Algorithm design and analysis 002

1. Sequence of basic exercises

Given a sequence of length N, the sequence is arranged from small to large. 1 < = n < = 200 input Description:
The first line is an integer n.
The second row contains n integers, which are the numbers to be sorted, and the absolute value of each integer is less than 10000.
Input example:
5
8 3 6 4 9
Output Description:
Output a row, and output the sorted sequence from small to large.
Output example:
3 4 6 8 9

Example code:

#include <iostream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int array[n];
    for (int i = 0; i < n; i++)
    {
        cin >> array[i];
    }
    sort(array, array + n);
    for (int i = 0; i < n; i++)
    {
        cout << array[i] << " ";
    }


}

2 basic practice time conversion

Given a time t in seconds, it is required to express the time in the format of "::. Represents time, minutes, and seconds. They are all integers without leading "0". For example, if t=0, the output should be "0:0:0"; If t=3661, output "1:1:1". Enter Description:
The input has only one line, which is an integer t (0 < = T < = 86399).
Input example:
0
Output Description:
The output has only one line, which is the time expressed in the format of "::" without quotation marks.
Output example:
0:0:0

Example code:

#include <iostream>
#include<algorithm>
using namespace std;
int main()
{
    unsigned int time;
    int hour = 0;
    int minute = 0;
    int second = 0;
    cin >> time;
    while(time >= 60)
    {
        minute++;
        time -= 60;
    }
    second = time;

    while(minute >= 60)
    {
        hour++;
        minute -= 60;
    }
    if (hour == 24)
        hour = 0;
    cout << hour << ":" << minute << ":" << second << endl;


}

3. Basic practice matrix multiplication

Given an N-order matrix A, output the M-power of a (M is a nonnegative integer)
For example:
　　A =
　　1 2
　　3 4
The second power of A
　　7 10
15.22 input Description:
The first row is a positive integer N, M (1 < = N < = 30, 0 < = M < = 5), representing the order of matrix A and the required idempotent
Next, N rows, N non negative integers with absolute values not exceeding 10 in each row, describe the value of matrix A
Input example:
2 2
1 2
3 4
Output Description:
Output a total of N rows, n integers per row, representing the matrix corresponding to the M-power of A. Adjacent numbers are separated by a space
Output example:
7 10
15 22

Code example:

#include<iostream>
#include<cstring>
using namespace std;
int start[50][105];
int tmp[50][105];
int last[50][105];
void Display(int a[][105], int n)
{
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
			cout << a[i][j] << " ";
		cout << endl;
	}
}
void Getmatrix(int start[][105], int tmp[][105], int last[][105], int n, int m)
{
	for (int i = 1; i < m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			for (int k = 0; k < n; k++)
			{
				int temp = 0;
				for (int x = 0; x < n; x++)
				{
					temp += start[j][x] * tmp[x][k];
					last[j][k] = temp;
				}
			}
		}
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				tmp[i][j] = last[i][j];
			}
		}
	}
}
int main()
{
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cin >> start[i][j];
		}
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			tmp[i][j] = start[i][j];
		}
	}
	if (m == 0)
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				if (i == j)
					last[i][j] = 1;
			}
		}
		Display(last, n);
	}
	else if (m == 1)
	{
		Display(start, n);
	}
	else
	{
		Getmatrix(start, tmp, last, n, m);
		Display(last, n);
	}
	return 0;
}

4. Division of algorithm training number

Divide the integer n into k parts, and each part cannot be empty, and any two parts cannot be the same (regardless of order).
For example: n=7, k=3, the following three methods are considered to be the same.
　　1，1，5; 1，5，1; 5，1，1;
Ask how many different methods there are. Enter Description:
　　n，k
Input example:
7 3
Output Description:
An integer, that is, a different division
Output example:
4 {four methods are: 1, 1, 5; 1, 2, 4; 1, 3, 3; 2, 2, 3;}

Code example:

#include<iostream>
#include<math.h>
#include<memory.h>
using namespace std;

int main()
{
    int n, k;
    cin >> n >> k;
    int dp[240][15];
    memset(dp,0, sizeof(dp));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= i; j++)
            if (j == 1)
                dp[i][j] = 1;
            else
                dp[i][j] = dp[i - j][j] + dp[i - 1][j - 1];

    cout << dp[n][k] << endl;
    return 0;
}

#include<iostream>
using namespace std;

int Dfs(int n, int k) {
    if (n == 0 || k == 0 || n < k) {
        return 0;
    }
    if (n == 1 || n == k) {
        return 1;
    }
    else
    {
        return Dfs(n - k, k) + Dfs(n - 1, k - 1);
    }
}

int main()
{
    int n, k;
    cin >> n >> k;
    cout << Dfs(n, k) << endl;
    return 0;
}