【SRM 717 div2 A】 NiceTable
Problem Statement
You are given a vector t that describes a rectangular table of zeroes and ones. Each character in t is either ‘0’ or ‘1’.
We say that a table is nice if there are two sequences x, y of zeroes and ones such that for each valid pair of indices i, j we have t[i][j] = x[i] xor y[j].
Some technical remarks:
The number of elements in x should be equal to the number of rows of the table, i.e., the number of elements in t.
The number of elements in y should be equal to the number of columns of the table, i.e., the length of each string in t.
The operation xor (exclusive or) is defined as follows: 0 xor 0 = 0, 1 xor 1 = 0, 0 xor 1 = 1, and 1 xor 0 = 1.
Verify whether the given table is nice. Return “Nice” if it is nice and “Not nice” otherwise. Note that the return value is case-sensitive.
Definition
Class:
NiceTable
Method:
isNice
Parameters:
vector
Returns:
string
Method signature:
string isNice(vector t)
(be sure your method is public)
Limits
Time limit (s):
2.000
Memory limit (MB):
512
Stack limit (MB):
512
Constraints
t will contain between 1 and 5 elements, inclusive.
Each element of t will contain between 1 and 5 characters, inclusive.
All elements of t will contain the same number of characters.
Each element of t will consist only of characters ‘0’ and ‘1’.
Examples
0)
{“01”,
“10”}
Returns: “Nice”
One valid choice is to choose x = y = {1, 0}.
1)
{“01”,
“11”}
Returns: “Not nice”
Assume that t is nice. The sequences x and y have to satisfy the following constraints:
x[0] xor y[0] = 0
x[0] xor y[1] = 1
x[1] xor y[0] = 1
x[1] xor y[1] = 1
From the first constraint we see that x[0] = y[0]. From the second and the third constraint we can then derive that we must also have x[1] = y[1]. But then the fourth constraint is not satisfied, which is a contradiction. Therefore, this t is not nice.
2)
{“0100”,
“1011”,
“0100”}
Returns: “Nice”
Here, one valid choice is x = {1, 0, 1} and y = {1, 0, 1, 1}.
3)
{“11”,
“10”,
“11”,
“11”,
“11”}
Returns: “Not nice”
【题目链接】:
【题意】
给你一个n*m的矩阵a;
然后问你是否存在一个长度为n的序列x和长度为m的序列y;
使得s[i]^s[j]=a[i][j]对于所有的i,j皆成立;
n,m<=5
a[i][j]中只会出现0和1
【题解】
暴力枚举x,y的每一位是0还是1就好;
【Number Of WA】
0
【反思】
一开始想错了,以为让x全都是0就好;
后来想,或许可以让x的第一位是0,然后其他位先不定;
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;
//head
int x[8],y[8],hang,lie,ok;
vector <string> a;
void dfs2(int now){
if (now > lie-1){
int temp = 1;
rep1(i,0,hang-1){
rep1(j,0,lie-1){
int k = a[i][j]-'0';
if (k!=(x[i]^y[j]))
temp = 0;
}
}
if (temp) ok = 1;
return;
}
y[now] = 0;
dfs2(now+1);
y[now] = 1;
dfs2(now+1);
}
void dfs1(int now){
if (now > hang-1){
dfs2(0);
return;
}
x[now] = 0;
dfs1(now+1);
x[now] = 1;
dfs1(now+1);
}
class NiceTable
{
public:
string isNice(vector <string> t)
{
a = t;
hang = t.size(),lie = t[0].size();
ok = 0;
dfs1(0);
if (ok)
return "Nice";
else
return "Not nice";
}
};
【SRM 717 div2 A】 NiceTable的更多相关文章
- 【SRM 717 DIV2 C】DerangementsDiv2
Problem Statement You are given two ints: n and m. Let D be the number of permutations of the set {1 ...
- 【SRM 717 div2 B】LexmaxReplace
Problem Statement Alice has a string s of lowercase letters. The string is written on a wall. Alice ...
