Lanterns Problem Description   Alice has received a beautiful present from Bob. The present contains n lanterns and m switches. Each switch controls some lanterns and pushing the switch will change the state of all lanterns it controls from off to on…

Lanterns 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/B 题目: Description Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with the beginning of the street corr…

B. Vanya and Lanterns time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with t…

Codeforces  492B   B. Vanya and Lanterns 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/B 题目: Description Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with…

Description Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point l. Then the i-th lanter…

题目链接:http://codeforces.com/problemset/problem/492/B #include <cstdio> #include <cstdlib> #include <iostream> #include <algorithm> #include <cstring> using namespace std; + ; double a[maxn]; double l[maxn], r[maxn]; int main()…

求矩阵的秩,及判断有无解 #include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include&l…

题意:有n个灯和m个开关,每个开关控制数个灯的状态改变,给出k条询问,问使灯的状态变为询问中的状态有多少种发法. 析:同余高斯消元法,模板题,将每个开关控制每个灯列成行列式,最终状态是结果列,同余高斯消元,如果无解就是0,否则结果就是1<<(自由变元的个数); 代码如下: #include<stdio.h> #include<algorithm> #include<iostream> #include<string.h> #include<…

题意:有n个灯笼,m个开关 每个开关可以控制k个灯笼, 然后分别列出控制的灯笼的编号(灯笼编号为1到n) 下面有Q个询问,每个询问会有一个最终状态(n个灯笼为一个状态)0代表关 1代表开 问到达这种状态,按开关的方法总数 解释一下案例: 3 2     (有3个灯笼, 2个开关) 2 1 2  (第一个开关控制2个灯笼,这两个灯笼的编号是1.2) 2 1 3  (第一个开关控制2个灯笼,这两个灯笼的编号是1.3) 2        (2个询问) 0 1 1  (这3个灯笼变为关.开.开 有几种按…

基本思路.首先构造一个n*(m+1)的矩阵,同时标记一个行数row,row从零开始,然后找出每一列第一个非零的数,和第row行互换, 然后对row到n行,异或运算.最终的结果为2^(m-row) #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int array[55][55],n,m,h[55][55]; int main() { int i,j,k,t,a,q…