Related
I'm trying to figure out, how can I calculate fft row by row of the following matrix (stored as an array) using fftw3:
double signal[9] = {
1, 2, 3,
4, 5, 6,
7, 8, 9
}
to obtain the matrix with rows (pseudo-code)
fft([signal[0], signal[1], signal[2]])
fft([signal[3], signal[4], signal[5]])
fft([signal[6], signal[7], signal[8]])
I can check the result using Python just applying np.fft.fft([[1, 2, 3], [4, 5, 6], [7, 8, 9]], axis=1) and I want to get the same result with fftw3.
My code:
#include <fftw3.h>
#include <iostream>
int main()
{
// it is needed to compute fft row-by-row of the following matrix
double signal[9] {
1, 2, 3,
4, 5, 6,
7, 8, 9
}; // 3x3 matrix
// so that the output will have the form (pseudo-code)
// fft(signal[0], signal[1], signal[2])
// fft(signal[3], signal[4], signal[5])
// fft(signal[6], signal[7], signal[8])
fftw_complex result[9]{ };
// prepare parameters
int rank = 1;
int n[] = { 3 };
int howmany = 3;
int idist = 3;
int odist = 3;
int istride = 1;
int ostride = 1;
auto plan_ = fftw_plan_many_dft_r2c(
rank, // 1D problem
n, // 1D transforms of length n[0] = 3
howmany, // 3 1D transforms
signal, // input matrix (array)
NULL, // will be assigned to n
istride, // distance between two elements in the same row
idist, // distance between the first elements of neighboring rows
result, // output
NULL, // will be assigned to n
ostride, // distance between two elements in the same row
odist, // distance between the firsn elements of neighboring rows
FFTW_ESTIMATE
);
fftw_execute(plan_);
for (int i = 0; i < 3; ++i)
{
for (int j = 0; j < 3; ++j)
{
std::cout << result[i * 3 + j][0] << " + 1j*" << result[i * 3 + j][1] << '\t';
}
std::cout << '\n';
}
fftw_destroy_plan(plan_);
return 0;
}
The corresponding explanations are in the code. Python gives the following answer:
array([[ 6. +0.j , -1.5+0.8660254j, -1.5-0.8660254j],
[15. +0.j , -1.5+0.8660254j, -1.5-0.8660254j],
[24. +0.j , -1.5+0.8660254j, -1.5-0.8660254j]])
My code gives the following
6 + 1j*0 -1.5 + 1j*0.866025 0 + 1j*0
15 + 1j*0 -1.5 + 1j*0.866025 0 + 1j*0
24 + 1j*0 -1.5 + 1j*0.866025 0 + 1j*0
I see that the output almost the same as with Python, but third elements of each row are zeros. Can someone help me?
Thank you.
For example, 16 can be expressed in these ways:
2*2*2*2
4*2*2
4*4
8*2
(excluding the trivial case 1*16)
36:
2*2*3*3
2*2*9
2*18
4*3*3
12*3
4*9
Therefore, I need to find all the representations of any given number just like above.
To start, I was able to come up with all the divisors of a given number with the codes below:
void FindDivisors(int n)
{
vector<int> my_vector;
for (int i = 2; i < n/2 + 1; i++)
{
if (n % i == 0)
my_vector.push_back(i); //store divisors in a vector.
}
}
After this, I got stuck. How do I get from all the divisors to all the factorizations wanted?
I can think of a recursion method that seems to work but do not exactly know how to implement it:
Take 16 as the example. Let's name the function to find all the representations of n to be "Factor(n)". All non-trivial divisors of 16 are {2,4,8}. We divide 16 by each of its divisor and get {8, 4, 2}. Then Factor(16) = the union of 2*Factor(8), 4*Factor(4) and 8*Factor(8). However, this is as far as I got. I am new to recursion and not sure if this route works out. I need some help on how to put things together.
As long as you don't know the number of factors, the most effective solution is recursion.
