To display the numbers in the series, only 7 numbers per line

To display a series of numbers as 1 2 3 4 .......... 100. Display only five integers per line.

#include<stdio.h> #include<conio.h> void main( ) { int j, p=0; for (j=1; j < = 100, j++) { printf("%d", j); p++; if (p==7) { printf("\n"); p=0; } } }

Hint:-

Method used in previous question can not be used here. That method can be used only if 80 divided by number of values per line gives you integer answer otherwise you have to use 2nd Method. 7 values per line = 80/7 = 11.43 and therefore you can not use the earlier method.

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Write a program in C to display the number from 1 to 200 in a series. Print 20 numbers on one line and use all the available columns on the screen.

Write a program in C to display the number from 1 to 200 in a series.
Print 20 numbers on one line and use all the available columns on the screen.

#include<stdio.h> #include<conio.h> main( ) { int i; clrscr( ); for (i=1; i < = 200; i++) { printf("%4d", i); } return(0); }

Hint:-

20 values per line = 80/20 = 4 = %4d 4 values per line = 80/4 = 20 = %20d 8 values per line = 80/8 = 10 = %10d

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Write a program in C to display the number form 1 2 3 4 .................100. Display only five integers per line.

Write a program in C to display the number form 1 2 3 4 .................100.
Display only five integers per line.

#include<stdio.h> #include<conio.h> main( ) { int i; clrscr( ); for (i=1; i < = 100; i++) { printf("%16d", i); } return(0); }

Hint:-

5 values per line = 80/5 = 16 = %16d 4 values per line = 80/4 = 20 = %20d 8 values per line = 80/8 = 10 = %10d

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