#include <iostream>
using namespace std;
int main() {
int n;
cout<<"The table of any number you want to print"<<endl;
cin>>n;
for (int i = 1; i <=10; ++i)
{
cout<<n<<"*"<<i<<"="<<n*n<<endl;
}
return 0;
}
I am finding the solution of the problem.
and i just want to knnow that why it is not printing the value.
I think the for loop that prints the multiplication table should use i to calculate the product rather than n
Try :
cout<<n<<"*"<<i<<"="<<n*i<<endl;
Related
I'm new to c++ but long story short , i want to write a c++ program that will accept input from user through an array and it will sum each array element input
#include <iostream>
#include <string>
#include <math.h>
using namespace std;
int main() {
int array[100];
int sum;
for(int i=0; i<100; i++){
cout<<"Insert element "<<i<<": ";
cin>>array[i];
sum = array[i]+ //summ with the next array input;
cout<<sum;
}
return 0;
}
means that if i enter any integer the program should be able to give the summation of the inputs in sequence from the first input to the last input
C++ standard library already has a function to do this which is std::accumulate:
#include <iostream>
#include <numeric>
int main() {
int array[5] = {1,2,3,4,5};
int total = std::accumulate(std::begin(array), std::end(array), 0);
return 0;
}
If you plan to use not a full array you should use std::begin(array), std::begin(array) + amount as ranges.
initialize your sum var to 0 initially and write sum+=array[i] instead of what you have written, there is also a limitation in this program as value of sum after all user input should be <=10^9 as int datatype store no. approximately upto 10^9 so take note of this fact also. Write cout<<sum<<endl; instead to be able to distinguish till previous input sum to the new input sum.
Hope this will help.
You can use a do while loop:
#include<iostream>
#include<string>
#include<math.h>
int main()
{
int array[100];
int sum=0;
int i=0;
std::cout<<"Insert element"<<" "<<i<<": ";
std::cin>>array[i];
do
{
sum=sum+array[i];
std::cout<<sum<<std::endl;
i++;
std::cout<<"Insert element"<<" "<<i<<": ";
}while(std::cin>>array[i]);
return 0;
}
If all you want is the sum then just compute the sum. No need to store anything:
#include <iostream>
#include <string>
#include <math.h>
int main() {
int sum = 0;
for(int i=0; i<100; i++){
std::cout << "Insert element " << i << ": ";
int t;
std::cin >> t;
sum += t
std::cout << sum;
}
}
#include <iostream>
using namespace std;
int main()
{
int n;
n=4;
int arr[n]={1,2,3,8};
int sum;
sum=5;
int curr=0;
cin>>sum;
for(int i=0;i<n;i++){
for(int j=i;j<n;j++){
curr+=arr[j];
if(curr==sum){
cout<<i;
}
cout<<curr<<endl;
}
}
}
For the given question I need to find the starting and ending index of such a subarray. I have tried the above code but am not able to get the right output. Please guide me.
I think your code only needs some minor modifications. You should add
some code to handle the case where your running sum is greater than the target sum, and you should also re-initialize your running sum correctly.
There may be some efficient solution that is faster than O(n^2), which I am not aware of yet. If someone knows of a solution with a better time complexity, please share with us.
Below is a simple algorithm that has the time complexity of O(n^2). (It may not have the most efficient time complexity for this problem).
This function prints out the 2 indices of the array. The sum of all elements between these 2 indices inclusively will equal the target sum.
void Print_Index_of_2_Elements(int array[], int total_element, int target)
{
// Use Brute force . Time complexity = O(n^2)
for (int i = 0; i < total_element; i++)
{
int running_sum = array[i];
// Second for loop
for (int j = (i + 1) ; j < total_element; j++)
{
if (running_sum == target)
{
cout << "Two indices are: " << i << " and " << j;
return; // Found Answer. Exit.
}
else if ( running_sum > target )
break;
else // running_sum < target
running_sum += array[j];
}
}
cout << " Sorry - no answer was found for the target sum.";
}
If you are someone that is a beginner in subarrays or arrays for the case. Then this code is for you:
#include <iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
int sum;
cin>>sum;
int curr=0;
for(int i=0;i<n;i++){
for(int j=i;j<n;j++){
if(curr==sum){
cout<<i+1<<" "<<j;
return 0;
}
else if (curr>sum){
curr=0;
}
else if(curr<sum){
curr+=arr[j];
}
}
}
return 0;
}
If you have any doubts regarding this, feel free to comment and let me know.
I have been trying to solve this simple problem in C++ but every time I submit, it says wrong answer. I am pretty sure I have got the logic right. Any help is appreciated.
Question: Find the sum of distances between the inputted numbers.
Ex. Input: 2 5 8 2 1
Distance=2+2+5+0
=9, (1 < n < 1000000)
PS: Input can't have the same number consecutively.
PSS: Subtask two is giving Wrong Answer
Code:
#include <iostream>
using namespace std;
int main() {
// your code goes here
int t,a[100000],n,sum=0;
cin>>t;
for(int i=0;i<t;i++)
{
cin>>n;
for(int j=0;j<n;j++)
{
cin>>a[j];
}
for(int j=0;j<n-1;j++)
{
if(a[j]!=a[j+1])
{
sum = sum + abs(a[j]-a[j+1])-1;
}
}
cout<<sum<<endl;
sum=0;
}
}
The problem with your code is that you are using int type for sum whose maximum value (1E11) can exceed the upper limit of int(if it's 32-bit or less). Use long long(atleast 64-bit) instead to store your sum.
