Movie Weekend Codechef Beginner Section : Runtime Error - c++

I have written the following code for the problem, ideone doesnt give me runtime error but the compiler for codechef is giving me one. The compiler in codechef is gcc 4.9.2
Problem [Link : https://www.codechef.com/problems/MOVIEWKN]
Little Egor is a huge movie fan. He likes watching different kinds of movies: from drama movies to comedy movies, from teen movies to horror movies. He is planning to visit cinema this weekend, but he's not sure which movie he should watch.
There are n movies to watch during this weekend. Each movie can be characterized by two integers Li and Ri, denoting the length and the rating of the corresponding movie. Egor wants to watch exactly one movie with the maximal value of Li × Ri. If there are several such movies, he would pick a one with the maximal Ri among them. If there is still a tie, he would pick the one with the minimal index among them.
Your task is to help Egor to pick a movie to watch during this weekend.
Expected Input
The first line of the input contains an integer T denoting the number of test cases.
The first line of the test case description contains an integer n(number of films).
The second line of the test case description contains n integers L1, L2, ...,Ln (L is the Length of the movie). The following line contains n integers R1, R2, ..., Rn(R is the Rating of the movie).
My Input
2
2
1 2
2 1
4
2 1 4 1
2 4 1 4
Expected Output
For each test case, output a single integer i denoting the index of the movie. Note that 1-based indexing is followed.
My Output
1
2
Explanation
In the first example case, both films have the same value of L × R, but the first film has a better rating.
In the second example case, the second and the fourth movies are equally good, but the second movie has a smaller index
My Solution
#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main() {
int T;
cin >> T;
cin.ignore();
if (T >= 1 && T <= 5) {
for (int q = 0; q < T; q++) {
int n;
size_t pos;
cin >> n;
cin.ignore();
if (n >= 1 && n <= 100) {
vector<int> L;
vector<int> R;
vector<int> C;
for (int u = 0; u < n; u++) {
int a;
cin >> a;
cin.ignore();
if (a >= 1 && a <= 100) {
L.push_back(a);
}
}
for (int u = 0; u < n; u++) {
int a;
cin >> a;
cin.ignore();
if (a >= 1 && a <= 100) {
R.push_back(a);
}
}
for (int u = 0; u < n; u++) {
int df = (L[u] * R[u]);
C.push_back(df);
}
for (size_t u = 0; u < C.size(); u++) {
int max = C[0];
if (max < C[u]) {
max = C[u];
pos = u+1;
}
if (max == C[u]) {
if (R[pos-1] < R[u]) { pos = u + 1; }
if (R[pos - 1] == R[u]) {
if (pos > u) { pos = u + 1; }
}
}
}
cout << pos << endl;
}
}
}
return 0;
}
After going through some of the questions on runtime error regarding codechef questions, i am guessing the problem is related to accessing an out of boundary element but i am not able to pinpoint the actual problem

In this line:
if (R[pos - 1] < R[u]) { pos = u + 1; }
The first time around the loop, pos is uninitialised, leading to undefined behaviour.
Simple fix:
add pos = 1 before the loop

I'm probably late but here is my simple solution:
t=int(input())
for j in range(0,t):
n=int(input())
a=[int(ele) for ele in input().split()]
b=[int(ele) for ele in input().split()]
c=[]
pos=0
max=-1
max2=-1
for i in range(0,n):
fl=0
c.append(a[i]*b[i])
if(max<c[i]):
max=c[i]
fl=1
if(fl==1):
pos=i+1
max2=b[i]
elif(max==c[i]):
if(max2<b[i]):
pos=i+1
max2=b[i]
print(pos)

