Lisas Workbook Hackerrank

Lisa just got a new math workbook. A workbook contains exercise problems, grouped into chapters.

There are chapters in Lisa’s workbook, numbered from 1 to n.
The i-th chapter has t_i problems, numbered from 1 to t_i.
Each page can hold up to k problems. There are no empty pages or unnecessary spaces, so only the last page of a chapter may contain fewer than k problems.
Each new chapter starts on a new page, so a page will never contain problems from more than one chapter.
The page number indexing starts at 1.
Lisa believes a problem to be special if its index (within a chapter) is the same as the page number where it’s located. Given the details for Lisa’s workbook, can you count its number of special problems?

Note: See the diagram in the Explanation section for more details.

Input Format

The first line contains two integers n and k— the number of chapters and the maximum number of problems per page respectively.
The second line contains n integers t₁, t₂,…, t_n, where t_i denotes the number of problems in the i-th chapter.

Constraints
1<=n, k, t_i<=100 Output Format Print the number of special problems in Lisa's workbook. Sample Input 5 3 4 2 6 1 10 Sample Output 4 Explanation The diagram below depicts Lisa's workbook with 5 chapters and a maximum of 3 problems per page. Special problems are outlined in red, and page numbers are in yellow squares.

There are 4 special problems and thus we print the number 4 on a new line.

Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
   int N,K,t;
cin>>N>>K;
int p[N];
for(int i=0;i<N;i++){
    cin>>t;
    p[i]=t;
}
int count=0,pi=0,pf=0,page=0,temp;  //pi,pf are intial problem ,final problem for a page
 for(int i=0;i<N;i++){
     temp=0;
    while(p[i]>0){            
        page++;            
        if(p[i]/K){                
            pf=K*(temp+1);
        }else{                
            pf=p[i]+K*(temp);
        }
        pi=temp*K+1;  
        if(page>=pi && page<=pf){
            count++;
        }            
        p[i]=p[i]-K;
        temp++;
    }         
 }    
cout<<count;
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    return 0;
}

II C++

#include<bits/stdc++.h>
using namespace std;
int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    int answer = 0;
    int page = 1;
    for(int chapter = 1; chapter <= n; ++chapter) {
        int problems;
        scanf("%d", &problems);
        for(int index = 1; index <= problems; ++index) {
            if(index == page)
                answer++;
            if(index == problems || index % k == 0)
                ++page;
        }
    }
    printf("%d\n", answer);
    return 0;
}

III C++

#include <bits/stdc++.h>
using namespace std;

int n, k;
int tab[107];
int res;
int a = 1, b;

int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &tab[i]);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= tab[i]; j++) {
            if (j == a) {
                res++;
            }
            b++;
            if (j == tab[i] || b == k) {
                b = 0;
                a++;
            }
        }
    }
    printf("%d\n", res);
    return 0;
}

# Enter your code here. Read input from STDIN. Print output to STDOUT
n, k = map(int, raw_input().split())
t = map(int, raw_input().split())
page, result = 0, 0
for i in xrange(n):
    ques = 0
    page += 1
    for j in xrange(t[i]):
        if ques == k:
            ques = 0
            page += 1
        ques += 1
        if j+1 == page:
            result += 1
print result