Codeforces Round #713 (Div. 3)AB题

2021-05-19

Codeforces Round #713 (Div. 3) Editorial

记录一下自己写的前二题本人比较菜

A. Spy Detected!

You are given an array a consisting of n (n≥3) positive integers. It is known that in this array, all the numbers except one are the same (for example, in the array [4,11,4,4] all numbers except one are equal to 4).

Print the index of the element that does not equal others. The numbers in the array are numbered from one.

Input
The first line contains a single integer t (1≤t≤100). Then t test cases follow.

The first line of each test case contains a single integer n (3≤n≤100) — the length of the array a.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤100).

It is guaranteed that all the numbers except one in the a array are the same.

Output
For each test case, output a single integer — the index of the element that is not equal to others.

样例
4
4
11 13 11 11
5
1 4 4 4 4
10
3 3 3 3 10 3 3 3 3 3
3
20 20 10

输出
2
1
5
3
题意:

就是有t组测试数据,每个测试数据给我们一个长度为n的数组,其中n>=3且数组中的元素值只有2个,而且其中1个元素在数组中只有一个,让我们求该元素的下标。

思路:

这个思路很简单,因为数组长度不超过100,且其中的元素也不超过100,那么我们可以开2个数组,p和s,其中p就是我们存的数组,而s就当哈希表一样存数组值对应的个数,然后我们再枚举一下s其中为1的就是我们要找的元素,然后返回下标就好了。

C++ 代码
#include <algorithm>
#include <cstring>
#include <iostream>
 
using namespace std;
 
const int N = 110;
int t, n;
int s[N];
int p[N]; 
int main()
{
    scanf("%d", &t);
 
    while (t--)
    {
        memset(s, 0, sizeof s); // 每次测试数据的时候都应该把s都置为0
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
        {
            cin >> p[i]; 
            s[p[i]]++; // 将p[i] 所对应的值映射到s中去,并且计数
        }
        // 这是枚举出来s[]所对应元素为1的就是我们要找的元素,然后返回就好了
        for (int i = 1; i <= n; i++)
        {
            if (s[p[i]] == 1)
            {
                printf("%d\n", i);
            }
        }
    }
    return 0;
}

B. Almost Rectangle

There is a square field of size n×n in which two cells are marked. These cells can be in the same row or column.

You are to mark two more cells so that they are the corners of a rectangle with sides parallel to the coordinate axes.

For example, if n=4 and a rectangular field looks like this (there are asterisks in the marked cells):

. . ∗ .

. . . .

∗ . . .

. . . .

Then you can mark two more cells as follows

∗ . ∗ .

. . . .

∗ . ∗ .

. . . .

If there are several possible solutions, then print any of them.

Input
The first line contains a single integer t (1≤t≤400). Then t test cases follow.

The first row of each test case contains a single integer n (2≤n≤400) — the number of rows and columns in the table.

The following n lines each contain n characters '.' or '*' denoting empty and marked cells, respectively.

It is guaranteed that the sums of n for all test cases do not exceed 400.

It is guaranteed that there are exactly two asterisks on the field. They can be in the same row/column.

It is guaranteed that the solution exists.

Output
For each test case, output n rows of n characters — a field with four asterisks marked corresponding to the statements. If there multiple correct answers, print any of them.

样例
6
4
..*.
....
*...
....
2
*.
.*
2
.*
.*
3
*.*
...
...
5
.....
..*..
.....
.*...
.....
4
....
....
*...
*...
输出
*.*.
....
*.*.
....
**
**
**
**
*.*
*.*
...
.....
.**..
.....
.**..
.....
....
....
**..
**..
题意

有t组测试数据,每个测试数据输入一个数字n表示接下来要输入一个n*n的字符矩阵,且矩阵中有且只有2个*然后我们需要填入2个*让这四个*能够对称,要是有多组的话就输出一组就好了。

思路

本人太菜了,刚看到这题目的时候都没想啥办法,就直接写了,然后写了好多判断不好读。我就说一下我看到的一个比较好的做法把。
我们可以用2个数组存2个*号的坐标。
这个时候我们可以发现有2中情况如下:
第一种:就是2个*号的横纵坐标都不相等,那么这个时候我们只需要将他们交错的坐标位置都用*号代替就好了。
第二种:就是其中的横坐标或者纵坐标相等(不可能横纵坐标都相等,因为必须要有2个*嘛)这个时候用了一个非常巧妙的办法就是,a[0] = a[1] == 0 这个判断让我们减少了所有的边界条件的判断,这是当横坐标相等的时候,如果纵坐标也相等的话我们就将a换成纵坐标的就好了。
接着我们看一下代码吧。

代码
// 题目地址
// https://codeforces.com/contest/1512/problem/B

#include <algorithm>
#include <cstring>
#include <iostream>

using namespace std;

const int N = 410;

int t, n;
char g[N][N];

int a[2];
int b[2];

int main()
{
    scanf("%d", &t);

    while (t--)
    {
        scanf("%d", &n);

        for (int i = 0; i < n; i++)
            scanf("%s", g[i]);

        int t = 0;
        // 找到2个*的下标位置
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
            {
                if (g[i][j] == '*')
                {
                    a[t] = i;
                    b[t++] = j;
                }
            }

        /*
        我们可以知道这个只有三种情况
        第一种:*的2个坐标都不等那么我们只需要
                直接将他们的交叉坐标点变为*就可以了。
        第二种: *的x坐标相等那么我们就可以通过
                a[0] = a[1] == 0的方式来处理
                如果a[1],等于0的话我们就让a[0]取1,
                如果不等的话就让a[0]取0,
                因为我们可以输出任意一种就好了。
        第三种:同第二种差不多*在y坐标上相等
        */
        if (a[0] == a[1])
            a[0] = a[1] == 0;
        if (b[0] == b[1])
            b[0] = b[1] == 0;

        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 2; j++)
                g[a[i]][b[j]] = '*';

        for (int i = 0; i < n; i++)
            printf("%s\n", g[i]);
    }

    //system("pause");
    return 0;
}

第一次写代码题解,记录下我做过的一些题目,要是大佬们发现有什么错误的地方,还望多多指出错误之处,orz orz orz。