Well, you can also optimize the code because you don't exactly need an array of 100000 integers and store the values in it. You can do so using only two variables.
Here is a modified implementation of your logic:
#include <iostream>
int main() {
int t, n, first, second;
long long sum; // or better use std::int_fast64_t sum;
std::cin >> t;
while (t--) {
sum = 0;
std::cin >> n >> first;
for (int i = 0; i < n - 1; ++i) {
std::cin >> second;
sum += std::abs(first - second) - 1;
first = second;
}
std::cout << sum << std::endl;
}
}
PS: In competitive coding checking the provided constraints like if(a[j]!=a[j+1]) is useless. The problem statement simply guarantees it that it will never be false.
I am writing a code where 2d matrix array is given and by choosing 1 element from each row you must output the smallest sums. Sums as in you must give n number of minimum sums
#include<iostream>
#include<math.h>
using namespace std;
int main() {
int n;
cin>>n;
int hist[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>hist[i][j];
}
}
int num=pow(n,n);
int sum[num];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
sum[i]=sum[i]+hist[i][j];
}
}
for(int i=0;i<n;i++){
cout<<sum[i]<<" ";
}
return 0;
}
example input would be:
3
1 8 5
9 2 5
10 7 6
The output will be
9 10 12
since 1+2+6=9; 1+2+7=10; 1+2+10
The main problem I am facing would be that I can't find the lowest sum or even the sums I tried to brute force it put it won't work.
Could you help me fix the code so that at least I could find the sums?
Many problems with your code (it's not even legal C++). But the problem that is causing your current question is that you must initialise sum to zero. at the moment you have garbage values in sum.
int sum[num] = {0};
Some other issues
int num=pow(n,n);
This calculates n to the power of n, but there are only n squared sums. So this would be better
int sum[n*n] = {0};
But the big issue, the issue that makes your code illegal C++, is that in C++ array dimensions must be compile time constants not variables. So this
int hist[n][n];
and this
int sum[num];
are not legal C++. They are legal in C, which is why your compiler is accepting them, but not every C++ compiler would. Since you are trying to write C++ code you should use a vector. Here's your code rewritten to use vectors.
#include <vector>
using std::vector;
...
vector<vector<int>> hist(n, vector<int>(n));
...
vector<int> sum(num, 0);
...
That's it nothing else needs to change.
Instead of brute forcing, why not realize that the smallest path is simply the smallest element of each row and the second smallest path is the smallest element of the first n-1 rows, and the second smallest element of n.
You can elegantly express this by sorting the rows of the matrix first and then keeping a counter of where you are at each row:
#include <algorithm>
#include <iostream>
#include <vector>
struct path {
path(int n) : n(n), indexes(n) {}
// Add one to last row index, then carry over to previous rows.
path& operator ++() {
indexes.back()++;
for (int i = n-1; i >= 0; i--) {
if (indexes[i] == n) {
indexes[i] = 0;
indexes[i-1]++;
} else {
break;
}
}
return this;
}
int n;
std::vector<int> indexes;
};
Now your problem is as simple as:
int main() {
int n;
cin>>n;
std::vector<std::vector<int>> hist(n, std::vector<int>(n));
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>hist[i][j];
}
// sort each row after reading
std::sort(hist[i].begin(), hist[i].end());
}
int num_minimum_sums = n;
path p(n);
while (num_minimum_sums-- > 0) {
int sum = 0;
for (int i = 0; i < n; i++) {
sum += hist[i][p.indexes[i]];
}
std::cout << sum << std::endl;
++p;
}
}
My code is showing TLE when there when number of entries in array(n) is 1 *10^5. What should i do? I saw the submission status and in all cases it ran fine except in the last case when n is 100 000, it shows time limit error.
Problem:: https://codeforces.com/contest/474/problem/B
My solution: https://pastebin.com/5RLBirpF
Code:
#include <iostream>
#include<bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t;
cin>>t;
int arr[t];
int arrfreq[t];
int sum=1;
for(int i=0;i<t;i++ )
{
cin>>arr[t];
sum+=arr[t];
arrfreq[i]=sum;
}
int m;
cin>>m;
int qsn[m];
int k;
for(int i =0;i<m;i++)
{
cin>>qsn[i];
}
for(int j =0;j<t;j++)
{
if(k<arrfreq[j])
{
cout<<j+1<<"\n";
break;
}
if(k==arrfreq[j])
{
cout<<j+2<<"\n";
break;
}
}
}
You are using one loop for calculating the number of pile. What you can do instead is find the answer in the same loop in which you are taking the input for qsn. Just take the input of qsn and find the pile number in that loop itself. That will reduce your code's time complexity and remove the TLE error.
I saw other solutions on the internet and found out that std::lower_bound function is what has to be used to avoid T.L.E.
But one thing, as mentioned on internet, this function returns a pointer to the lower bound, then , why are we subtracting c from it and then +1 also(in the 2nd last cout line)??(refer to the code.)
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,i,m,j,cnt=0,ans,x,sum=0;
cin>>n;
int a[n],c[n];
for(i=0; i<n; i++){
cin>>a[i];
sum+=a[i];
c[i]=sum;
}
cin>>m;
int b[m];
for(i=0; i<m; i++) cin>>b[i];
for(j=0; j<m; j++)
{
cout<<lower_bound(c,c+n,b[j])-c+1<<endl;
}
return 0;
}