You will need a vector to store the current path:
vector<int> path; // it will store all current factors
and a recursive function.
For every iteration you try to divide current remainder and remember:
void OutputRec(int value)
{
if (value == 1) // We have divided the whole number
{
for (int i = 0; i < path.size(); i++) cout << path[i] << " ";
cout << endl;
return;
}
for (int i = value; i > 1; i--)
{
if (value % i == 0) { // if it is divisible
path.push_back(i); // add to path
OutputRec(i, value / i); // continue working
path.pop_back(); // remove from path
}
}
}
void Output(int value)
{
cout << "Result for " << value << ": " << endl;
path = vector<int>();
OutputRec(value);
}
int main() {
Output(4);
Output(16);
Output(36);
Output(256);
return 0;
}
For example, let we have 8. Here is how it works:
OutputRec(8), path = []:
i = 8: divisible (8 / 1 = 1)
OutputRec(1), path = [8], value == 1 > output "8".
i = 7, 6, 5: not divisible
i = 4: divisible (8 / 4 = 2)
OutputRec(2), path = [4]:
i2 = 2: divisible (2 / 2 = 1)
OutputRec(1), path = [4 2], value == 1 > output "4 2".
i = 3: not divisible
i = 2: divisible (8 / 2 = 4)
OutputRec(4), path = [2]:
i3 = 4: divisible (4 / 4 = 1)
OutputRec(1), path = [2 4], value == 1 > output "2 4".
i3 = 3: not divisible
i3 = 2: divisible (4 / 2 = 2)
OutputRec(2), path = [2 2]:
i4 = 2: divisible (2 / 2 = 1)
OutputRec(1), path = [2 2 2], value == 1 > output "2 2 2"
Result:
8,
4 * 2,
2 * 4,
2 * 2 * 2
Now, you see the problem: [2 4] and [4 2] is duplicated answer. How can we avoid duplicating answers? The easiest approach is to output values only in ascending (descending, does not matter) order.
Let's add max variable and store the last number inserted into a path and find only those dividers which are less or equal to it.
For example, if current path is [2], then the recursion shouldn't go for [2 4], but should go for [2 2]. The initial value of max is equal to value as the first number in the path can be any.
void OutputRec(int max, int value)
{
if (value == 1) {
for (int i = 0; i < path.size(); i++) cout << path[i] << " ";
if (path.size() == 1) cout << "1";
cout << endl;
return;
}
for (int i = max; i > 1; i--) // loop from max. Min(max, value) would be better
{
if (value % i == 0) {
path.push_back(i);
OutputRec(i, value / i);
path.pop_back();
}
}
}
void Output(int value)
{
cout << "Result for " << value << ": " << endl;
path = vector<int>();
OutputRec(value, value);
}
Here is how it works now:
OutputRec(8), path = []:
i = 8: divisible (8 / 8 = 1)
OutputRec(1), path = [8], value == 1 > output "8".
i = 7, 6, 5: not divisible
i = 4: divisible (8 / 4 = 2)
OutputRec(2), path = [4]:
i2 = 2: divisible
OutputRec(1), path = [4 2], value == 1 > output "4 2".
i = 3: not divisible
i = 2: divisible (8 / 2 = 4)
OutputRec(4), path = [2]:
// notice i3 starts from 2, because last inserted is 2
i3 = 2: divisible (4 / 2 = 2)
OutputRec(2), path = [2 2]:
i4 = 2: divisible (2 / 2 = 1)
OutputRec(1), path = [2 2 2], value == 1 > output "2 2 2"
Result:
8,
4 * 2,
2 * 2 * 2
Everything works fine! The only thing which looks incorrect is "8". It probably should be "8 * 1". You can add a line like if (path.size() == 1) cout << "1"; to the output.
Result:
8 * 1,
4 * 2,
2 * 2 * 2
Here is the working Ideone demo.
I hope that I didn't confuse you even more. It is really difficult to explain what recursion is at the first time.
Recursion is like swimming or cycling - it looks difficult first. Then, you try it and learn it, and once you understand it - you will wonder why couldn't you do this before :) Good luck.
Can anybody find any potentially more efficient algorithms for accomplishing the following task?:
For any given permutation of the integers 0 thru 7, return the index which describes the permutation lexicographically (indexed from 0, not 1).
For example,
The array 0 1 2 3 4 5 6 7 should return an index of 0.
The array 0 1 2 3 4 5 7 6 should return an index of 1.
The array 0 1 2 3 4 6 5 7 should return an index of 2.
The array 1 0 2 3 4 5 6 7 should return an index of 5039 (that's 7!-1 or factorial(7)-1).
The array 7 6 5 4 3 2 1 0 should return an index of 40319 (that's 8!-1). This is the maximum possible return value.
My current code looks like this:
int lexic_ix(int* A){
int value = 0;
for(int i=0 ; i<7 ; i++){
int x = A[i];
for(int j=0 ; j<i ; j++)
if(A[j]<A[i]) x--;
value += x*factorial(7-i); // actual unrolled version doesn't have a function call
}
return value;
}
I'm wondering if there's any way I can reduce the number of operations by removing that inner loop, or if I can reduce conditional branching in any way (other than unrolling - my current code is actually an unrolled version of the above), or if there are any clever bitwise hacks or filthy C tricks to help.
I already tried replacing
if(A[j]<A[i]) x--;
with
x -= (A[j]<A[i]);
and I also tried
x = A[j]<A[i] ? x-1 : x;
Both replacements actually led to worse performance.
And before anyone says it - YES this is a huge performance bottleneck: currently about 61% of the program's runtime is spent in this function, and NO, I don't want to have a table of precomputed values.
Aside from those, any suggestions are welcome.
Don't know if this helps but here's an other solution :
int lexic_ix(int* A, int n){ //n = last index = number of digits - 1
int value = 0;
int x = 0;
for(int i=0 ; i<n ; i++){
int diff = (A[i] - x); //pb1
if(diff > 0)
{
for(int j=0 ; j<i ; j++)//pb2
{
if(A[j]<A[i] && A[j] > x)
{
if(A[j]==x+1)
{
x++;
}
diff--;
}
}
value += diff;
}
else
{
x++;
}
value *= n - i;
}
return value;
}
I couldn't get rid of the inner loop, so complexity is o(n log(n)) in worst case, but o(n) in best case, versus your solution which is o(n log(n)) in all cases.
Alternatively, you can replace the inner loop by the following to remove some worst cases at the expense of another verification in the inner loop :
int j=0;
while(diff>1 && j<i)
{
if(A[j]<A[i])
{
if(A[j]==x+1)
{
x++;
}
diff--;
}
j++;
}
Explanation :
(or rather "How I ended with that code", I think it is not that different from yours but it can make you have ideas, maybe)
(for less confusion I used characters instead and digit and only four characters)
abcd 0 = ((0 * 3 + 0) * 2 + 0) * 1 + 0
abdc 1 = ((0 * 3 + 0) * 2 + 1) * 1 + 0
acbd 2 = ((0 * 3 + 1) * 2 + 0) * 1 + 0
acdb 3 = ((0 * 3 + 1) * 2 + 1) * 1 + 0
adbc 4 = ((0 * 3 + 2) * 2 + 0) * 1 + 0
adcb 5 = ((0 * 3 + 2) * 2 + 1) * 1 + 0 //pb1
bacd 6 = ((1 * 3 + 0) * 2 + 0) * 1 + 0
badc 7 = ((1 * 3 + 0) * 2 + 1) * 1 + 0
bcad 8 = ((1 * 3 + 1) * 2 + 0) * 1 + 0 //First reflexion
bcda 9 = ((1 * 3 + 1) * 2 + 1) * 1 + 0
bdac 10 = ((1 * 3 + 2) * 2 + 0) * 1 + 0
bdca 11 = ((1 * 3 + 2) * 2 + 1) * 1 + 0
cabd 12 = ((2 * 3 + 0) * 2 + 0) * 1 + 0
cadb 13 = ((2 * 3 + 0) * 2 + 1) * 1 + 0
cbad 14 = ((2 * 3 + 1) * 2 + 0) * 1 + 0
cbda 15 = ((2 * 3 + 1) * 2 + 1) * 1 + 0 //pb2
cdab 16 = ((2 * 3 + 2) * 2 + 0) * 1 + 0
cdba 17 = ((2 * 3 + 2) * 2 + 1) * 1 + 0
[...]