Related

C++ Largest number in array. Positive and negative

I have a task to print maximum int of matrix second line.
Example input:
3 2 (n, m)
-1 -2 <- 1 line
4 5 <- 2 line
2 6 <- 3 line
Max int in second line is 5. My program prints it. But if second line would be -100 -150, it not works. Sure it is because I have max = 0, but I don't know how to use it properly. I'm a student. Thanks in advance.
It is my code:
#include <iostream>
using namespace std;
int main() {
int n, m, max = 0;
cin >> n >> m;
int matrix[10][10];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> matrix[i][j];
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (matrix[1][j] > max) {
max = matrix[1][j];
}
}
}
if (max == 0 || n == 1) {
cout << "No";
} else {
cout << max;
}
}
And code works pretty good, unless there are negative numbers in second line
You are correct to suspect max = 0;. Why is that a problem? Well, first, perhaps you should try to explain to your rubber duck why it is correct. As you try to do so, you are likely to express an intent along the lines of "this value will not make it through the checks" or "this value will be replaced in the first iteration of the loop". Why? "Because matrix[1][j] > max will be true, so... Hold on, wasn't the problem when matrix[1][j] > 0 is false? So when max is 0, um... problem?"
The overall strategy is valid, but there is a requirement that max be initialized to a low enough value. There are two common strategies I can think of at the moment.
Use a value that is as low as possible for the type you are using. That is:
int max = std::numeric_limits<int>::lowest();
Use the value from the first iteration of the loop. No need to provide a value that is just going to be replaced anyway. There are some caveats for this, though. The most relevant for your example can be expressed as a question: what if there is no first iteration? (Perhaps there is only one row? Perhaps there are no columns?) Also, you would need to initialize max between your loops, after the matrix has been given values.
int max = (n > 1 && m > 0) ? matrix[1][0] : /* what value do you want here? */;

Find the number of pairs of positive integers satisfying the inequality

I'm trying to solve a programming problem where I have to display the number of positive integer solutions of the inequality x² + y² < n, where n is given by the user. I've already written a code that seems to work but not as fast as I'd like it to. Is there any way to speed it up?
My current code:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
long long n, i, r, k, p, a;
cin >> k;
while (k--)
{
r = 0;
cin >> n;
p = sqrt(n);
for (i = 1; i <= p; i++)
{
a = sqrt(n - (i * i));
r += a;
if ((((i * i) + (a * a)) == n) && (a > 0))
{
r--;
}
}
cout << r << "\n";
}
return 0;
}
Edit:
This is a solution for this task.
The task in English:
Find the number of natural solutions (x≥1, y≥1) of the inequality x²+y² < n, where 0 < n < 2147483647. For example, for n=10 there are 4 solutions: (1,1), (1,2), (2,1), (2,2).
Input
In the first line of input the number of test cases k is given. In the next k lines, there are the n values given.
Output
In the output, you have to display in separate lines the number of natural solutions of the inequality.
Example
Input:
2
10
11
Output:
4
6
Your solution seems fast already. The main possibility to reduce the time spent is to suppress the call to sqrtin the loop. This is obtained by considering that the value a = sqrt(n - (i * i)) does not vary very much from one iteration to the next one.
Here is the code:
r = 0;
p = sqrt(n);
if ((p*p) == n) p--;
a = p;
for (long long i = 1; i <= p; i++)
{
while ((n-i*i) <= a*a) {
--a;
}
r += a;
}

Finding the minimal product

How do you find the minimal product from an array? This is the problem I have and the attempted solution isn't working. What have I done wrong?
https://www.codechef.com/problems/CHRL4
After visiting a childhood friend, Chef wants to get back to his home. Friend lives at the first street, and Chef himself lives at the N-th (and the last) street. Their city is a bit special: you can move from the X-th street to the Y-th street if and only if 1 <= Y - X <= K, where K is the integer value that is given to you. Chef wants to get to home in such a way that the product of all the visited streets' special numbers is minimal (including the first and the N-th street). Please, help him to find such a product.
Input
The first line of input consists of two integer numbers - N and K - the number of streets and the value of K respectively. The second line consist of N numbers - A1, A2, ..., AN respectively, where Ai equals to the special number of the i-th street.
Output
Please output the value of the minimal possible product, modulo 1000000007.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^5
1 ≤ K ≤ N
Example
Input:
4 2
1 2 3 4.
Output:
8
#include <iostream>
using namespace std;
int P(int A[], int N, int K) {
if (N == 1) return A[0];
int m = A[0], prod = m;
for (int i = 1; i < N; ++i) {
if (1 <= A[i]-m && A[i]-m <= K) {
prod *= A[i];
}
}
return prod;
}
int main() {
int A[] = {1, 2, 3, 4};
cout << P(A, 4, 2);
}
I get 6 instead of 8.
Such problems can typically be solved by Dynamic Programming:
Construct an appropriate state variable: Let the state be S = current street. Let the factor at street S be calledC_S
For each state S, collect the possible actions: a(S) = {go to any street T for which : 1 <= C_T - C_S <= K, T <=N }, a(N) = {}.
Introduce a value function V(S) = minimal product to get from S to N. Set V(N) = C_N.
Having all this together, one can now solve the Bellman equation backwards from N, where particularly the value V(0) is sought:
V(S) = min_{allowed T} { V(T)*C_S }
Example implementation:
int main()
{
int N = 4;
int K = 2;
std::vector<int> C{1,2,3,4};
std::vector<int> V(N);
V.back() = C.back();
for(int i = N - 2; i>= 0; --i)
{
int min = std::numeric_limits<int>::max(); //possible overflow here,
//better change that
for(int j=i+1; j< N; ++j)
{
double DeltaC = C[j] - C[i];
if(DeltaC <= K && DeltaC >= 1)
{
double vt = V[j] * C[i];
if(vt < min)
{
min = vt;
}
}
}
V[i] = min;
}
std::cout<<V[0]<<std::endl;
}
DEMO
The output is 8.
Please understand the code, test it and then use it with a good conscience (whatever that means).

