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display the first n natural numbers
+2
votes
asked
Oct 10, 2019
by
Shaurya Deep
(
140
points)
Your answer
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7 Answers
0
votes
answered
Oct 11, 2019
by
Harsha
(
380
points)
#include<stdio.h>
int main()
{
int i;
for(i=1;i<i+1;i++)
{
printf("%d\n",i);
}
}
commented
Oct 11, 2019
by
Harsha
(
380
points)
if you want first 10 natural numbers put 11 in place of i+1
have any doubts message me insta id(gutsboy_harsha)
commented
Oct 27, 2019
by
Molero Raffi
(
540
points)
You could have just done this:
int i = 0;
while (1) printf("%i\n", ++i);
and just replaced "true" with "i <= n" where n is the last natural number to print
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0
votes
answered
Oct 12, 2019
by
saiteja
(
290
points)
#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
printf("%d\t",i);
}
}
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0
votes
answered
Oct 14, 2019
by
karthik srinivas
(
190
points)
#include<stdio.h>
void main()
{
int i,x;
printf("Enter the n value");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
printf("%d, ",i);
}
getch();
}
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0
votes
answered
Oct 16, 2019
by
G Abhijit
#include<stdio.h>
void main()
{
int i,n=0;
printf("Enter the value of n = ");
scanf("%d",&n);
for(i=1;i<=n-1;i++)
{
printf("%d, ",i);
}
printf("%d.",n);
}
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0
votes
answered
Oct 22, 2019
by
santoshverma
(
140
points)
void main()
{
int n;
cout<<"enter the number till you want the natural number to get printed";
cin>>n;
for(int i=0;i<n-1;i++)
{
cout<<" natural numbers are";
cout<<i;
}
}
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0
votes
answered
Nov 7, 2019
by
dheeraj
(
1,090
points)
a=int(input('enter upto u want natural numbers '))
for i in range(1,a+1):
print(i,end=' ')
output: @python
enter upto u want natural numbers 100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
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0
votes
answered
Nov 9, 2019
by
anonymous
#include<stdio.h>
void main()
{
int n;int i;
for(i=0;i<=n;i++)
{
printf("%d\n",i);
}
}
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