Codeforces 400 A. Inna and Choose Options

本文介绍了一个有趣的游戏挑战,玩家需要将一串包含'X'和'O'的卡片排列成特定大小的矩阵,以确保至少有一列全部为'X'。通过模拟的方法,文章详细解释了如何找出所有可能的矩阵尺寸组合。

A. Inna and Choose Options

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
/*题目大意 给定一个长度为12的字符串包含两种字符'o'和'x',问将这个字符串转换为长宽分别为多少的矩阵会让矩阵的某一列都为'x',输出数量和所有可能的长宽.*/
/*模拟即可*/
#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define pr pair<int,int>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Pair pair<int,int>
const int INF = 0x3f3f3f3f;
using namespace std;

char arr[15];

bool cal(int x,int y){
    for(int i=0;i<y;i++){
        bool f = true;
        for(int j=i;j<12;j+=y){
            if(arr[j] == 'O'){
                f = false; break;
            }
        }
        if(f) return true;
    }
    return false;
}

int main(){
    int t;
    scanf("%d",&t);
    while(t-- && scanf("%s",arr)){
        queue<pr>Q;
        for(int i=1;i<=12;i++)
            for(int j=1;j<=12;j++)
                if(i * j == 12 && cal(i,j))
                    Q.push(make_pair(i,j));
        printf("%d",Q.size());
        while(!Q.empty()){
            printf(" %dx%d",Q.front().first,Q.front().second);Q.pop();
        }
        printf("\n");
    }
    return 0;
}




### Codeforces 1732A Bestie 题目解析 对于给定的整数数组 \(a\) 和查询次数 \(q\),每次查询给出两个索引 \(l, r\),需要计算子数组 \([l,r]\) 的最大公约数(GCD)。如果 GCD 结果为 1,则返回 "YES";否则返回 "NO"[^4]。 #### 解决方案概述 为了高效解决这个问题,可以预先处理数据以便快速响应多个查询。具体方法如下: - **预处理阶段**:构建辅助结构来存储每一对可能区间的 GCD 值。 - **查询阶段**:利用已有的辅助结构,在常量时间内完成每个查询。 然而,考虑到内存限制以及效率问题,直接保存所有区间的结果并不现实。因此采用更优化的方法——稀疏表(Sparse Table),它允许 O(1) 时间内求任意连续子序列的最大值/最小值/GCD等问题,并且支持静态RMQ(Range Minimum Query)/RANGE_GCD等操作。 #### 实现细节 ##### 构建稀疏表 通过动态规划的方式填充二维表格 `st`,其中 `st[i][j]` 表示从位置 i 开始长度为 \(2^j\) 的子串的最大公约数值。初始化时只需考虑单元素情况即 j=0 的情形,之后逐步扩展至更大的范围直到覆盖整个输入序列。 ```cpp const int MAXN = 2e5 + 5; int st[MAXN][20]; // Sparse table for storing precomputed results. vector<int> nums; void build_sparse_table() { memset(st,-1,sizeof(st)); // Initialize the base case where interval length is one element only. for(int i = 0 ;i < nums.size(); ++i){ st[i][0]=nums[i]; } // Fill up sparse table using previously computed values. for (int j = 1;(1 << j)<=nums.size();++j){ for (int i = 0;i+(1<<j)-1<nums.size();++i){ if(i==0 || st[i][j-1]!=-1 && st[i+(1<<(j-1))][j-1]!=-1) st[i][j]=__gcd(st[i][j-1],st[i+(1<<(j-1))][j-1]); } } } ``` ##### 处理查询请求 当接收到具体的 l 和 r 参数后,可以通过查找对应的 log₂(r-l+1) 来定位合适的跳跃步长 k ,进而组合得到最终答案。 ```cpp string query(int L,int R){ int K=(int)(log2(R-L+1)); return __gcd(st[L][K],st[R-(1<<K)+1][K])==1?"YES":"NO"; } ``` 这种方法能在较短时间内完成大量查询任务的同时保持较低的空间开销,非常适合本题设定下的性能需求。
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