c++:Hackerank:Error in taking input

This is a part of my question.I tried many times but couldn't get the answer
Problem Statement
You are given a list of N people who are attending ACM-ICPC World Finals. Each of them are either well versed in a topic or they are not. Find out the maximum number of topics a 2-person team can know. And also find out how many teams can know that maximum number of topics.
Note Suppose a, b, and c are three different people, then (a,b) and (b,c) are counted as two different teams.
Input Format
The first line contains two integers, N and M, separated by a single space, where N represents the number of people, and M represents the number of topics. N lines follow.
Each line contains a binary string of length M. If the ith line's jth character is 1, then the ith person knows the jth topic; otherwise, he doesn't know the topic.
Constraints
2≤N≤500
1≤M≤500
Output Format
On the first line, print the maximum number of topics a 2-person team can know.
On the second line, print the number of 2-person teams that can know the maximum number of topics.
Sample Input
4 5
10101
11100
11010
00101
Sample Output
5
2
Explanation
(1, 3) and (3, 4) know all the 5 topics. So the maximal topics a 2-person team knows is 5, and only 2 teams can achieve this.
this is a a part of my work.Any clue how can i get this to work
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, m, max = 0, max1 = 0, count = 0;
cin >> n >> m; //for input of N and M
int a[n][m];
for (int i = 0; i<n; i++) //for input of N integers of digit size M
for (int j = 0; j<m; j + >>
cin >> a[i][j];
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
max = 0;
for (int k = 0; k<m; k++)
{
if (a[i][k] == 1 || a[j][k] == 1) max++;
cout << k;
if (k = m - 1 && max>max1) max1 = max;
if (k == m - 1 && max == max1) count++;;
}
}
}
cout << max1 << endl << count;
return 0;
}
I think the way of taking my input logic is wrong.could you please help me out.I am stuck in this question from 5 days.
PLease only help me on how should i take input and how to read the digit of integer.
Don't have a compiler with me so there's probably a syntax boner or two in there, but the logic walks through on paper.
Builds the storage:
std::cin >> n >> m; //for input of N and M
std::vector<std::vector<bool>>list(n,std::vector<bool>(m, false));
Loads the storage:
char temp;
for (int i = 0; i < n; i++) //for input of N integers of digit size M
{
for (int j = 0; j < m; j++)
{
std::cin >> temp;
if (temp == 1)
{
list[i][j] = true;
}
}
}
Runs the algorithm
for (int a = 0; a < n; a++)
{
for (int b = a+1; b < n; b++)
{
int knowcount = 0;
for (int j = 0; j < m; j++)
{
if (list[a][j] | list[b][j])
{
knowcount ++;
}
}
if (knowcount > max)
{
groupcount = 1;
max = know;
}
else if(knowcount == max)
{
groupcount ++;
}
}
}
Your method of input is wrong. According to your method, the input will have to be given like this (with spaces between individual numbers):
1 0 1 0 1
1 1 1 0 0
1 1 0 1 0
0 0 1 0 1
Only then it makes sense to create a matrix. But since the format in the question does not contain any space between a number in the same row, thus this method will fail. Taking into consideration the test case, you might be tempted to store the 'N' numbers in a single dimensional integer array, but keep in mind the constraints ('M' can be as big as 500 and int or even unsigned long long int data type cannot store such a big number).