dcba 23 = ((3 * 3 + 2) * 2 + 1) * 1 + 0
First "reflexion" :
An entropy point of view. abcd have the fewest "entropy". If a character is in a place it "shouldn't" be, it creates entropy, and the earlier the entropy is the greatest it becomes.
For bcad for example, lexicographic index is 8 = ((1 * 3 + 1) * 2 + 0) * 1 + 0 and can be calculated that way :
value = 0;
value += max(b - a, 0); // = 1; (a "should be" in the first place [to create the less possible entropy] but instead it is b)
value *= 3 - 0; //last index - current index
value += max(c - b, 0); // = 1; (b "should be" in the second place but instead it is c)
value *= 3 - 1;
value += max(a - c, 0); // = 0; (a "should have been" put earlier, so it does not create entropy to put it there)
value *= 3 - 2;
value += max(d - d, 0); // = 0;
Note that the last operation will always do nothing, that's why "i
First problem (pb1) :
For adcb, for example, the first logic doesn't work (it leads to an lexicographic index of ((0* 3+ 2) * 2+ 0) * 1 = 4) because c-d = 0 but it creates entropy to put c before b. I added x because of that, it represents the first digit/character that isn't placed yet. With x, diff cannot be negative.
For adcb, lexicographic index is 5 = ((0 * 3 + 2) * 2 + 1) * 1 + 0 and can be calculated that way :
value = 0; x=0;
diff = a - a; // = 0; (a is in the right place)
diff == 0 => x++; //x=b now and we don't modify value
value *= 3 - 0; //last index - current index
diff = d - b; // = 2; (b "should be" there (it's x) but instead it is d)
diff > 0 => value += diff; //we add diff to value and we don't modify x
diff = c - b; // = 1; (b "should be" there but instead it is c) This is where it differs from the first reflexion
diff > 0 => value += diff;
value *= 3 - 2;
Second problem (pb2) :
For cbda, for example, lexicographic index is 15 = ((2 * 3 + 1) * 2 + 1) * 1 + 0, but the first reflexion gives : ((2 * 3 + 0) * 2 + 1) * 1 + 0 = 13 and the solution to pb1 gives ((2 * 3 + 1) * 2 + 3) * 1 + 0 = 17. The solution to pb1 doesn't work because the two last characters to place are d and a, so d - a "means" 1 instead of 3. I had to count the characters placed before that comes before the character in place, but after x, so I had to add an inner loop.
Putting it all together :
I then realised that pb1 was just a particular case of pb2, and that if you remove x, and you simply take diff = A[i], we end up with the unnested version of your solution (with factorial calculated little by little, and my diff corresponding to your x).
So, basically, my "contribution" (I think) is to add a variable, x, which can avoid doing the inner loop when diff equals 0 or 1, at the expense of checking if you have to increment x and doing it if so.
I also checked if you have to increment x in the inner loop (if(A[j]==x+1)) because if you take for example badce, x will be b at the end because a comes after b, and you will enter the inner loop one more time, encountering c. If you check x in the inner loop, when you encounter d you have no choice but doing the inner loop, but x will update to c, and when you encounter c you will not enter the inner loop. You can remove this check without breaking the program
With the alternative version and the check in the inner loop it makes 4 different versions. The alternative one with the check is the one in which you enter the less the inner loop, so in terms of "theoretical complexity" it is the best, but in terms of performance/number of operations, I don't know.
