The program must accept N integers as the input. The program must traverse the given integers and if the integer 5 is encountered then the following integers must be added till the integer 10 occurs. Finally, the program must print the total sum as the output.
Boundary Condition(s):
1 <= N <= 1000
Input Format:
The first line contains N.
The second line contains N integers separated by space(s).
Output Format:
The first line contains the sum as per the given conditions.
Example Input/Output 1:
Input:
9
2 5 3 2 10 3 10 5 4
Output:
9
Explanation:
The integer 5 occurs at position 2 and the immediate next integer value 10 occurs at position 5. There are two integers (3 and 2) between them which are added to 5.
The next integer value 5 occurs at position 8 the integers after 5 is 4 which is also added to the result.
Hence the output is 9
Program
#include<stdio.h>
#include <stdlib.h>
int main()
{
int N,A[1000],i,j=0,temp=0;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%d",&A[i]);
for(i=0;j<N;)
{
if(A[i]==5)
{
while(A[++i]!=10)
{
temp+=A[i];
if (i>=N)
break;
}
}
j=++i;
}
printf("%d",temp);
}
Output
Input:
8
1 5 5 12 10 5 6 10
Output:
23
1 <= N <= 1000
Input Format:
The first line contains N.
The second line contains N integers separated by space(s).
Output Format:
The first line contains the sum as per the given conditions.
Example Input/Output 1:
Input:
9
2 5 3 2 10 3 10 5 4
Output:
9
Explanation:
The integer 5 occurs at position 2 and the immediate next integer value 10 occurs at position 5. There are two integers (3 and 2) between them which are added to 5.
The next integer value 5 occurs at position 8 the integers after 5 is 4 which is also added to the result.
Hence the output is 9
Program
#include<stdio.h>
#include <stdlib.h>
int main()
{
int N,A[1000],i,j=0,temp=0;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%d",&A[i]);
for(i=0;j<N;)
{
if(A[i]==5)
{
while(A[++i]!=10)
{
temp+=A[i];
if (i>=N)
break;
}
}
j=++i;
}
printf("%d",temp);
}
Output
Input:
8
1 5 5 12 10 5 6 10
Output:
23