Basic questions (I)

1、 Analyze the following functions

#include <stdio.h>
int main()
{
    
	int a, b, c;
	a = 5;
	c = ++a;// First ++, Reuse a (c=6,a=6)
	b = ++c, c++, ++a, a++;// Be careful (b = ++c) It's an expression  (c=8,b=7,a=8)
	b += a++ + c;//+= The priority is very low , So first count the second  a=9,b=23
	printf("a = %d b = %d c = %d\n:", a, b, c);
	return 0;
}

The output is ：

2、
Write a function to return the parameter binary 1 The number of .

such as ： 15 0000 1111 4 individual 1

// Method 1 ：
// Version of a   defects , Negative numbers cannot be calculated 
// Version 2   Just one change , Change the formal parameter to unsigned type 
#include <stdio.h>
int stat1(int n)// Version of a 
int stat1(unsigned int n)// Version 2 
{
    
	int count = 0;
	while (n)
	{
    
		if ((n % 2) == 1)
		{
    
			count++;
		}
		n /= 2;
	}
	return count;
}

int main()
{
    
	int num = 0;
	scanf("%d", &num);
	int n = stat1(num);
	printf("%d\n", n);
	return 0;
}

// Method 2 
#include <stdio.h>
int stat1(int n)
{
    
	int i = 0;
	int count = 0;
	for (i = 0; i < 32; i++)
	{
    
		if (((n >> i) & 1) == 1)
		{
    
			count++;
		}
	}
	return count;
}

int main()
{
    
	int num = 0;
	scanf("%d", &num);
	int n = stat1(num);
	printf("%d\n", n);
	return 0;
}

// Method 3 
#include <stdio.h>
int stat1(int n)
{
    
	int count = 0;
	while(n)
	{
    
		n = n & (n - 1);
		count++;
	}
	return count;
}

int main()
{
    
	int num = 0;
	scanf("%d", &num);
	int n = stat1(num);
	printf("%d\n", n);
	return 0;
}

3、 Programming to realize ：
Two int（32 position ） Integers m and n In the binary representation of , How many bits (bit) Different ？

Input example :
1999 2299
Output example :7

// Method 1 ：
#include <stdio.h>
int count_diff_bit(int m, int n)
{
    
	int count = 0;
	int i = 0;
	for (i = 0; i < 32; i++)
	{
    
		if (((m >> i) & 1) != ((n >> i) & 1))
		{
    
			count++;
		}
	}
	return count;
}
int main()
{
    
	int m = 0;
	int n = 0;
	scanf("%d %d", &m, &n);
	int ret = count_diff_bit(m, n);
	printf("%d\n", ret);
	return 0;
}

// Method 2 ：
#include <stdio.h>
int count_diff_bit(int m, int n)
{
    
	int count = 0;
	int ret = m ^ n;// ^  XOR operator   Same as 0, Dissimilarity is 1
	while (ret)
	{
    
		ret = ret & (ret - 1);
		count++;
	}
	return count;
}
int main()
{
    
	int m = 0;
	int n = 0;
	scanf("%d %d", &m, &n);
	int ret = count_diff_bit(m, n);
	printf("%d\n", ret);
	return 0;
}

4、 Get all even and odd bits in an integer binary sequence , Print out the binary sequence separately

//10
//0000000000000000000000000001010
// Specify the last as the first 
#include <stdio.h>
int main()
{
    
	int i = 0;
	int num = 0;
	scanf("%d", &num);

	// Get odd digit numbers 
	for (i = 30; i >= 0; i -= 2)
	{
    
		printf("%d ",(num >> i) & 1);
	}
	printf("\n");

	// Get even digits 
	for (i = 31; i >= 1; i -= 2)
	{
    
		printf("%d ", (num >> i) & 1);
	}
	return 0;
}

5、 Analyze the following code

#include <stdio.h>
int i;// Global variables 
// Global variables , Static variables are placed in the static area 
// Global variables , Static variables are not initialized , The default value is 0
// local variable , Put it in the stack area , Not initialized , The default value is random 
int main()
{
    
    i--;//-1
    //sizeof The result of this operator calculation is size_t Type of , It's an unsigned integer 
    if (i > sizeof(i))// When they compare , There's an integer boost , Convert to unsigned integer , here -1 Is a very positive number 
    {
    
        printf(">\n");
    }
    else
    {
    
        printf("<\n");
    }
    return 0;
}

Global variables , When no initial value is given , The compiler initializes it to... By default 0.
i The initial value of 0,i– result -1,i For shaping ,sizeof(i) seek i The type size is 4, According to this analysis , The results should be printed <, however sizeof The return value type of is actually an unsigned integer , Therefore, the compiler will automatically set the left i Automatically convert to unsigned data ,-1 The corresponding unsigned integer is a very large number , exceed 4 perhaps 8, So it should actually be printed >

6、
Input description ： Multi group input , An integer （2~20）, Represents the number of rows output , It also means composition “x” The length of the backslash and the backslash of .
Output description ： Enter... For each line , For output “*” Composed of X Shape patterns .

#include <stdio.h>
int main()
{
    
	int n = 0;
	while(scanf("%d", &n) == 1)
	{
    
		int j = 0;
		int i = 0;
		for (i = 0; i < n; i++)
		{
    
			for (j = 0; j < n; j++)
			{
    
				if (i == j)
					printf("*");
				else if (i + j == n - 1)
					printf("*");
				else
					printf(" ");
			}
			printf("\n");
		}
	}
	return 0;
}

7、
Input description ：
Multi group input , There are two integers in a row , Represents year and month respectively , Space off .
Output description ：
Enter... For each group , Output as a line , An integer , It means how many days there are in this month of the year .

Example ：
Input ：2008 2
Output ：29

// Method 1 
#include <stdio.h>
int main()
{
    
	int year = 0;
	while (scanf("%d", &year) == 1)
	{
    
		int n = 0;
		scanf("%d", &n);
		if (((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
		{
    
			if (n == 2)
				printf("29");
			else if (n == 4 || n == 6 || n == 9)
				printf("30");
			else
				printf("31");
		}
		else
		{
    
			if (n == 2)
				printf("28");
			else if (n == 2 || n == 4 || n == 6 || n == 9)
				printf("30");
			else
				printf("31");
		}
	}
	return 0;
}

// Method 2 
#include <stdio.h>

int is_leap_year(int y)// Determine if it's a leap year 
{
    
	return(((y % 4 == 0) && (y % 100 != 0)) || (y % 400 == 0));
}

int main()
{
    
	int y = 0;// year 
	int m = 0;// month 
	int d = 0;// Days 
	int data[13] = {
     0,31,28,31,30,31,30,31,31,30,31,30,31 };// Array start complement 0, Make the array subscript correspond to the month 
	while (scanf("%d%d", &y, &m) == 2)//scanf Function received several values , Will return a few 
	{
    
		int d = data[m];
		if ((is_leap_year(y) == 1) && (m == 2))// To determine whether it is a leap year 2 month 
		{
    
			d++;
		}
		printf("%d\n", d);
	}
	return 0;
}