- 【Codeforces #312 div2 A】Lala Land and Apple Trees
# [Codeforces #312 div2 A]Lala Land and Apple Trees 首先,此题的大意是在一条坐标轴上,有\(n\)个点,每个点的权值为\(a_{i}\),第一次从原 ...
- 【TP SRM 703 div2 250】AlternatingString
Problem Statement A string of zeros and ones is called an alternating string if no two adjacent char ...
- 【TP SRM 703 div2 500】 GCDGraph
Problem Statement You are given four ints: n, k, x, and y. The ints n and k describe a simple undire ...
- 【cf 483 div2 -C】Finite or not?(数论)
链接:http://codeforces.com/contest/984/problem/C 题意 三个数p, q, b, 求p/q在b进制下小数点后是否是有限位. 思路 题意转化为是否q|p*b^x ...
- 【市场调研与分析】Intel发力移动安全领域——By Me at 20140613
[市场调研与分析]Intel发力移动安全领域 ...
- 【疯狂造轮子-iOS】JSON转Model系列之二
[疯狂造轮子-iOS]JSON转Model系列之二 本文转载请注明出处 —— polobymulberry-博客园 1. 前言 上一篇<[疯狂造轮子-iOS]JSON转Model系列之一> ...
- 【疯狂造轮子-iOS】JSON转Model系列之一
[疯狂造轮子-iOS]JSON转Model系列之一 本文转载请注明出处 —— polobymulberry-博客园 1. 前言 之前一直看别人的源码,虽然对自己提升比较大,但毕竟不是自己写的,很容易遗 ...
随机推荐
- MingW和cygwin的区别(转)
个人总结:读完这段文字需要5分钟 总结: MingW https://zh.wikipedia.org/wiki/MinGW Cygwin https://zh.wikipedia.org/wiki/ ...
- Springboot - -web应用开发-Servlets, Filters, listeners
一.Web开发使用 Controller 基本上可以完成大部分需求,但是我们还可能会用到 Servlet. Filter. Listener等等 二.在spring boot中的三种实现方式 方法一: ...
- SpringBoot实战(二)Restful风格API接口
在上一篇SpringBoot实战(一)HelloWorld的基础上,编写一个Restful风格的API接口: 1.根据MVC原则,创建一个简单的目录结构,包括controller和entity,分别创 ...
- 【BZOJ 1196】[HNOI2006]公路修建问题
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 二分最后选的边中的最大值是多少. mid 则所有边权小于等于mid的边都可以用了. 那么我们要怎么选择呢? ->优先选择一级的 ...
- 【BZOJ 1059】[ZJOI2007]矩阵游戏
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 最后要求对于所有的i,a[i][i]=1 那么,如果第i行的第j列为1. 就说明我们可以把这个第i行换到第j行. 因为这样的话,a[ ...
- Python 安装 httplib2
简述 httplib2 是一个使用 Python 写的支持的非常全面的 HTTP 特性的库.需要 Python2.3 或更高版本的运行环境,0.5.0 版及其以后包含了对 Python3 的支持. 简 ...
- 蓝牙压力測试报抛android.os.TransactionTooLargeException异常分析总结
1.从main日志中找到异常点,例如以下: 08-20 11:05:19.754 5023 5023 E AndroidRuntime: FATAL EXCEPTION: main 08-20 11: ...
- Android实战技巧之三十七:图片的Base64编解码
通经常使用Base64这样的编解码方式将二进制数据转换成可见的字符串格式,就是我们常说的大串.10块钱一串的那种,^_^. Android的android.util包下直接提供了一个功能十分完备的Ba ...
- CorePlot学习六---点击scatterPlot中的symbol点时弹出对应的凝视
因为项目须要用到用户点击 symbol时,弹出对应的具体信息,发现国内解说的比較少,经过一番搜索验证最终解决,先看效果图: 详细须要改动的代码例如以下: 首先要引用托付方法:CPTScatterPlo ...
- hdu_4430,二分
注意处理溢出 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm ...