codeforces Inna and Choose Options
2015-10-30 20:54
369 查看
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:
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:
[code]#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; }
相关文章推荐
- 2015多校第九,十场总结
- sicily 1438. Shopaholic
- 网 上 找 了 些 练 习 题 试 试 手
- Centos6.5 搭建 SVN 服务器 及 钩子同步更新网站代码
- Tomcat的安装与第一个servlet程序的部署
- linux笔记:权限管理命令chmod,chown,chgrp,umask
- Linux的chattr与lsattr命令
- apache用户认证配置
- 网站源码 网站模板 扁平化后台管理 Bootstrap、HTML5、CSS3 Java
- Linux下常见性能分析工具
- (转)支付宝牛逼的原因:来看内部架构剖析
- (转)论架构师的自我修养
- 《学习opencv》笔记——矩阵和图像操作——cvSetIdentity,cvSolve,cvSplit,cvSub,cvSubS and cvSubRS
- opencv solvePnP
- VS2008 中OpenGL的安装 (我的安装)
- linux目录扫描
- How To Build CyanogenMod Android (oneplus/bacon) On Linux
- apache配置解析php
- nginx和keepalive共存亡
- LVM逻辑卷扩容