Hope all of this helps (since the question is rather old, and I didn't read all the answers in details). If not, I still had fun doing it. Sorry for the long post. Also I'm new on Stack Overflow (as a member), and not a native speaker, so please be nice, and don't hesitate to let me know if I did something wrong.
Linear traversal of memory already in cache really doesn't take much times at all. Don't worry about it. You won't be traversing enough distance before factorial() overflows.
Move the 8 out as a parameter.
int factorial ( int input )
{
return input ? input * factorial (input - 1) : 1;
}
int lexic_ix ( int* arr, int N )
{
int output = 0;
int fact = factorial (N);
for ( int i = 0; i < N - 1; i++ )
{
int order = arr [ i ];
for ( int j = 0; j < i; j++ )
order -= arr [ j ] < arr [ i ];
output += order * (fact /= N - i);
}
return output;
}
int main()
{
int arr [ ] = { 11, 10, 9, 8, 7 , 6 , 5 , 4 , 3 , 2 , 1 , 0 };
const int length = 12;
for ( int i = 0; i < length; ++i )
std::cout << lexic_ix ( arr + i, length - i ) << std::endl;
}
Say, for a M-digit sequence permutation, from your code, you can get the lexicographic SN formula which is something like: Am-1*(m-1)! + Am-2*(m-2)! + ... + A0*(0)! , where Aj range from 0 to j. You can calculate SN from A0*(0)!, then A1*(1)!, ..., then Am-1 * (m-1)!, and add these together(suppose your integer type does not overflow), so you do not need calculate factorials recursively and repeatedly. The SN number is a range from 0 to M!-1 (because Sum(n*n!, n in 0,1, ...n) = (n+1)!-1)
If you are not calculating factorials recursively, I cannot think of anything that could make any big improvement.
Sorry for posting the code a little bit late, I just did some research, and find this:
http://swortham.blogspot.com.au/2011/10/how-much-faster-is-multiplication-than.html
according to this author, integer multiplication can be 40 times faster than integer division. floating numbers are not so dramatic though, but here is pure integer.
int lexic_ix ( int arr[], int N )
{
// if this function will be called repeatedly, consider pass in this pointer as parameter
std::unique_ptr<int[]> coeff_arr = std::make_unique<int[]>(N);
for ( int i = 0; i < N - 1; i++ )
{
int order = arr [ i ];
for ( int j = 0; j < i; j++ )
order -= arr [ j ] < arr [ i ];
coeff_arr[i] = order; // save this into coeff_arr for later multiplication
}
//
// There are 2 points about the following code:
// 1). most modern processors have built-in multiplier, \
// and multiplication is much faster than division
// 2). In your code, you are only the maximum permutation serial number,
// if you put in a random sequence, say, when length is 10, you put in
// a random sequence, say, {3, 7, 2, 9, 0, 1, 5, 8, 4, 6}; if you look into
// the coeff_arr[] in debugger, you can see that coeff_arr[] is:
// {3, 6, 2, 6, 0, 0, 1, 2, 0, 0}, the last number will always be zero anyway.
// so, you will have good chance to reduce many multiplications.
// I did not do any performance profiling, you could have a go, and it will be
// much appreciated if you could give some feedback about the result.
//
long fac = 1;
long sn = 0;
for (int i = 1; i < N; ++i) // start from 1, because coeff_arr[N-1] is always 0
{
fac *= i;
if (coeff_arr[N - 1 - i])
sn += coeff_arr[N - 1 - i] * fac;
}
return sn;
}
int main()
{
int arr [ ] = { 3, 7, 2, 9, 0, 1, 5, 8, 4, 6 }; // try this and check coeff_arr
const int length = 10;
std::cout << lexic_ix(arr, length ) << std::endl;
return 0;
}
This is the whole profiling code, I only run the test in Linux, code was compiled using G++8.4, with '-std=c++11 -O3' compiler options. To be fair, I slightly rewrote your code, pre-calculate the N! and pass it into the function, but it seems this does not help much.
