Arithmetic operations and related exercises in C language

Write it at the front ：
Reading this article can solve the following problems ： How to use it C Language performs arithmetic operations ？ How to use bit operation to speed up your code ？ How to skillfully use bit operation in actual combat code ？ Multiple ways to realize the same function .

1. Operator

Operator	explain	Example
=	Assignment operator	a = b;
+ 、- 、* 、/、()	Basic four principles	a = (b + c) * d;
%	The remainder operator	a = b % 2;
&、 |、 ^、 ~	An operation ( A very important class )	a = ~b | c;
<<、>>	Move left and right	a = b >>2;

1.1 An operation

Bit operations are done in binary .

1.1.1 Bitwise AND (&)

All binary bits are 1 The result is 1, There is one for 0, Then for 0. When the number of digits is insufficient , Front complement 0.
Such as ：

• 2&3 --> 10 & 11 --> 10, Convert decimal to zero 2
• 8 & 3 --> 1000 & 11 --> 1000 & 0011–>0, Convert decimal to zero 0

Use scenarios ：
Judge the numbers n Can you divide 2,n & 1 by 1 You can't divide , by 0 Then you can divide

1.1.2 Press bit or (|)

One of the binary bits is 1, Then for 1, All for 0 Then for 0. When the number of digits is insufficient , Front complement 0.
Such as ：

• 2|3 --> 10 | 11 --> 11, Convert decimal to zero 3
• 8 | 3 --> 1000 | 11 --> 1000 | 0011 -->1011, Convert decimal to zero 11

1.1.3 bitwise exclusive or operator (^)

The same binary bit is 0, Different for 1. When the number of digits is insufficient , Front complement 0.
Such as ：

• 2^3 --> 10 ^ 11 --> 01, Convert decimal to zero 1
• 8 ^ 3 --> 1000 ^ 11 --> 1000 ^ 0011–>1011, Convert decimal to zero 11.

Use scenarios ：

• ^ yes ^ The inverse operation of .
• a ^ b = c, be c ^ b = a and c ^ a = b.
• a ^ a = 0.
• 0 ^ a = a.

1.1.4 According to the not (~)

Binary digit 0 Turn into 1,1 Turn into 0

Such as :

• ~3 --> ~11 -->11(30 individual 1)00[ Complement code ], Convert decimal to zero -4

Complement code = ~ Original code + 1
When the original code is reversed , The sign bits remain the same
give an example ：
1 The original code of ：0…0(31 individual 0)1, ~ Original code : 01…1(30 individual 1)0, Complement code :01…1(31 individual 1)
-1 The original code of : 10…0(30 individual 0)1, ~ Original code : 1…1(31 individual 1)0, Complement code ：1…1(32 individual 1)
-1 Complement =1 All original codes are reversed +1

1.2 Move left and right

1.2.1 Move left

Move left ： Low complement 0
Move left 1 position , It's equivalent to doubling .
Such as : 2 << 1, Convert decimal to zero 4.

1.2.2 Move right

Move right ： High complement sign bit
Such as ：3 >> 1, Convert decimal to zero 1.

2. Data type conversion

There are two types of data conversion ： Strong conversion and weak conversion
Strong transformation ： Use int,double And so on , Will lose precision .
Weak transformation : Use similar 3*1.0/2 Method , take int Type to double.

3. Hexadecimal conversion

3.1 Binary to decimal

10010=12**4+12**1=18

3.2 Hexadecimal to decimal

8F=816**1+1516**0=143

3.3 Decimal to binary

Method 1： Use short division , Until 0, Then write from bottom to top ;
Method 2： Make up
27=16+8+2+1=24+23+21+20=11010

4.math Common functions

Write it at the front ：
Use the following functions , Import required math library ;
At compile time , Use the command

gcc test.c -lm

4.1 pow function ： Exponential function

Prototype ： double pow(double a, double b);
Example ：pow(4,2)=16.0000

4.2 sqrt function ： Square root function

Prototype ： double sqrt(double X);
Example ：pow(16)=4.0000

4.3 ceil function ： Round up the function

Prototype ： double ceil(double X);
Example ：ceil(2.79)=3.0000

4.4 floor function ： Round down the function

Prototype ： double floor(double X);
Example ：floor(2.79)=2.0000

4.5 abs function ： Integer absolute value function

Prototype ：int abs(int X);
Example ：abs(-2)=2

4.6 fabs function ： Real absolute value function

Prototype ：double fabs(double X);
Example ：fabs(-2.6)=2.6

4.7 log function ： With e Is the base logarithm function

Prototype ：double log(double X);
Example ：log(4)=1.386294

4.8 log10 function ： With e Is the base logarithm function

Prototype ：double log10(double X);
Example ：log10(4)= 0.602060

4.9 acos function ：arccos function

Prototype ：double acos(double X);
X Is the radian value of the angle
Example ：acos(-1)=3.1415926

5. Fixed width integer type

#include <stdio.h>
#include <inttypes.h>
 
int main(){
     
    printf("%zu\n", sizeof(int64_t));  // Check the memory size 
    printf("%s\n", PRId64);  // Format macro constants 
    printf("INT32_MIN : %" PRId32 ", INT32_MAX : %" PRId32 "\n", INT32_MIN, INT32_MAX);  //32 Bit shaping min max 
    printf("INT64_MIN : %" PRId64 ", INT64_MAX : %" PRId64 "\n", INT64_MIN, INT64_MAX);  //64 Bit shaping min max 

    int64_t n = 7;
    printf("%+"PRId64"\n", n);
    return 0;
}

6. Exercises

subject 1

Enter a number a, Output its cube root . Click to see the answer

subject 2：

seek π Value . Click to see the answer

subject 3：

Enter an angle value , And convert it to radian value . Click to see the answer

subject 4：

Loop in two numbers a,b, And exchange two values . Click to see the answer

Refer to the answer

subject 1 answer

//  subject 1
#include<stdio.h>
#include<math.h>
int main(){
     
    double a;
    scanf("%lf", &a);
    // solution 1, Understand thoughts 
    printf("%lf\n", pow(a, 1.0 / 3));
    // solution 2
    printf("%lf\n", cbrt(a));
    return 0;
}

subject 2 answer

#include<stdio.h>
#include<math.h>
#define pi acos(-1)
// #define pi 3.14
int main(){
     
    double a;
    scanf("%lf", &a);
    printf("%lf\n", a * pi / 180);
    return 0;
}

subject 3 answer

#include<stdio.h>
#include<math.h>
#define pi acos(-1)
// #define pi 3.14
int main(){
     
    double a;
    scanf("%lf", &a);
    printf("%lf\n", a * pi / 180);
    return 0;
}

subject 4 answer

//  Method 1
#include<stdio.h>
int main(){
     
    int a, b;
    while (~scanf("%d%d", &a, &b)){
     
        printf("a is %d, b is %d\n", a, b);
        int c;
        c = a;
        a = b;
        b = c;
        printf("after change: a is %d, b is %d\n", a, b);
    }
    return 0;
}
//  Method 2
#include<stdio.h>
int main(){
     
    int a, b;
    while (~scanf("%d%d", &a, &b)){
     
        printf("a is %d, b is %d\n", a, b);
        a = a + b;
        b = a - b;
        a = a - b;
        printf("after change: a is %d, b is %d\n", a, b);
    }
    return 0;
}
//  Method 3
#include<stdio.h>
int main(){
     
    int a, b;
    while (~scanf("%d%d", &a, &b)){
     
        printf("a is %d, b is %d\n", a, b);
        b ^= a;
        a ^= b;
        printf("after change: a is %d, b is %d\n", a, b);
    }
    return 0;
}