Dr. monkey class C language learning notes play 3

<1> if statement

if(expression) { . . } else { . . }

Condition usage: if so, how about it, otherwise

Look at a programming problem first

Input an integer, if the number is greater than 60, then output "Dr. monkey is handsome";
If the number is no more than 60, the output "Dr. monkey is really handsome".

Let's take a look at the fixed format first

#include<stdio.h> int main() { return 0; }

Let's set up some conditions, and pay attention to some points here!
<1> If there is only one statement after if and else, you can leave out the braces
<2> else and if are not followed by semicolons!
<3> Each branch statement needs to be wrapped (\ n)

#include<stdio.h> int main() { int a; //Consolidate int (integer) scanf("%d",&a); //Format of scanf statement if(a>60) //If there is only one statement after if and else, you can leave out the braces { printf("Dr. monkey is so handsome\n"); } else //if and else without semicolon { printf("Dr. monkey is really handsome\n"); //All branches need to be followed by \ n (line feed) } return 0; }

Final output format

#include<stdio.h> int main() { int a; //Consolidate int (integer) scanf("%d",&a); //Format of scanf statement if(a>60) //If there is only one statement after if and else, you can leave out the braces printf("Dr. monkey is so handsome\n"); else //if and else without semicolon printf("Dr. monkey is really handsome\n"); //All branches need to be followed by \ n (line feed) return 0; }

Input an integer, if the number is greater than 60, then output "Dr. monkey Shuai".

#include<stdio.h> int main() { int a; scanf("%d",&a); if(a>60) printf("Dr. monkey is so handsome\n"); return 0; }

Let's take another look

#include<stdio.h> int main() { int x,y; //Integers define two variables x,y with int scanf("%d",&x); //scanf statement if(x<0) { //Braces can be omitted if there is only one statement y=-1; } return 0; }

Continue after omission

#include<stdio.h> int main() { int x,y; //Integers define two variables x,y with int scanf("%d",&x); if(x<0) y=-1; //Braces can be omitted if there is only one statement else { if(x>0) //Insert another if statement y=1; else //There is no semicolon after the if and else statements! y=0; } printf("x=%d,y=%d\n",x,y); //%d (integer) \ n (line feed) return 0; }

Let's simplify at last

#include<stdio.h> int main() { int x,y; scanf("%d",&x); if(x<0) y=-1; //Braces can be omitted if there is only one statement else //Let's take a closer look at the else statement. When the if condition is met, only one statement will be executed / / all the parentheses can be removed if(x>0) //So we can remove the brackets y=1; else y=0; printf("x=%d,y=%d\n",x,y); //%d (integer) \ n (line feed) return 0; }

We can also nest if and else in if

#include<stdio.h> int main() { int x,y; scanf("%d",&x); if(x>=0) if(x>0) y=1; else y=0; else y=-1; printf("x=%d,y=%d\n",x,y); return 0; }

Another question
Input two real numbers a and b, and output them in the order of numerical value from small to large.

#include<stdio.h> int main() { double a,b,t; //Note that the topic is real number. Add another variable here scanf("%lf,%lf",&a,&b); if(a>b) //Condition when a > b (subject requirement a < b) { //If a=5,b=3 t=a; //t=a t=5 a=b; //a=b b=3 a=b=3 b=t; //b=t b=5 } printf("%lf,%lf\n",a,b); return 0; }

Input three real numbers a, b and c, and output them in the order of numerical value from small to large.

#include<stdio.h> int main() { double a,b,c,d,t;//Because I'm not sure if there's a number with a decimal point, double is used here scanf("%lf,%lf,%lf,%lf",a,b,c,d);//scanf statement double corresponding to% lf forgotten students, please take a look at the previous one if(a>b)//If a > b, suppose a = 5, B = 1 (easy to understand) { t=a;//Let t=a, t=5 a=b;//a=b,a=1 b=t;//b=t, because t=5,b=t, so b=5, so a < B, OK, let's continue, Gan! } if(a>c) { t=a; a=c; c=t; } if(a>d) { t=a; a=d; d=t; } if(b>c)//If a > b, b=2,c=1 { t=b;//t=2 b=c;//b=c, assign 2 to c,c=2 c=t;//c=t,c=2 } if(b>d) { t=b; b=d; d=t; } if(c>d) { t=c; c=d; d=t; } printf("%lf,%lf,%lf,%lf",a,b,c,d); return 0; }

Common expressions
The relationship between one variable and another number

<1> > greater than
<2> > = greater than or equal to
<3> < less than
<4> < = less than or equal to
<5> = = equal to (use as follows)//
Note: a=2, only the value of a is assigned to this number, = = is equal to

#include<stdio.h> int main() { int x,y; scanf("%d",&x); if(x<0) y=-1; else if(x==0) //y=0 when x=0, otherwise y=-1 y=0; else y=-1; printf("x=%d,y=%d\n",x,y); return 0; }

<6> ! = not equal to
<7> & & satisfied on both sides (example as follows)
To wash and brush your teeth
Key points: meet the requirements of brushing, washing and wearing masks
<8> Both sides of the scrotum satisfy one (example as follows)
Ci Bai wakes up hungry, orders a takeout and cooks by himself
Key point: meet one of them, either cook or order takeout

Connect the two sentences

If you cook by yourself, you need to brush your teeth, wash your teeth and put on a mask to go out and buy food to cook, or you can order a take away at home

Classic example
Please make a program to judge whether a year is a leap year. (Note: if the current year share is not a multiple of 100 and a multiple of 4, the year is a leap year; if the current year share is a multiple of 100 and a multiple of 400, the year is also a leap year.)