The performance profiling for N = 9 (362,880 permutations) is:
Time durations are: 34, 30, 25 milliseconds
Time durations are: 34, 30, 25 milliseconds
Time durations are: 33, 30, 25 milliseconds
The performance profiling for N=10 (3,628,800 permutations) is:
Time durations are: 345, 335, 275 milliseconds
Time durations are: 348, 334, 275 milliseconds
Time durations are: 345, 335, 275 milliseconds
The first number is your original function, the second is the function re-written that gets N! passed in, the last number is my result. The permutation generation function is very primitive and runs slowly, but as long as it generates all permutations as testing dataset, that is alright. By the way, these tests are run on a Quad-Core 3.1Ghz, 4GBytes desktop running Ubuntu 14.04.
EDIT: I forgot a factor that the first function may need to expand the lexi_numbers vector, so I put an empty call before timing. After this, the times are 333, 334, 275.
EDIT: Another factor that could influence the performance, I am using long integer in my code, if I change those 2 'long' to 2 'int', the running time will become: 334, 333, 264.
#include <iostream>
#include <vector>
#include <chrono>
using namespace std::chrono;
int factorial(int input)
{
return input ? input * factorial(input - 1) : 1;
}
int lexic_ix(int* arr, int N)
{
int output = 0;
int fact = factorial(N);
for (int i = 0; i < N - 1; i++)
{
int order = arr[i];
for (int j = 0; j < i; j++)
order -= arr[j] < arr[i];
output += order * (fact /= N - i);
}
return output;
}
int lexic_ix1(int* arr, int N, int N_fac)
{
int output = 0;
int fact = N_fac;
for (int i = 0; i < N - 1; i++)
{
int order = arr[i];
for (int j = 0; j < i; j++)
order -= arr[j] < arr[i];
output += order * (fact /= N - i);
}
return output;
}
int lexic_ix2( int arr[], int N , int coeff_arr[])
{
for ( int i = 0; i < N - 1; i++ )
{
int order = arr [ i ];
for ( int j = 0; j < i; j++ )
order -= arr [ j ] < arr [ i ];
coeff_arr[i] = order;
}
long fac = 1;
long sn = 0;
for (int i = 1; i < N; ++i)
{
fac *= i;
if (coeff_arr[N - 1 - i])
sn += coeff_arr[N - 1 - i] * fac;
}
return sn;
}
std::vector<std::vector<int>> gen_permutation(const std::vector<int>& permu_base)
{
if (permu_base.size() == 1)
return std::vector<std::vector<int>>(1, std::vector<int>(1, permu_base[0]));
std::vector<std::vector<int>> results;
for (int i = 0; i < permu_base.size(); ++i)
{
int cur_int = permu_base[i];
std::vector<int> cur_subseq = permu_base;
cur_subseq.erase(cur_subseq.begin() + i);
std::vector<std::vector<int>> temp = gen_permutation(cur_subseq);
for (auto x : temp)
{
x.insert(x.begin(), cur_int);
results.push_back(x);
}
}
return results;
}
int main()
{
#define N 10
std::vector<int> arr;
int buff_arr[N];
const int length = N;
int N_fac = factorial(N);
for(int i=0; i<N; ++i)
arr.push_back(N-i-1); // for N=10, arr is {9, 8, 7, 6, 5, 4, 3, 2, 1, 0}
std::vector<std::vector<int>> all_permus = gen_permutation(arr);
std::vector<int> lexi_numbers;
// This call is not timed, only to expand the lexi_numbers vector
for (auto x : all_permus)
lexi_numbers.push_back(lexic_ix2(&x[0], length, buff_arr));
lexi_numbers.clear();
auto t0 = high_resolution_clock::now();
for (auto x : all_permus)
lexi_numbers.push_back(lexic_ix(&x[0], length));
auto t1 = high_resolution_clock::now();
lexi_numbers.clear();
auto t2 = high_resolution_clock::now();
for (auto x : all_permus)
lexi_numbers.push_back(lexic_ix1(&x[0], length, N_fac));
auto t3 = high_resolution_clock::now();
lexi_numbers.clear();
auto t4 = high_resolution_clock::now();
for (auto x : all_permus)
lexi_numbers.push_back(lexic_ix2(&x[0], length, buff_arr));
auto t5 = high_resolution_clock::now();
std::cout << std::endl << "Time durations are: " << duration_cast<milliseconds> \
(t1 -t0).count() << ", " << duration_cast<milliseconds>(t3 - t2).count() << ", " \
<< duration_cast<milliseconds>(t5 - t4).count() <<" milliseconds" << std::endl;
return 0;
}
I've been trying to put the total income calculated for that particular day into an array. So that at the end so that I can later total all the values in the array for a grand total.
