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Program to count number of consecutive lists whose sum is n in C++
Suppose we have a number n, we have to find the number of lists of positive consecutive values that sum up to n.
So, if the input is like n = 15, then the output will be 4, as The possible lists are: [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], and [15].
To solve this, we will follow these steps:
- begin := 1, end := 1, x := (n + 1)
- sum := 0
- while end <= x, do:
- sum := sum + end
- while sum >= n, do:
- if sum is same as n, then:
- (increase count by 1)
- sum := sum - begin
- (increase begin by 1)
- if sum is same as n, then:
- (increase end by 1)
- return count + 1
Let us see the following implementation to get better understanding:
Example
#include
using namespace std;
int solve(int n) {
int begin=1,end=1,x=(n+1)/2,count=0;
long int sum=0;
while(end <= x){
sum += end;
while(sum >= n){
if(sum == n)
count++;
sum -= begin;
begin++;
}
end++;
}
return count+1;
}
main(){
cout << (solve(15));
}Input
15
Output
4
Updated on: 26-Nov-2020
279 Views
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