CodeForces 711A A. Bus to Udayland

最新推荐文章于 2023-11-26 15:12:17 发布

shyoldman

最新推荐文章于 2023-11-26 15:12:17 发布

阅读量370

点赞数

CC 4.0 BY-SA版权

分类专栏：就当它是水题吧

本文链接：https://blog.youkuaiyun.com/qq_37943488/article/details/77851569

就当它是水题吧专栏收录该内容

93 篇文章

订阅专栏

解决两位朋友在巴士上寻找相邻空座位的问题，通过扫描每排座位来确定是否有符合条件的座位。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The IOI bus has nrows of seats. There are 4 seats in each row, and the seats are separated into pairs by a walkway. When ZS and Chris came, some places in the bus was already occupied.

ZS and Chris are good friends. They insist to get a pair of neighbouring empty seats. Two seats are considered neighbouring if they are in the same row and in the same pair. Given the configuration of the bus, can you help ZS and Chris determine where they should sit?

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of rows of seats in the bus.

Then, n lines follow. Each line contains exactly 5 characters, the first two of them denote the first pair of seats in the row, the third character denotes the walkway (it always equals '|') and the last two of them denote the second pair of seats in the row.

Each character, except the walkway, equals to 'O' or to 'X'. 'O' denotes an empty seat, 'X' denotes an occupied seat. See the sample cases for more details.

Output

If it is possible for Chris and ZS to sit at neighbouring empty seats, print "YES" (without quotes) in the first line. In the next n lines print the bus configuration, where the characters in the pair of seats for Chris and ZS is changed with characters '+'. Thus the configuration should differ from the input one by exactly two charaters (they should be equal to 'O' in the input and to '+' in the output).

If there is no pair of seats for Chris and ZS, print "NO" (without quotes) in a single line.

If there are multiple solutions, you may print any of them.

Examples

input
6
OO|OX
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX

output
YES
++|OX
XO|XX
OX|OO
XX|OX
OO|OO
OO|XX

input
4
XO|OX
XO|XX
OX|OX
XX|OX

output
NO

input
5
XX|XX
XX|XX
XO|OX
XO|OO
OX|XO

output
YES
XX|XX
XX|XX
XO|OX
XO|++
OX|XO

Note

Note that the following is an incorrect configuration for the first sample case because the seats must be in the same pair.

O+|+X

XO|XX

OX|OO

XX|OX

OO|OO

OO|XX

题目大意：有两个人坐车，他们必须挨在一起，图中有圈的就是空位置，有X的就是有人坐了，中间的竖线是分割线，如果两个人能挨着坐，那么就在它们坐的地方输出++

水题

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char str[1010][10];
int main(){
    //freopen("in.txt","r",stdin);
    int n;
    while(cin>>n){
        getchar();
        for(int i=0;i<n;i++){
            scanf("%s",str[i]);
        }
        int flag;
        for(int i=0;i<n;i++){
            flag=1;
            for(int j=0;j<2;j++){
                if(str[i][j]!='O'){
                    flag=0;
                }
            }
            if(flag){
                for(int j=0;j<2;j++){
                    str[i][j]='+';
                }
                break;
            }
            else{
                flag=1;
                for(int j=3;j<5;j++){
                    if(str[i][j]!='O'){
                        flag=0;
                    }
                }
            }
            if(flag){
                for(int j=3;j<5;j++){
                    str[i][j]='+';
                }
                break;
            }
        }
        if(flag){
            cout<<"YES"<<endl;
            for(int i=0;i<n;i++){
                printf("%s\n",str[i]);
            }
        }
        else cout<<"NO"<<endl;
    }
}