I've got 2 arrays so far that have the demand for pies and number of apples picked. To calculate the income from pies, apple trays and total income for that day I've put it into a for loop.
So far I've got this: (this is for inputting the calculated value in the array)
float total[30];
int i, incmPie, numPie, rApples, applesLeft, FTray, incmFTray, PFTray;
float totalincm, incmApples, incmRApples, incmPFTray, totalincome;
**float total[30];**
int pieDemand[30]={4, 4, 2, 7, 1, 6, 7, 8, 9, 12, 2,13,13, 5, 3, 9, 9, 9, 8, 7,
12, 1, 3, 3,10,12, 3, 6, 9, 3};
int applesPicked[30]={330,123,110,245,321,999,0,100,77,89,100,200,300,390,700,20,701,6,800,90,
600,45,690,700,719,790,800,1000,66,666};
int date[30] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30};
printf("\n==============================================================================");
printf("\n Date Income from Pie Income from Apples Total income");
printf("\n==============================================================================");
for (i = 0 ; i <30; i++)
{
if (applesPicked[i] == 0)
{
incmPie = 0;
incmApples = 0;
totalincm = 0;
**total[i] = totalincm;**
}
else if (applesPicked[i] < (pieDemand[i]*8))
{
numPie = applesPicked[i]/8;
incmPie = numPie * 14;
rApples = applesPicked[i]%8;
incmRApples = rApples * 0.5;
incmApples = incmRApples;
totalincm = incmPie + incmRApples;
**total[i] = totalincm;**
}
else
{
incmPie = pieDemand[i] * 14;
applesLeft = applesPicked[i] - (pieDemand[i]*8);
FTray = applesLeft/20;
incmFTray = FTray * 12;
PFTray = applesLeft%20;
incmPFTray = PFTray * 0.5;
incmApples = incmFTray + incmPFTray;
totalincm = incmApples + incmPie;
**total[i] = totalincm;**
}
**totalincome** = total[1] + total[2] + total[3] + total[4] + total[5] + total[6] + total[7] + total[8] + total[9] + total[10] + total[11] + total[12] + total[13] + total[14] + total[15] + total[16] + total[17] + total[18] + total[19] + total[20] + total[21] + total[22] + total[23] + total[24] + total[25] + total[26] + total[27] + total[28] + total[29] + total[30];
printf("\n"); //prints onto the next line.
printf("%d/04/2013",date[i]); // prints the date.
printf("%15d", incmPie); // prints the income from pies for each value in the arrays.
printf("%20g", incmApples); // prints the income from apples from both full trays and remaining apples for each value in the arrays.
printf("%28g", totalincm);
}
printf("\n==============================================================================");
**printf("\n Total income for the entire month: $%g", totalincome);**
printf("\n------------------------------------------------------------------------------");
_getch();
}
and i'm using this code to sum the total of the array:
totalincome = total[1] + total[2] + ... + total[30];
Any help will be appreciated! :)
In C++ (almost all programming languages), array index starts at 0, not 1! Check out Zero-based numbering for more info.
