UVA 10518 How Many Calls?】的更多相关文章

题目链接:uva 10518 - How Many Calls? 公式f(n) = 2 * F(n) - 1, F(n)用矩阵快速幂求. #include <stdio.h> #include <string.h> long long n; int b; struct state { int s[2][2]; state(int a = 0, int b = 0, int c = 0, int d = 0) { s[0][0] = a, s[0][1] = b, s[1][0] =…
题意:一个递推式第n项%b是多少. 递推式: 构造矩阵: #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> using namespace std; long long MOD; long long n; struct Matrix { ][]; int R, C; Matrix operator*(Matrix b);…
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…
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Cellular Network Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1456 A cellular network is a radio network made up of a number of cells each served by a base station located in the cell. The base sta…
Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life. A new idea Bill has recently developed assigns a non-negat…
Feel Good Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 16786   Accepted: 4627 Case Time Limit: 1000MS   Special Judge Description Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated…
就是给出非常多点,要求分成两个集合,在同一个集合里的点要求随意两个之间的距离都大于5. 求一个集合.它的点数目是全部可能答案中最少的. 直接从随意一个点爆搜,把它范围内的点都丢到跟它不一样的集合里.不断这样搞即可了. 由于可能有非常多相离的远,把每次搜索得到的那个最小的数目加起来就可以. 因为全部点都格点上,所以仅仅须要枚举一个点可以包括的点是否在数据中存在就可以. 当然也能够用一棵树直接去找.这我并不会.. 时间复杂度是81nlogn 湖大的OJ机器太老.. .还要开栈. . .UVA LIV…