Codeforces Round #663 (Div. 2)

Codeforces Round #663 (Div. 2) 传送门

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A - Suborrays

Description

A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).
For a positive integer n, we call a permutation p of length n good if the following condition holds for every pair i and j (1≤i≤j≤n) —
(pi OR pi+1 OR … OR pj−1 OR pj)≥j−i+1, where OR denotes the bitwise OR operation.
In other words, a permutation p is good if for every subarray of p, the OR of all elements in it is not less than the number of elements in that subarray.
Given a positive integer n, output any good permutation of length n. We can show that for the given constraints such a permutation always exists.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤100). Description of the test cases follows.
The first and only line of every test case contains a single integer n (1≤n≤100).

Output

For every test, output any good permutation of length n on a separate line.

input

3
1
3
7

output

1
3 1 2
4 3 5 2 7 1 6

Note

For n=3, [3,1,2] is a good permutation. Some of the subarrays are listed below.
3 OR 1=3≥2 (i=1,j=2)
3 OR 1 OR 2=3≥3 (i=1,j=3)
1 OR 2=3≥2 (i=2,j=3)
1≥1 (i=2,j=2)
Similarly, you can verify that [4,3,5,2,7,1,6] is also good.

Code

#include
#include
#include
using namespace std;
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		if(n == 1)
		{
			cout<<n<<endl;
		}else
		{
			for(int i = 1 ; i <= n ; i++)
			{
				cout<<i;
				if(i != n)
				{
					cout<<" ";
				}
			}
			cout<<endl;
		}
	}
	return 0;
}

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B - Fix You

Description

Consider a conveyor belt represented using a grid consisting of n rows and m columns. The cell in the i-th row from the top and the j-th column from the left is labelled (i,j).
Every cell, except (n,m), has a direction R (Right) or D (Down) assigned to it. If the cell (i,j) is assigned direction R, any luggage kept on that will move to the cell (i,j+1). Similarly, if the cell (i,j) is assigned direction D, any luggage kept on that will move to the cell (i+1,j). If at any moment, the luggage moves out of the grid, it is considered to be lost.
There is a counter at the cell (n,m) from where all luggage is picked. A conveyor belt is called functional if and only if any luggage reaches the counter regardless of which cell it is placed in initially. More formally, for every cell (i,j), any luggage placed in this cell should eventually end up in the cell (n,m).
This may not hold initially; you are, however, allowed to change the directions of some cells to make the conveyor belt functional. Please determine the minimum amount of cells you have to change.
Please note that it is always possible to make any conveyor belt functional by changing the directions of some set of cells.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤10). Description of the test cases follows.
The first line of each test case contains two integers n,m (1≤n≤100, 1≤m≤100) — the number of rows and columns, respectively.
The following n lines each contain m characters. The j-th character in the i-th line, ai,j is the initial direction of the cell (i,j). Please note that an,m= C.

Output

For each case, output in a new line the minimum number of cells that you have to change to make the conveyor belt functional.

input

4
3 3
RRD
DDR
RRC
1 4
DDDC
6 9
RDDDDDRRR
RRDDRRDDD
RRDRDRRDR
DDDDRDDRR
DRRDRDDDR
DDRDRRDDC
1 1
C

output

1
3
9
0

Note

In the first case, just changing the direction of (2,3) to D is enough.
You can verify that the resulting belt is functional. For example, if we place any luggage at (2,2), it first moves to (3,2) and then to (3,3).
In the second case, we have no option but to change the first 3 cells from D to R making the grid equal to RRRC.

解题思路

因为C是在(n,m)处，所以直接找第m列和第n行进行判断和修改，标记后输出即可

Code

#include
#include
#include
using namespace std;
char map[128][128];
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int t;
	cin>>t;
	while(t--)
	{
		int n,m;
		cin>>n>>m;
		for(int i = 1 ; i <= n ; i++)
		{
			for(int j = 1 ; j <= m ; j++)
			{
				cin>>map[i][j];
			}
		}
		int Cnt = 0;
		for(int i = 1 ; i <= n-1 ; i++)
		{
			if(map[i][m] == 'R')
			{
				map[i][m] = 'D';
				Cnt++;
			}
		}
		for(int i = 1 ; i <= m-1 ; i++)
		{
			if(map[n][i] == 'D')
			{
				map[i][1] = 'R';
				Cnt++;
			}
		}
		cout<<Cnt<<endl;
	}
	return 0;
}