Change it to
totalincome = total[0] + total[1] + ... + total[29];
Or simply, to make your life much easier, use a loop:
totalincome = 0;
for (int i = 0; i < sizeof(total)/sizeof(total[0]); ++i)
totalincome += total[i];
totalincome = 0;
for (int i = 0; i < sizeof(total)/sizeof(total[0]); ++i)
totalincome += total[i];
For static arrray, this will work. If the array is dynamically allocated or passed as a pointer, you have to keep track of the number of elements.
totalincome = 0;
for (int i = 0; i < numelements; ++i)
totalincome += total[i];
You need to put totalincome out of the loop.
#include <iostream>
#include <vector>
using namespace std;
int main(void)
{
int i, s, g;
vector<int> a;
cin >> s;
for(i=1;i<=s;i++)
{
g = s;
if(g<10) a.push_back(g);
else {
vector<int> temp;
while(g > 0)
{
int k = g % 10;
g = g / 10;
temp.push_back(g);
}
for(int j=temp.size();j>0;j--)
{
a.push_back(temp[j]);
}
}
}
cout << a[s-1] << endl;
return 0;
}
What is wrong with the code above ? It doesn't give me the appropriate results.
The vector a is supposed to hold the values from 1, 2, 3...up to s such that a = 12345..910111213... and print to output a[s]. Ex if s=15 a=123456789101112131415 and a[15] = 2 .
If someone could tell me what's the problem
for(int j=temp.size();j>0;j--)
{
a.push_back(temp[j]);
}
Here the values of j include temp.size and exclude 0. Since vectors (like basically everything else with integer indices) are 0-indexed, this will access a out of bounds on the first iteration (i.e. when you access temp[temp.size()]).
You have the line g = s; when I think you want g = i;. As written, for s = 5, this will be your vector a: 5 5 5 5 5 Which is not at all what you want.
[Edit] Your handling of numbers > 10 is also off. For example, what happens in your code currently for the number 12? Well, temp will be 1 0 instead of 1 2, and then this will be pushed into a as 0 1, which is again not what you want.
To fix this, think about what k is supposed to do.
Corrected code:
int i, s, g;
vector<int> a;
cin >> s;
for(i=1;i<=s;i++)
{
g = i; //Why was it s?
if(g<10) a.push_back(g);
else {
vector<int> temp;
while(g > 0)
{
int k = g % 10;
g = g / 10;
temp.push_back(k); //You need to push the remainder
}
for(int j=temp.size()-1;j>=0;j--) //Out of bounds error
{
a.push_back(temp[j]);
}
}
}
cout << a[s-1] << endl;
return 0;
And a looks like this when s = 15 - is this what you were looking for?
a[0] = 1
a[1] = 2
a[2] = 3
a[3] = 4
a[4] = 5
a[5] = 6
a[6] = 7
a[7] = 8
a[8] = 9
a[9] = 1
a[10] = 0
a[11] = 1
a[12] = 1
a[13] = 1
a[14] = 2
a[15] = 1
a[16] = 3
a[17] = 1
a[18] = 4
a[19] = 1
a[20] = 5
I think there's a few problems here.
First, as MBennett said above, you should have done g = i; not g = s; to begin with.
Second, I think your inner loop also has an error, where you should be pushing back k not g as you are now.
Third, you should be doing push_front() not push_back() as you are now. Think of it this way, if you only had that loop, and had the number 162, if you push BACK (not front) every time, then it pushes 2, 6, 1, and so the sequence will have that, and not 1, 6, 2, in the order you want. Your copy after that seems OK, though there's more efficient ways of doing it.
I think that's it. Make those changes and it should function, though I haven't compiled it myself, I'm just solving in my head.