Problem
I need to compute a function of an array of integers. For every three-element subset (or triplet) of the array, I need to compute the term floor((sum of triplet)/(product of triplet)). Then I need to return the sum of all such terms.
Example
Input (length; array):
5
1 2 1 7 3
Output:
6
Explanation
The following triplets exist in the given array:
1 2 1
1 2 7
1 2 3
1 1 7
1 1 3
1 7 3
2 1 7
2 1 3
2 7 3
1 7 3
Considering these triplets from the sample input:
1 2 1 contributes 2, because floor((1+2+1)/(1*2*1)) = floor(4/2) = 2
1 2 3 contributes 1
1 1 7 contributes 1
1 1 3 contributes 1
2 1 3 contributes 1
All other triplets contribute 0 to the sum.
Hence the answer is (2+1+1+1+1)=6.
My Solution
What I tried is complexity O(n^3). Code is given below:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
long t,n[300005],sum=0,mul=1,i,j,k,res=0;
cin >> t;
for(i=0;i<t;i++)
cin >>n[i];
for(i=0;i<t-2;i++)
for(j=i+1;j<t-1;j++)
for(k=j+1;k<t;k++)
{
sum = n[i]+n[j]+n[k];
mul = n[i]*n[j]*n[k];
res += floor(sum/mul);
}
cout << res << endl;
return 0;
}
Is there any hint of better optimization?
While still O(n^3), you could save some operations by caching the redundant calculations between n[i] and n[j] as you iterate over n[k].
For example:
long sum_ij,mul_ij;
for(i=0;i<t-2;i++) {
for(j=i+1;j<t-1;j++) {
sum_ij = n[i]+n[j];
mul_ij = n[i]*n[j];
for(k=j+1;k<t;k++)
{
sum = sum_ij+n[k];
mul = mul_ij*n[k];
res += floor(sum/mul);
}
}
}
Related
How can I find the prime numbers in a one-dimensional array in C++ in a simple way ??
{
int list[5];
int i,sum = 0;
for (i = 0; i < 5; i++)
{
cout << "Enter The List [" << i << "]: "; cin >> list[i];
sum = sum + list[i];
}
cout << endl;
cout << "The Sum Is:" << sum << endl;
}
Emphasizing on the comment of #john:
Create a function (say bool is_prime(int n)).
Now check if the number n is a prime or not.
So, you need to check if each of the positive integers more than 1 before n divides n or not without leaving any remainder. There's a shorter workaround, which will greatly reduce the computational cost. Just checking till the square root of the number n will do. Hence the function sqrt() is used.
So now, our is_prime() function is pretty easy to build as you can see:
bool is_prime(int n)
{
int i,p=0;
for(i=2;i<=sqrt(n);i++)
{
if(n%i==0)
{
p=1;
break; //even if one integer divides the number, then it is composite.
}
}
if(p==1)
return false; //The number is a composite.
else
return true; //The number is a prime.
}
Now, you just need to pass every value of the array into this function, and your job will be done.
Also, this program can be made even better if you check for the special case of 1 which is neither composite nor prime. A suggestion is, check your array element if it is 1 or not. If not, then pass the value in the function, else just print that it is a 1.
NOTE: The sqrt() function is available in the cmath library in C++ so you need to include that in your program too.
You can use sieve of Eratosthenes. Simply how it works is it iterates (from 2) through an boolean array and if arr[i] is prime (is true, i is the given number), sets every multiplicity to false.
Start with an array filled with true
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
first you have to set arr[0] and arr[1] to false, because these are not prime numbers
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Now, you go to 2 and set every multiplication of it to false.
in this case 4, 6, 8, 10, 12, 14 16...
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0
Then do it for 3
so 6, 9, 12, 15
Numbers 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
is prime 0 0 1 1 0 1 0 1 0 0 0 1 0 1 0 0 0
4 is not prime so you skip it
5 is prime so you do the same as for 2 and 3 (10 -> false, 15 -> false etc.)
after you use it, you can simply check if n is prime
if (arr[n] == true)
cout << n << " is prime";
else
cout << n << " is not prime";
you can find it easily on internet, for example here (there are some optimizations you can add, too)
You have n coins with certain values. Your task is to find all the money sums you can create using these coins.
Input
The first input line has an integer n: the number of coins.
The next line has n integers x1,x2,…,xn: the values of the coins.
Output
First print an integer k: the number of distinct money sums. After this, print all possible sums in increasing order.
