Inna and Choose Options

一道中等题目 题目不难 读懂题目是关键

题意：多实例 输入一个12位只含有O X的字符串 注意输入T后再输入字符串前要getchar()

把12位字符串分成a*b的二介矩阵形式 如果每行都有X那么就输出a*b 如果没有 就输出0.

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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<math.h>
#include<time.h>
using namespace std;
int main()
{
int x[101];
int y[101];
int t;
char s[101];
int ans;
int k;
scanf("%d",&t);
getchar();
while(t--)
{
ans=0;
k=0;
gets(s);
int find=0;
for (int i=0;i<12;i++)
{
if (s[i]=='X')
{
find=1;
break;
}
}
if (find==0)
{
printf("0\n");
continue;
}
else
{
ans++;
x[k]=1;
y[k]=12;
k++;
}
for (int i=0;i<6;i++)
{
if (s[i]=='X'&&s[i+6]=='X')
{
ans++;
x[k]=2;
y[k]=6;
k++;
break;
}
}
for (int i=0;i<4;i++)
{
if (s[i]=='X'&&s[i+4]=='X'&&s[i+8]=='X')
{
ans++;
x[k]=3;
y[k]=4;
k++;
break;
}
}
for (int i=0;i<3;i++)
{
if (s[i]=='X'&&s[i+3]=='X'&&s[i+6]=='X'&&s[i+9]=='X')
{
ans++;
x[k]=4;
y[k]=3;
k++;
break;
}
}
for (int i=0;i<2;i++)
{
if (s[i]=='X'&&s[i+2]=='X'&&s[i+4]=='X'&&s[i+6]=='X'&&s[i+8]=='X'&&s[i+8]=='X'&&s[i+10]=='X')
{
ans++;
x[k]=6;
y[k]=2;
k++;
break;
}
}
find=0;
for (int i=0;i<=11;i++)
{
if (s[i]=='O')
{
find=1;
break;
}
}
if (find==0)
{
ans++;
x[k]=12;
y[k]=1;
k++;
}
printf("%d",ans);
for (int i=0;i<k;i++)
{
printf(" %dx%d",x[i],y[i]);
}
printf("\n");
}
return 0;
}