# #C Programming (I) - 6 ## Nested loop English: Lei KuoLiang nicolaslouis@mail.fcu.edu.tw Chinese(TW): Wang M.H --- ### This week's course catalog 0. Review 1. Nested loop 2. flag application --- ## Last week's course review ---- ### Loop * while * do while * for ---- #### while ```c while ( Judgment sentence ) { Narrative sentence 1; Narrative sentence 2; ...... } ``` ```flow st=>start: Enter the while loop e=>end: Leave the while loop stat=>operation: Narrative sentence cond=>condition: Judgment sentence st->cond(yes,right)->stat(top)->cond cond(no)->e ``` ---- #### do while ```c do { Narrative sentence 1; Narrative sentence 2; ...... } while ( 判斷句 ); //Attention!!!!Semicolon Is Needed(;)!!! ``` ```flow st=>start: Enter the do while loop e=>end: Leave the while loop stat=>operation: Narrative sentence cond=>condition: Judgment sentence st->stat->cond cond(yes, right)->stat cond(no)->e ``` ---- #### for ```c for ( Enter the loop narrative sentence; judgment sentence; loop narrative sentence ){ Narrative sentence 1; Narrative sentence 2; ... } ``` ```flow st=>start: Enter for loop e=>end: Leave for the loop init=>operation: Enter the loop narrative loop=>operation: Circular narrative stat=>operation: Narrative sentence cond=>condition: Judgment sentence st->init->cond cond(no,bottom)->e cond(yes)->stat(right)->loop(right)->cond ``` --- ## Nested loop Because the bird's nest is wrapped in a circle, the nested circle means that the layer of the circle is surrounded by a layer of loops. ---- A code that can print two rectangular triangles of 5. ```c for(int i = 1; i <= 5; i++){ for(int j = 0; j < i; j++){ printf("*"); } printf("\n"); } ``` Results of the ``` * ** *** **** ***** ``` ---- For the counter of the for loop, the outermost layer is i, the second layer is j, the third layer is k, and the fourth layer is m. --- ### Exercise 1 Enter 1 integer N to print an isosceles triangle with a base of (N * 2 - 1) and a height of N ``` 5 * *** ***** ******* ********* ``` --- ### Exercise 2 Use the nested for loop to print the 99 multiplication table (note the number alignment) ``` 1*1= 1 2*1= 2 3*1= 3 4*1= 4 5*1= 5 6*1= 6 7*1= 7 8*1= 8 9*1= 9 1*2= 2 2*2= 4 3*2= 6 4*2= 8 5*2=10 6*2=12 7*2=14 8*2=16 9*2=18 1*3= 3 2*3= 6 3*3= 9 4*3=12 5*3=15 6*3=18 7*3=21 8*3=24 9*3=27 1*4= 4 2*4= 8 3*4=12 4*4=16 5*4=20 6*4=24 7*4=28 8*4=32 9*4=36 1*5= 5 2*5=10 3*5=15 4*5=20 5*5=25 6*5=30 7*5=35 8*5=40 9*5=45 1*6= 6 2*6=12 3*6=18 4*6=24 5*6=30 6*6=36 7*6=42 8*6=48 9*6=54 1*7= 7 2*7=14 3*7=21 4*7=28 5*7=35 6*7=42 7*7=49 8*7=56 9*7=63 1*8= 8 2*8=16 3*8=24 4*8=32 5*8=40 6*8=48 7*8=56 8*8=64 9*8=72 1*9= 9 2*9=18 3*9=27 4*9=36 5*9=45 6*9=54 7*9=63 8*9=72 9*9=81 ``` --- ## Flag application In the loop, flag can be used as the basis for jumping out of the loop. ---- Below is a piece of code that "prints the first 10 prime numbers within 1000" <!--  --> ```c int count_prime = 0; for (int i = 2; i < 1000; i++){ int is_prime = 1; //State initialization for (int j = 2; j < i; j++) if (i % j == 0){ is_prime = 0; break; } //If it is a prime number, it will be printed and counted. if(is_prime){ printf("%d", i); count_prime++; //If 10 numbers have been printed, the loop will be jumped out (and the end will not be blank), otherwise continue if(count_prime >= 10) break; printf(" "); } } printf("\n"); ``` ---- In the code of a round battle, record the flag of the current round state: ```c int char1_hp = 30, char1_atk = 8; int char2_hp = 33, char2_atk = 7; int now_turn = 1; printf("Character 1\nHP：%d\tATK：%d\n" "Character 2\nHP：%d\tATK：%d\n\n", char1_hp, char1_atk, char2_hp, char2_atk); while(char1_hp > 0 && char2_hp > 0){ if (now_turn == 1){ char2_hp -= char1_atk; printf("Character 1 launches an attack on character 2, causing %d damage.\n" "Role 2 remaining %d point HP\n\n", char1_atk, char2_hp); now_turn = 2; //Offensive and defensive exchange } else if (now_turn == 2){ char1_hp -= char2_atk; printf("Character 2 launches an attack on character 1, causing %d damage.\n" "Role 1 remaining %d point HP\n\n", char2_atk, char1_hp); now_turn = 1; //Offensive and defensive exchange } } printf("Role %d loses combat ability\n", now_turn); ``` --- ## Homework ---- One day, Xiaomei wanted to design the pattern on the napkin, so she asked his engineer friend Xiao Ming to write a program that could output a bunch of diamonds to make Xiaomei photocopy. Let the user enter the diagonal length of the diamond, the number of columns, and the number of rows, and print a matching pattern after a blank line. ---- * Diagonal length input range of diamonds is 3, 5, 7, 9 * The number of rows (columns) and the number of columns (rows) is 2~10. * If the user enters an error, he/she needs to prompt for an error and then re-enter * Each left and right adjacent diamonds need to be separated by a blank key * Each of the upper and lower adjacent diamonds needs to be separated by one line ----   --- ###### tags: `108 Ai-Mod-Eng-LKL`