1、http://codeforces.com/problemset/problem/400/A
2、题目大意：
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abcdefghijkl，划分成3*4的就是这样子的
abcd
efgh
ijkl
如果有某一列全部是’X',那么就输出这样的a和b
3、题目：
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 sizea?×?b from the cards he put on the table as follows: the firstb cards form the first row of the table, the secondb cards form the second row of the table and so on, the lastb 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 numbersa and b to choose. Help her win the game: print to her all the possible ways of numbersa,?b that she can choose and win.
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 thet tests on a separate line.
The description of each test is a string consisting of 12 characters, each character is either "X", or "O". Thei-th character of the string shows the character that is written on thei-th card from the start.
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 paira,?b. Next, print on this line the pairs in the formataxb. Print the pairs in the order of increasing first parameter (a). Separate the pairs in the line by whitespaces.
4 OXXXOXOOXOOX OXOXOXOXOXOX XXXXXXXXXXXX OOOOOOOOOOOO
3 1x12 2x6 4x3 4 1x12 2x6 3x4 6x2 6 1x12 2x6 3x4 4x3 6x2 12x1 0
4、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char s[15];
struct node
{int x;int y;
} a[15];
int k=0;
int main()
{int t;scanf("%d",&t);while(t--){int sum=0;k=0;scanf("%s",s);int flag=0;for(int i=0; i<12; i++){if(s[i]=='X'){flag=1;break;}}if(flag==0){printf("0\n");continue;}else{sum++;a[k].x=1;a[k].y=12;k++;}for(int i=0; i<6; i++){if(s[i]=='X' && s[i+6]=='X'){sum++;a[k].x=2;a[k].y=6;k++;break;}}for(int i=0; i<4; i++){if(s[i]=='X' && s[i+4]=='X' && s[i+8]=='X'){sum++;a[k].x=3;a[k].y=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'){sum++;a[k].x=4;a[k].y=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+10]=='X' ){sum++;a[k].x=6;a[k].y=2;k++;break;}}flag=0;for(int i=0; i<=11; i++){if(s[i]=='O'){flag=1;break;}}if(flag==0){sum++;a[k].x=12;a[k].y=1;k++;}printf("%d",sum);for(int i=0; i<k; i++){printf(" %dx%d",a[i].x,a[i].y);}printf("\n");}return 0;
}