# Input

There are n + 1 lines, the first line gives only one number `n`, the number of jobs. Each of the following lines gives three intergers S_i, T_i, V_i separated by space.

1 <= S_i < T_i <= 10000

1 <= V_i <= 10000

1 <= n <= 1000

Your code should read the input from standard input (e.g.
using functions `input()/raw_input()` in Python and `cin/scanf` in C++).

# Output

One number in a line, the total weight of the maximum weight compatible job subset.

Your code should write the output to standard output (e.g. using functions `print` in Python and `cout/printf` in C++).

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# Input

The graph has `n` vertices and `m` edges.
There are m + 1 lines, the first line gives three numbers `n`,`m` and `s`(1 <= s <=n), describing the number of vertices, the number of edges and the source vertex. Each of the following lines contains three integers `a`, `b` and `c`, meaning there is an edge (a,b) belong to E and the length of the edge is c. All the numbers in a line are separated by space. (`1 <= a,b <= n`)

1 <= `n` <= 1000, 1 <= `m` <= 5000, and the length of `edge` is in the range of [-1000, 1000]. You can assume that all the vertices are reachable from `s`.

Your code should read the input from standard input (e.g.
using functions `input()/raw_input()` in Python and `cin/scanf` in C++).

# Output

One line, if there is a negative cycle, output should be ‘True’. If not, output should be ‘False’.

Your code should write the output to standard output (e.g. using functions `print` in Python and `cout/printf` in C++).

The post CMPSC465-Assignment 4 Part 1 Find Negative Cycle appeared first on Mantutor.

# Input

There are n + 1 lines, the first line gives only one number `n`, the number of lines. Each of the following lines contains two floating numbers `a` and `b` separated by space which represent a line as `y = a * x + b`.

You can assume that 0 <= n <= 1000, and that each floating numbers
are in the range of [-100.0, 100.0].

Your code should read the input from standard input (e.g.
using functions `input()/raw_input()` in Python and `cin/scanf` in C++).

# Output

Two numbers in a line separated by space, the number of lines in the upper-envelop and the number lines in the lower-envelop.

Your code should write the output to standard output (e.g. using functions `print` in Python and `cout/printf` in C++).

# Requirement

Your algorithm should run in O(nlogn) time.

Time limtation: 5 seconds.

Memory limitation: 1.0 GB

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Two lines, the first line gives only one number `n`, the number of integers in the array. And the second line gives `n` numbers which are
the `n` integers in the array. All numbers are separated by space.

You can assume that 0 <= n <= 10000, and that each number in the arrarys
are in the range of [-2147483648, 2147483647].

Your code should read the input from standard input (e.g. using functions `input()/raw_input()` in Python and `cin/scanf` in C++).

# Output

One line, describing the sorted array, in ascending order.
There should be `n` integers in a line give all numbers in the sorted array (in ascending order).
All numbers should be separated with space.
(You don’t need to output the size of the array `n`.)

Your code should write the output to standard output (e.g. using functions `print` in Python and `cout/printf` in C++).

# Requirement

Your algorithm should run in O(nlogn) time. You are not allowed to call any in-built sort function.

Time limtation: 5 seconds.

Memory limitation: 1.0 GB.

# Environment

Your code will be running on Ubuntu 18.04.5.

Now only accept C++ and Python2/Python3 code, g++ version 7.5.0 and Python versions are Python 2.7.17 and Python 3.6.9.

# Examples and Testing

Some examples (e.g., input-x.txt and output-x.txt, x = 1, 2) are provided.
For Python code, try the following to test your code
“`
python ./solution.py < input-x.txt > my-output-x.txt
“`
For C++ code, try the following to test your code
“`
g++ -o mybinary solution.cpp
./mybinary < input-x.txt > my-output-x.txt
“`

Your output `my-output-x.txt` needs to be *match exactly* to the given `output-x.txt`.
On Unix-based systems you can use `diff` to compare them:
“`
diff my-output-x.txt output-x.txt
“`
On Windows you can use `fc` to compare them:
“`
fc my-output-x.txt output-x.txt
“`

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