阵:把xi列的bij都标为1. 例如样例二: #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; ; int n, m; + ]; struct Matrix { + ][ + ]; int R, C; Matr…
题目链接:https://vjudge.net/problem/UVA-11551 题意: 给定一列数,每个数对应一个变换,变换为原先数列一些位置相加起来的和,问r次变换后的序列是多少 题解: 构造矩阵:要加的位置值为1,其余位置为0.然后用快速幂计算. 代码如下: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector…
Alice is given a list of integers by Bob and is asked to generate a new list where each element in the new list is the sum of some other integers in the original list. The task is slightly more involved, as Bob also asks Alice to repeat this several…
KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…
[kuangbin带你飞]专题1-23 专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil Deposit…
专题一 简单搜索 POJ 1321 棋盘问题POJ 2251 Dungeon MasterPOJ 3278 Catch That CowPOJ 3279 FliptilePOJ 1426 Find The MultiplePOJ 3126 Prime PathPOJ 3087 Shuffle'm UpPOJ 3414 PotsFZU 2150 Fire GameUVA 11624 Fire!POJ 3984 迷宫问题HDU 1241 Oil DepositsHDU 1495 非常可乐HDU 26…
Description Misha trains several ACM teams at the university. He is an experienced coach, and he does not underestimate the meaning of friendly and collaborative atmosphere during training sessions. It used to be that way, but one of the teams happen…
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UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…