Inna and Choose Options

Crawling in process...
Crawling failed
Time Limit:1000MS    
Memory Limit:262144KB    
64bit IO Format:
%I64d & %I64u

Description

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 Input

Input
4
OXXXOXOOXOOX
OXOXOXOXOXOX
XXXXXXXXXXXX
OOOOOOOOOOOO
Output
3 1x12 2x6 4x3
4 1x12 2x6 3x4 6x2
6 1x12 2x6 3x4 4x3 6x2 12x1
0
将12张卡片分成矩阵,如果矩阵中有一列全都是X就输出这种排列方式,输出方式看样例,这里要注意输出的是x是小写的,还有对应关系,因为这还wa了两次
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
char str[15];
int vis[15];
memset(vis,0,sizeof(vis));
memset(str,'\0',sizeof(str));
cin>>str;
int ans=0,cnt=0;
for(int i=0;i<12;i++)
if(str[i]=='X')
cnt++;
if(cnt==0)
cout<<0<<endl;
else
{
vis[0]=1;
ans++; int f=false;
for(int i=0;i<6;i++)//2*6
if(str[i]=='X'&&str[i+6]=='X')
f=true;
if(f)
vis[1]=1,ans++; f=false;
for(int i=0;i<3;i++)//4*3
if(str[i]=='X'&&str[i+3]=='X'&&str[i+6]=='X'&&str[i+9]=='X')
f=true;
if(f)
vis[2]=1,ans++; f=false;
for(int i=0;i<4;i++)//3*4
if(str[i]=='X'&&str[i+4]=='X'&&str[i+8]=='X')
f=true;
if(f)
vis[3]=1,ans++; f=false;
for(int i=0;i<2;i++)//6*2
if(str[i]=='X'&&str[i+2]=='X'&&str[i+4]=='X'&&str[i+6]=='X'&&str[i+8]=='X'&&str[i+10]=='X')
f=true;
if(f)
vis[4]=1,ans++;
if(cnt==12) ans++;
printf("%d",ans);
if(vis[0]) printf(" 1x12");
if(vis[1]) printf(" 2x6");
if(vis[3]) printf(" 3x4");
if(vis[2]) printf(" 4x3");
if(vis[4]) printf(" 6x2");
if(cnt==12) printf(" 12x1");
printf("\n");
}
}
return 0;
}

最新文章

  1. Android指纹解锁
  2. WinForm三级联动
  3. AngularJS 特性—SinglePage、template、Controller
  4. JSP如何保存页面上众多的复选状态
  5. libusb(.NET)开源项目使用小结
  6. Java学习-013-文本文件读取实例源代码(两种数据返回格式)
  7. Python list去重及找出,统计重复项
  8. 自己写的Dapper通用数据访问层
  9. linux系统安装iprouter
  10. 【LeetCode】144. Binary Tree Preorder Traversal
  11. 计蒜客模拟赛D1T3 蒜头君的坐骑:用dfs转移dp
  12. Ubuntu 16.04下Samba服务器搭建和配置(配截图)
  13. JavaScript中解决jQuery和Prototype.js同时引入冲突问题
  14. 2016年蓝桥杯省赛A组c++第1题
  15. Java -- 深入浅出GC自动回收机制
  16. php cli命令 自定义参数传递
  17. DagScheduler 和 TaskScheduler
  18. iOS面试3
  19. 20145105 《Java程序设计》第4周学习总结
  20. Python学习---django知识补充之CBV

热门文章

  1. 当From窗体中数据变化时,使用代码获取数据库中的数据然后加入combobox中并且从数据库中取得最后的结果
  2. gym101673G. A Question of Ingestion (DP)
  3. luogu P4137 Rmq Problem / mex 主席树 + 思维
  4. CSDN怎么转载别人的博客
  5. MySQL Docker方式安装
  6. Python面向对象,析构继承多态
  7. Python爬虫入门教程第七讲: 蜂鸟网图片爬取之二
  8. HTML-js 压缩上传的图片方法(默认上传的是file文件)
  9. 【Codeforces 242C】King&#39;s Path
  10. -- &gt; define的用法与学习(1)