Finding Longest Increasing Sub Sequence in a round table of numbers

I was recently working on the following problem.
http://www.codechef.com/problems/D2
The Chef is planning a buffet for the DirectiPlex inauguration party, and everyone is invited. On their way in, each guest picks up a sheet of paper containing a random number (this number may be repeated). The guests then sit down on a round table with their friends.
The Chef now decides that he would like to play a game. He asks you to pick a random person from your table and have them read their number out loud. Then, moving clockwise around the table, each person will read out their number. The goal is to find that set of numbers which forms an increasing subsequence. All people owning these numbers will be eligible for a lucky draw! One of the software developers is very excited about this prospect, and wants to maximize the number of people who are eligible for the lucky draw. So, he decides to write a program that decides who should read their number first so as to maximize the number of people that are eligible for the lucky draw. Can you beat him to it?
Input
The first line contains t, the number of test cases (about 15). Then t test cases follow. Each test case consists of two lines:
The first line contains a number N, the number of guests invited to the party.
The second line contains N numbers a1, a2, ..., an separated by spaces, which are the numbers written on the sheets of paper in clockwise order.
Output
For each test case, print a line containing a single number which is the maximum number of guests that can be eligible for participating the the lucky draw.
Constraints
1 ≤ N ≤ 10000
You may assume that each number number on the sheet of paper; ai is randomly generated, i.e. can be with equal probability any number from an interval [0,U], where U is some upper bound (1 ≤ U ≤ 106).
Example
Input:
3
2
0 0
3
3 2 1
6
4 8 6 1 5 2
Output:
1
2
4
On checking the solutions I found this code:
#include <iostream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#define LIMIT 37
using namespace std;
struct node {
int val;
int index;
};
int N;
int binary(int number, vector<int>& ans) {
int start = 0;
int n = ans.size();
int end = n - 1;
int mid;
if (start == end)
return 0;
while (start != end) {
mid = (start + end) / 2;
if (ans[mid] == number)
break;
if (ans[mid] > number)
end = mid;
else
start = mid + 1;
}
mid = (start + end) / 2;
return mid;
}
void display(vector<int>& list) {
cout << endl;
for (int i = 0; i < list.size(); i++)
cout << list[i] << " ";
cout << endl;
}
int maxsubsequence(vector<int>& list) {
vector<int> ans;
int N = list.size();
ans.push_back(list[0]);
int i;
// display(list);
for (i = 1; i < N; i++) {
int index = binary(list[i], ans);
/*if(index+1<ans.size())
continue;*/
if (list[i] < ans[index])
ans[index] = list[i];
if (list[i] > ans[index])
ans.push_back(list[i]);
// display(ans);
}
return ans.size();
}
int compute(int index, int* g) {
vector<int> list;
list.push_back(g[index]);
int itr = (index + 1) % N;
while (itr != index) {
list.push_back(g[itr]);
itr = (itr + 1) % N;
}
return maxsubsequence(list);
}
int solve(int* g, vector<node> list) {
int i;
int ret = 1;
for (i = 0; i < min(LIMIT, (int)list.size()); i++) {
// cout<<list[i].index<<endl;
ret = max(ret, compute(list[i].index, g));
}
return ret;
}
bool cmp(const node& o1, const node& o2)
{ return (o1.val < o2.val); }
int g[10001];
int main() {
int t;
cin >> t;
while (t--) {
cin >> N;
vector<node> list;
int i;
for (i = 0; i < N; i++) {
node temp;
cin >> g[i];
temp.val = g[i];
temp.index = i;
list.push_back(temp);
}
sort(list.begin(), list.end(), cmp);
cout << solve(g, list) << endl;
}
return 0;
}
Can someone explain this to me. I am well aware of calculating LIS in nlog(n).
What I am not able to understand is this part:
int ret = 1;
for (i = 0; i < min(LIMIT, (int)list.size()); i++) {
// cout<<list[i].index<<endl;
ret = max(ret, compute(list[i].index, g));
}
and the reason behind sorting
sort(list.begin(),list.end(),cmp);
This algorithm is simply guessing at the starting point and computing the LIS for each of these guesses.
The first value in a LIS is likely to be a small number, so this algorithm simply tries the LIMIT smallest values as potential starting points.
The sort function is used to identify the smallest values.
The for loop is used to check each starting point in turn.
WARNING
Note that this algorithm may fail for certain inputs. For example, consider the sequence
0,1,2,..,49,9900,9901,...,99999,50,51,52,...,9899
The algorithm will try just the first 37 starting points and miss the best starting point at 50.
You can test this by changing the code to:
int main() {
int t;
t=1;
while (t--) {
N=10000;
vector<node> list;
int i;
for (i = 0; i < N; i++) {
node temp;
if (i<50)
g[i]=i;
else if (i<150)
g[i]=9999-150+i;
else
g[i]=i-100;
temp.val = g[i];
temp.index = i;
list.push_back(temp);
}
sort(list.begin(), list.end(), cmp);
cout << solve(g, list) << endl;
}
return 0;
}
This will generate different answers depending on whether LIMIT is 37 or 370.
In practice, for randomly generated sequences it will have a good chance of working (although I don't know how to compute the probability exactly).