Given
an array and a threshold value k where k is used to divide each element
of the array. Find the total number of divisions we get after dividing
each element of the array by k.
for example:
A[ ] = 5 8 10 13 6 2 and k is 3
Output will be 17
Explanation:
Number Parts counts
5 {3,2} 2
8 {3,3,2} 3
10 {3,3,3,1} 4
13 {3,3,3,3,1} 5
6 {3,3} 2
2 {2} 1
The result thus will be 2+3+4+5+2+1 = 17
for example:
A[ ] = 5 8 10 13 6 2 and k is 3
Output will be 17
Explanation:
Number Parts counts
5 {3,2} 2
8 {3,3,2} 3
10 {3,3,3,1} 4
13 {3,3,3,3,1} 5
6 {3,3} 2
2 {2} 1
The result thus will be 2+3+4+5+2+1 = 17
Input:
The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N and threshold value k, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
The second line of each test case contains N space separated integers denoting array elements.
Output:
Corresponding to each test case, in a new line, print the total count.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 500
1 ≤ A[i] ≤ 1000
1 ≤ Threshold value(k) ≤ 20
1 ≤ N ≤ 500
1 ≤ A[i] ≤ 1000
1 ≤ Threshold value(k) ≤ 20
Example:
Input
1
6 3
5 8 10 13 6 2
1
6 3
5 8 10 13 6 2
Output
17
17
C-Solution
#include <stdio.h>
int main() {
int t,n,s,k=0,i;
scanf("%d",&t);
while(t>0)
{
k=0;
scanf("%d %d",&n,&s);
int a[n];
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]%s==0)
k+=a[i]/s;
else
{
k+=(a[i]/s)+1;
}
}
printf("%d\n",k);
t--;
}
return 0;
}
int main() {
int t,n,s,k=0,i;
scanf("%d",&t);
while(t>0)
{
k=0;
scanf("%d %d",&n,&s);
int a[n];
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]%s==0)
k+=a[i]/s;
else
{
k+=(a[i]/s)+1;
}
}
printf("%d\n",k);
t--;
}
return 0;
}
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