time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There always is something to choose from! And now, instead of "Noughts and Crosses", Inna choose a very unusual upgrade of this game. The rules of the game are given below:

There is one person playing the game. Before the beginning of the game he puts 12 cards in a row on the table. Each card contains a character: "X" or "O".
Then the player chooses two positive integers a and
b (a·b = 12), after that he makes a table of size
a × b from the cards he put on the table as follows: the first
b cards form the first row of the table, the second
b cards form the second row of the table and so on, the last
b cards form the last (number
a) row of the table. The player wins if some column of the table contain characters "X" on all cards. Otherwise, the player loses.

Inna has already put 12 cards on the table in a row. But unfortunately, she doesn't know what numbers
a and b to choose. Help her win the game: print to her all the possible ways of numbers
a, b that she can choose and win.

Input

The first line of the input contains integer
t (1 ≤ t ≤ 100). This value shows the number of sets of test data in the input. Next follows the description of each of the
t tests on a separate line.

The description of each test is a string consisting of 12 characters, each character is either "X", or "O". The
i-th character of the string shows the character that is written on the
i-th card from the start.

Output

For each test, print the answer to the test on a single line. The first number in the line must represent the number of distinct ways to choose the pair
a, b. Next, print on this line the pairs in the format
axb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by
whitespaces.

Sample test(s)
Input
4
OXXXOXOOXOOX
OXOXOXOXOXOX
XXXXXXXXXXXX
OOOOOOOOOOOO
Output
3 1x12 2x6 4x3
4 1x12 2x6 3x4 6x2
6 1x12 2x6 3x4 4x3 6x2 12x1
0

给仅仅含‘X'和’O'字符的长度为12的字符串,将它写入矩阵a*b(a*b=12)中。问存在某一列全为‘X'的矩阵的个数。

用的万能的打表。

23333


#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector> using namespace std; const int N = 20;
int n ;
char s[N];
char str[N][N];
int flag;
int ans; int tt(int a, int b)
{
int i, j;
for(i=0; i<a; i++)
for(j=0; j<b; j++)
{
str[i][j] = s[ i*b+j ];
}
for(i=0; i<b; i++)
{
for(j=0; j<a; j++)
{
if( str[j][i]!='X' )
break;
}
if(j==a)
return 1;
} return 0;
} int main()
{
scanf("%d", &n);
while( n-- )
{
ans = 0;
scanf("%s", &s);
if( tt(1, 12) )
ans++;
if( tt(2, 6) )
ans++;
if( tt(3, 4) )
ans++;
if( tt(4, 3) )
ans++;
if( tt(6, 2) )
ans++;
if( tt(12, 1) )
ans++; printf("%d", ans); if( tt(1, 12) )
printf(" 1x12");
if( tt(2, 6) )
printf(" 2x6");
if( tt(3, 4) )
printf(" 3x4");
if( tt(4, 3) )
printf(" 4x3");
if( tt(6, 2) )
printf(" 6x2");
if( tt(12, 1) )
printf(" 12x1"); printf("\n");
} return 0;
}

Codeforces Round #234 (Div. 2) :A. Inna and Choose Options的更多相关文章

  1. Codeforces Round #234 (Div. 2):B. Inna and New Matrix of Candies

    B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes i ...

  2. Codeforces Round #234 (Div. 2) A. Inna and Choose Options

    A. Inna and Choose Options time limit per test 1 second memory limit per test 256 megabytes input st ...

  3. Codeforces Round #234 (Div. 2) A. Inna and Choose Options 模拟题

    A. Inna and Choose Options time limit per test 1 second memory limit per test 256 megabytes input st ...

  4. Codeforces Round #234 (Div. 2)

    A. Inna and Choose Options time limit per test 1 second memory limit per test 256 megabytes input st ...

  5. Codeforces Round #234 (Div. 2) B. Inna and New Matrix of Candies

    B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes i ...

  6. Codeforces Round #234 (Div. 2) B. Inna and New Matrix of Candies SET的妙用

    B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes i ...

  7. Codeforces Round #482 (Div. 2) : Kuro and GCD and XOR and SUM (寻找最大异或值)

    题目链接:http://codeforces.com/contest/979/problem/D 参考大神博客:https://www.cnblogs.com/kickit/p/9046953.htm ...

  8. Codeforces Round #482 (Div. 2) :B - Treasure Hunt

    题目链接:http://codeforces.com/contest/979/problem/B 解题心得: 这个题题意就是三个人玩游戏,每个人都有一个相同长度的字符串,一共有n轮游戏,每一轮三个人必 ...

  9. Codeforces Round #532 (Div. 2):F. Ivan and Burgers(贪心+异或基)

    F. Ivan and Burgers 题目链接:https://codeforces.com/contest/1100/problem/F 题意: 给出n个数,然后有多个询问,每次回答询问所给出的区 ...

随机推荐

  1. Git:常用命令(一)

    取得项目的Git 仓库 从当前目录初始化 git init 初始化后,在当前目录下会出现一个名为.git 的目录,所有Git 需要的数据和资源都存放在这个目录中.不过目前,仅仅是按照既有的结构框架初始 ...

  2. IOS判断用户的网络类型(2/3/4G、wifi)

    直接贴代码吧,ios7之后是获取的较为准确,7以下我拿iphone5测试的是无法区分3g/2g.连iphone4都能升到7.1.4,而且目前主流的设备7以下的系统已经很少了,这个方案尽管不太完美,但影 ...

  3. 手游开发Android平台周边工具介绍

    1.渠道接入 主要是需要接入各平台的登录.充值接口,各家SDK又不统一,Android渠道都是鱼龙混杂,就算小渠道你看不上,但量多了,加起来也还可观,所以大家都拿出吃奶的尽去铺渠道.国内几大主要的An ...

  4. 调用人人网API

    大致步骤与上篇调用新浪微博API类似.只是感觉新浪微博的做的更好一些,人人网的非常多要手动操作 与新浪微博类似,先在人人网开放平台http://dev.renren.com/注冊站内应用, 把该填的填 ...

  5. 用SimpleAdapter来设置ListView的内容

    Mainactivit.java package com.kale.listview; import java.util.ArrayList; import java.util.HashMap; im ...

  6. C# byte[]和文件FileStream相互转化

    , pReadByte.Length);             }            catch            {                return false;        ...

  7. List集合中的数据按照某一个属性进行分组

    有的时候,我们需要在java中对集合中的数据进行分组运算.例如:Bill对象有money(float)和type(String)属性,现有个集合List<Bill>,需要按照Bill的ty ...

  8. MSSQL 数据库语句原来是区分大小写的啊

    一直以来我们都认为数据库语句是不区分大小写,其实这是错误的认识,之所以不区分是因为数据库语言不区分大小写.这里我们以mssql2005中自带的AdventureWorksDW数据库为例. 执行以下语句 ...

  9. [leetcode]Candy @ Python

    原题地址:https://oj.leetcode.com/problems/candy/ 题意: There are N children standing in a line. Each child ...

  10. jquery 控制css样式

    一.CSS 1.css(name) 访问第一个匹配元素的样式属性. 返回值 String 参数 name (String) : 要访问的属性名称 示例: $("p").css(&q ...