Constraints
1≤n≤100
1≤xi≤1000
Example
Input:
4
4 2 5 2
Output:
9
2 4 5 6 7 8 9 11 13
I have written a code which works perfectly for the small inputs but gives the wrong answer to the large inputs. Please help to find the mistake and how do I correct it.
my code is:
#include <bits/stdc++.h>
using namespace std;
set<long long> s;
// Prints sums of all subsets of array
void subsetSums(long long arr[], long long n)
{
// There are totoal 2^n subsets
long long total = 1 << n;
// Consider all numbers from 0 to 2^n - 1
for (long long i = 0; i < total; i++)
{
long long sum = 0;
// Consider binary reprsentation of
// current i to decide which elements
// to pick.
for (long long j = 0; j < n; j++)
if (i & (1 << j))
sum += arr[j];
// Print sum of picked elements.
if (sum)
s.insert(sum);
}
}
// Driver code
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long n;
cin >> n;
long long arr[n];
for (long long i = 0; i < n; i++)
{
cin >> arr[i];
}
subsetSums(arr, n);
cout << s.size() << "\n";
for (auto it = s.begin(); it != s.end(); ++it)
cout << *it << " ";
return 0;
}
for example, it gives the wrong answer for
50
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
as
18
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
the correct output should be:
50
1 2 3 4 ... 50
your code is simply too slow 2^n subsets gives 1,267,650,600,228,229,401,496,703,205,376 subsets in the worst case (when n=100) while C++ does on average about 1000,000,000 operations per second.
This problem can be solved with dynamic programming, consider having an array dp of size 100001, so that dp[x] denotes if sum of x is possible to achieve.
Base case is easy - sum of 0 is possible without using any coins: dp[0]=1
Then for each coin we can try to increase existing sums by coins value to fill up our table:
for each coinValue:
for coinSum = 100000 - coinValue; coinSum >=0; coinSum--)
if(dp[coinSum])
dp[coinSum + coinValue]=1
Notice that we are looping backwards, this is done on purpose so that each coin gets used only once.
Complexity: O(n^2*maxCoinValue)
Your algorithm is poor, but the reason you're getting wrong results is because you're overflowing int. long long total = 1<<n; shifts an int left by n places, and the fact you're assigning the result to a long long is irrelevant.
You can find problems like this using ubsan. Here's a reproduction of your problem, including warning messages from ubsan:
$ clang++ -fsanitize=undefined a.cpp -o a && ./a
50
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
a.cpp:11:25: runtime error: shift exponent 50 is too large for 32-bit type 'int'
a.cpp:22:24: runtime error: shift exponent 32 is too large for 32-bit type 'int'
18
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
Having difficulty finding an explanation to this.
What does this code do? I understand it creates an array of vector but that's about it.
How can I print the vector array and access elements to experiment with it?
#define MAXN 300009
vector<int>dv[MAXN];
int main()
{
for(int i=1;i<MAXN;i++)
for(int j=i;j<MAXN;j+=i)
dv[j].push_back(i);
}
The code is easy enough to instrument. The reality of what it ends up producing is a very simple (and very inefficient) Sieve of Eratosthenes. Understanding that algorithm, you'll see what this code does to produce that ilk.
Edit: It is also a factor-table generator. See Edit below.
Instrumenting the code and dumping output afterward, and reducing the number of loops for simplification we have something like the following code. We use range-based-for loops for enumerating over each vector in the array of vectors:
#include <iostream>
#include <vector>
#define MAXN 20
std::vector<int>dv[MAXN];
int main()
{
for(int i=1;i<MAXN;i++)
{
for(int j=i;j<MAXN;j+=i)
dv[j].push_back(i);
}
for (auto const& v : dv)
{
for (auto x : v)
std::cout << x << ' ';
std::cout << '\n';
}
}
The resulting output is:
1
1 2
1 3
1 2 4
1 5
1 2 3 6
1 7
1 2 4 8
1 3 9
1 2 5 10
1 11
1 2 3 4 6 12
1 13
1 2 7 14
1 3 5 15
1 2 4 8 16
1 17
1 2 3 6 9 18
1 19
Now, note each vector that only has two elements (1 and an additional number). That second number is prime. In our test case those two-element vectors are:
1 2
1 3
1 5
1 7
1 11
1 13
1 17
1 19
In short, this is a very simple, and incredibly inefficient way of finding prime numbers. A slight change to the output loops to only output the second element of all vectors of length-two-only will therefore generate all the primes lower than MAXN. Therefore, using:
for (auto const& v : dv)
{
if (v.size() == 2)
std::cout << v[1] << '\n';
}
We will get all primes from [2...MAXN)
Edit: Factor Table Generation
If it wasn't obvious, each vector has an ending element (that not-coincidentally also lines up with the subscripts of the outer array). All preceding elements make up the positive factors of that number. For example:
1 2 5 10
is the dv[10] vector, and tells you 10 has factors 1,2,5,10. Likewise,
1 2 3 6 9 18
is the dv[18] vector, and tells you 18 has factors 1,2,3,6,9,18.
