codeforces Inna and Choose Options

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本文介绍了一款升级版的井字棋游戏，玩家需将12张包含'X'或'O'的卡片排列成特定矩阵大小以赢得游戏。通过分析不同矩阵配置，文章提供了一种判断胜利条件的方法，并给出了相应的实现代码。

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

题意：给出12张牌，这12张牌只有两种字符‘X’,’O’,有a*b=12,将这12张牌组成a*b的矩阵，其中前b张作为第一行，再后边的b张作为第二行，依次继续，例如123456789000，划分成3*4的就是这样子的
1234
5678
9000

代码很low非常low，省的操心想：

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);
    for(int k=0; k<T; k++)
    {
        char a[12];
        int sum=0;
        cin>>a;
        int x,y,flag=0,b[12],s=0;
        for(int i=0; i<12; i++)
        {
            if(a[i]=='X')
                flag++;
        }
        if(flag==0)
        {
            cout<<"0"<<endl;
            continue;
        }
        b[s++]=1;
        if(flag==12)
        {
            printf("6 1x12 2x6 3x4 4x3 6x2 12x1\n");
            continue;
        }
        if((a[0]==a[6])&&a[0]=='X'||(a[1]==a[7])&&a[1]=='X'||(a[2]==a[8])&&a[2]=='X'||(a[3]==a[9])&&a[3]=='X'||(a[4]==a[10])&&a[4]=='X'||(a[5]==a[11])&&a[5]=='X')
            b[s++]=2;
        if((a[0]==a[4])&&(a[4]==a[8])&&(a[0]=='X')||(a[1]==a[5])&&(a[5]==a[9])&&(a[1]=='X')||(a[2]==a[6])&&(a[6]==a[10])&&(a[2]=='X')||(a[3]==a[7])&&(a[7]==a[11])&&(a[3]=='X'))
            b[s++]=3;
        if((a[0]==a[3])&&(a[3]==a[6])&&(a[6]==a[9])&&(a[0]=='X')||(a[1]==a[4])&&(a[4]==a[7])&&(a[7]==a[10])&&(a[1]=='X')||(a[2]==a[5])&&(a[5]==a[8])&&(a[8]==a[11])&&(a[2]=='X'))
            b[s++]=4;
        if((a[0]==a[2])&&(a[2]==a[4])&&(a[4]==a[6])&&(a[6]==a[8])&&(a[8]==a[10])&&(a[0]=='X')||(a[1]==a[3])&&(a[3]==a[5])&&(a[5]==a[7])&&(a[7]==a[9])&&(a[9]==a[11])&&(a[1]=='X'))
            b[s++]=6;
        printf("%d ",s);
        for(int i=0; i<s; i++)
        {
            if(i!=s-1)
                printf("%dx%d ",b[i],12/b[i]);
            else
                printf("%dx%d\n",b[i],12/b[i]);
        }
    }
    return 0;
}