When the current year's share is not a multiple of 100 and a multiple of 4, the current year's share is a multiple of 400 (certainly a multiple of 100)

(year% 100! = 0 & & year% 40) | year% 4000

#include<stdio.h> int main() { int a; //Integer variable a printf("Please enter the year\n"); scanf("%d",a); if((a%100!=0&&%4==0)||a%400==0) printf("%d Year is leap year\n",a); else printf("%d Year is not a leap year\n",a); return 0; }

Example 2 input a character to determine whether it is an uppercase letter. If it is, convert it to a lowercase letter. If it is not, do not convert it. Then output the final character.

#include<stdio.h> int main() { char echo; //char for letters, echo for assignment variables scanf("%c",a); //scanf statement letter corresponds to% c if(echo>='A'&&echo<='Z') //If the variable satisfies capital A-Z letters echo=echo+32 //+32 to lowercase, similarly - 32 to uppercase printf("%c\n",echo); //Output without lowercase, without else return 0; }

Expression 1? Expression 2: expression 3
Judge expression 1. If expression 1 holds, execute expression 2; otherwise, execute expression 3 (similar to if statement)
for instance:
Do you wear a mask? Wear a mask to go out, only at home

#include<stdio.h> int main() { char echo; //char for letters, echo for assignment variables scanf("%c",a); //scanf statement letter corresponds to% c echo=(echo>='A'&&echo<='Z')?(echo+32):echo //If the variable satisfies capital A-Z letters return 0; //Just execute echo+32 and make it lowercase, otherwise it will remain unchanged }

Input two real numbers a and b, and output them in the order of numerical value from small to large.

#include<stdio.h> int main() { double a,b; scanf("%lf,%lf",&a,&b); a>b?printf("%lf,%lf\n",b,a):printf("%lf,%lf\n",a,b); return 0; }

switch Statements

Grade A, B, C and D are the four grades of a course. Now we need to convert them to the score segment of the hundred point system. The rules are: Grade A is converted to 85-100, grade B is converted to 70-84, grade C is converted to 60-69, and grade D is converted to < 60. Please make a program, grade input by keyboard, output score segment.

#include<stdio.h> int main { double echo; scanf("%f",echo); switch(echo) //Format switch (the variable we want to judge), the variable is generally a letter or an integer { case'A':printf("The student's score is 85~100\n");break;//Let's write this sentence, and then add a semicolon; add a break; semicolon case'B':printf("The student's score is 70~84\n");break; case'C':printf("The student's score is 60~69\n");break; case'A':printf("The student's score is<60\n");break; default:printf("The grade of this student is wrong!\n");break; } return 0; }

The grade of a course was originally a, B, C and D. now we need to turn it into the score segment of the hundred point system. The rule is: Grade A, grade B into 70-100, grade C, grade D into < 70. Please make a program, grade input by keyboard, output score segment.

#include<stdio.h> int main() { char echo; scanf("%c",echo); switch(echo) { case'A': //If the upper and lower results are the same, the upper ones can not be written (remember to write the lower ones) case'B':printf("The student's score is 70~100\n");break; case'C': //Remember to wear a colon!:) case'D':printf("The student's score is<70\n");break; defaul:printf("The grade of the student is not right!\n");break; } return 0; }

The former grade system of A certain course is to convert it into A grade. The rules are: A score above 90, B score 80-89, C score 70-79, D score 60-69 and E score below 60. Please make A program, score input by keyboard, output level.

#include<stdio.h> int main() { double echo; //For example, 74.5 scanf("%lf",echo); //scanf statement double corresponds to% lf switch((int)(echo/10)) //For example, chestnut 74/10=7.4 int rounding 7 { case 10: //More than 90 points is A, so the first sentence can be left blank case 9:printf("The student's grade is A\n");break; case 8:printf("The student's grade is B\n");break; case 7:printf("The student's grade is C\n");break; case 6:printf("The student's grade is D\n");break; case 5: //Similarly, the score below 60 is E, so these sentences can not be written case 4: case 3: case 2: case 1: case 0:printf("The student's grade is E\n");break; default:printf("The score entered does not meet the requirements 23333\n");break; //No 120! So the writing does not meet the requirements } return 0; }

Come on! Crabs and crabs

10 June 2020, 23:02 | Views: 6750