In short, if someone wanted to know all the factors of some number N that is < MAXN, this would be a way of putting all that info into tabular form.
I am given this problem:
We are going over recursion in my class and I do not quite understand it, I was wondering if someone can help me with this problem
let c(n) be the number of different group integers that can be chosen from the integers 1 through n-1, so that the integers in each group add up to n (for example, n=4=[1+1+1+1]=[1+1+2]=[2+2]). Write a recursive definition for c(n) under the following variations:
a) You count permutations. For example, 1,2,1 and 1,1,2 are two groups that each add up to 4
b)you ignore permutations
I know permutations is how many ways you can arrange a set of numbers, so is my code below correct? I get an answer of 7?
Here is my code for part a:
int recurse (int n);
int main(){
int a=4;
int sum_perm;
sum_perm=recurse(a);
cout<<sum_perm-1<<endl;
//Can I do -1 here because it should be from a group of integers from 1 to n-1?
return 0;
}
int recurse(int n)
{
int sum = 1;
if (n == 1){
return 1;
}
for(int i = 1; i < n; i++){
sum += recurse(n - i);
}
return sum;
}
For part B, if I am not counting permutations, what am I counting?
Here is my code for part b:
int without (int n, int max);
int main(){
int a=4, b =3;
int sum_without;
sum_without=without(a,b);
cout<<sum_without<<endl;
system("Pause");
return 0;
}
int without(int n, int max)
{
if(n == 1 || max == 1){
return 1;
}
else if (n == max){
return 1 + without(n, n-1);
}
else{
return without(n,max-1) + without(n-max, max);
}
}
You don't show any code to generate the combinations of numbers that produce a sum. Link to wiki article about partitions .
In this case, the goal is to count the number of combinations and/or permutations, which might be possible without actually generating a set of combinations. Not sure if recursion helps here, but you can convert any loop into recursion if you pass enough variables.
Example "partitions"
1 combination that sums to 1:
1
2 combinations that sum to 2:
1 1
2
3 combinations that sum to 3:
1 1 1
1 2
3
5 combinations that sum to 4:
1 1 1 1
1 1 2
1 3
2 2
4
7 combinations that sum to 5:
1 1 1 1 1
1 1 1 2
1 1 3
1 2 2
1 4
2 3
5
11 combinations of numbers that sum to 6:
1 1 1 1 1 1
1 1 1 1 2
1 1 1 3
1 1 2 2
1 1 4
1 2 3
2 2 2
1 5
2 4
3 3
6
I would recommend combinations directly being considered. While it seems like the more difficult case a simple rule makes it trivial.
Numbers calculated are in decreasing order
This requires you to track the last number, but ensures you don't calculate 1 5 and then 5 1, as the former is impossible.
I'm trying to do the following in CUSP:
A=[
1,1,0,0;
2,2,2,0;
0,3,3,3;
0,0,4,4];
B=[1,1,1,1]';
disp(mldivide(A,B));
which is
X=[0.9167,0.0833,-0.5000,0.7500]
On the other hand I get a strange answer from CUSP
#include <cusp/dia_matrix.h>
#include <cusp/krylov/cg.h>
#include <cusp/print.h>
int main()
{
cusp::dia_matrix<int,float,cusp::host_memory> A(4,4,10,3);
A.diagonal_offsets[0] = -1;
A.diagonal_offsets[1] = 0;
A.diagonal_offsets[2] = 1;
for (int i = 0;i <3;i++)
{
for (int q = 0 ;q < A.num_cols;q++)
{
A.values(q,i)=q+1;
}
}
//copy
cusp::dia_matrix<int,float,cusp::device_memory> AA = A;
cusp::array1d<float,cusp::device_memory> BB(A.num_rows,1);
cusp::array1d<float,cusp::device_memory> XX(A.num_rows,0);
cusp::print(AA);
cusp::print(XX);
cusp::print(BB);
cusp::krylov::cg(AA,XX,BB);\
cusp::print(XX);
return 0;
}
The result looks like
sparse matrix <4, 4> with 10 entries
0 0 1
0 1 1
1 0 2
1 1 2
1 2 2
2 1 3
2 2 3
2 3 3
3 2 4
3 3 4
array1d <4>
0
0
0
0
array1d <4>
1
1
1
1
array1d <4>
-39.9938
-53.436
87.9025
-30.1429
The last one doesn't look quite right. Anybody know what I'm doing wrong? Am I using the code wrong or are we supposed to have a really good guessed solution + use a preconditioner?
The conjugate gradient method is only valid for use in symmetric positive definite matrices. Your matrix isn't symmetric. That is why it isn't (and cannot) producing a valid solution. Either use an appropriate, well conditioned SPD matrix, or use a different numerical method.