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Why do I get this output: 0 1 2 3 0 1 2 3 0 1 after running the code below? Doesn't the modulus operation finds the remainder after division of one number by another?
#include <iostream>
using namespace std;
int main ()
{
for (int i=0; i< 10; ++i)
cout << i % 4 << " ";
}
The answer is correct. '%' mean "reminder". The % operator is remainder operator. The A % B operator actually answer the question “If I divided A by B using integer arithmetic, what would the remainder be?”
dividend = quotient * divisor + remainder
0 % 4 = 0
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
.....
etc..
For negative number...
1 % (-4) = 1
(-2) % 4 = -2
(-3) % (-4) = -3
With a remainder operator, the sign of the result is the same as the sign of the dividend
you can read more at What's the difference between “mod” and “remainder”?
Yes, that's how modulus works. The output is correct.
0 % 4 = 0
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
...
Take the number, remove as many 4's as you can. Whatever is left over is the modulus.
It does.0 / 4 = 0 remainder 01 / 4 = 0 remainder 1and so on.
Modulus operator returns the remainder after dividing the first number with the second one.
0 % 4 = 0
1 % 4 = 1
2 % 4 = 2
3 % 4 = 3
4 % 4 = 0
5 % 4 = 1
6 % 4 = 2
7 % 4 = 3
8 % 4 = 0
9 % 4 = 1
Related
I'm writing a multi-precision library in C++, using a base of 2^64, and currently I'm working on the mod operation. I'm using Algorithm D described in Donald E. Knuth's 1998 edition of "The Art Of Computer Programming" Vol. 2, Section 4.3.1, for division, which yields a quotient and a remainder. For the mod operation, I'm performing a division, throwing away the quotient in the end. Although Knuth's Algorithm D is very fast if implemented in C++ with some ASM enhancements for the partial division and the concurrent multi-precision multiplication/subtraction in each step, I'm not sure if there is a better way, since throwing away a painstakingly computed result doesn't seem efficient to me.
Unfortunately, it's not possible to get rid of the partial division in Algorithm D, because the partial quotient is required to compute the remainder, by subtracting the product of the partial quotient and the divisor from the dividend iteratively.
I've searched the Internet for alternative solutions, and found the influential papers written by Paul Barrett and Peter L. Montgomery. However, the fancy tricks they use seem to pay off only if lots of mod operations are performed in a row with the same modulus, since they involve heavy precomputations. This is the case in complex operations like modular exponentiation, where the mod of several squares and products is required for a single modulus. Barrett starts with the basic definition of the remainder, r = a - b * (a / b), and changes the division to a multiplication with the reciprocal of b. Then he presents an efficient way to compute this multiplication, which pays off if the reciprocal is computed once for several similar computations. Montgomery transforms the operands into a completely different residue system, where modular arithmetic is cheap, but for the price of transformations to and fro.
Additionally, Both algorithms introduce some restrictions, which need to be met for correct operation. Montgomery, for instance, usually requires the operands to be odd, which is the case in RSA calculations with primes, but which cannot be assumed in the general case. Outside these restrictions, even more overhead for normalizations is required.
So what I need, is an efficient one-shot mod function without overhead and special restrictions. Hence my question is: Is it possible to compute a remainder without computing the quotient in the first place, in a way that is more efficient than division?
One suggestion would be to write a simple function that would calculate A%B=C and store the A, B and C values into an array, then store all the results into a vector. Then print them out to see the relationships of all of the inputs and output values.
There is one thing that can be done to simplify some of this work and that is to know some of the properties of the mod function. These two statements will help you out with the function.
0 mod N = 0
N mod 0 = undefined
Since 0 mod N = 0 we can put a test case for A and if it is 0 we can just use that to populate our array. Likewise if B = 0 we can populate our array's C value with -1 just to represent undefined because you can not perform A mod 0 as the compilation will fail due to division by 0.
I wrote this function to do just that; then I run it through a loop for both A & B from [0,15].
#include <array>
#include <vector>
#include <iostream>
std::array<int, 3> calculateMod(int A, int B) {
std::array<int, 3 > res;
if (A == 0) {
res = std::array<int, 3>{ 0, B, 0 };
}
else if (B == 0) {
res = std::array<int, 3>{ A, 0, -1 };
}
else {
res = std::array<int, 3>{ A, B, A%B };
}
return res;
}
int main() {
std::vector<std::array<int, 3>> results;
int N = 15;
for (int A = 0; A <= N; A++) {
for (int B = 0; B <= N; B++) {
results.push_back(calculateMod(A, B));
}
}
// Now print out the results in a table form:
int i = 0; // Index for formatting output
for (auto& res : results) {
std::cout << res[0] << " % " << res[1] << " = " << res[2] << '\n';
// just for formatting output data to make it easier to read.
i++;
if ( i > N ) {
std::cout << '\n';
i = 0;
}
}
return 0;
}
Here is it's output:
0 % 0 = 0
0 % 1 = 0
0 % 2 = 0
0 % 3 = 0
0 % 4 = 0
0 % 5 = 0
0 % 6 = 0
0 % 7 = 0
0 % 8 = 0
0 % 9 = 0
0 % 10 = 0
0 % 11 = 0
0 % 12 = 0
0 % 13 = 0
0 % 14 = 0
0 % 15 = 0
1 % 0 = -1
1 % 1 = 0
1 % 2 = 1
1 % 3 = 1
1 % 4 = 1
1 % 5 = 1
1 % 6 = 1
1 % 7 = 1
1 % 8 = 1
1 % 9 = 1
1 % 10 = 1
1 % 11 = 1
1 % 12 = 1
1 % 13 = 1
1 % 14 = 1
1 % 15 = 1
2 % 0 = -1
2 % 1 = 0
2 % 2 = 0
2 % 3 = 2
2 % 4 = 2
2 % 5 = 2
2 % 6 = 2
2 % 7 = 2
2 % 8 = 2
2 % 9 = 2
2 % 10 = 2
2 % 11 = 2
2 % 12 = 2
2 % 13 = 2
2 % 14 = 2
2 % 15 = 2
3 % 0 = -1
3 % 1 = 0
3 % 2 = 1
3 % 3 = 0
3 % 4 = 3
3 % 5 = 3
3 % 6 = 3
3 % 7 = 3
3 % 8 = 3
3 % 9 = 3
3 % 10 = 3
3 % 11 = 3
3 % 12 = 3
3 % 13 = 3
3 % 14 = 3
3 % 15 = 3
4 % 0 = -1
4 % 1 = 0
4 % 2 = 0
4 % 3 = 1
4 % 4 = 0
4 % 5 = 4
4 % 6 = 4
4 % 7 = 4
4 % 8 = 4
4 % 9 = 4
4 % 10 = 4
4 % 11 = 4
4 % 12 = 4
4 % 13 = 4
4 % 14 = 4
4 % 15 = 4
5 % 0 = -1
5 % 1 = 0
5 % 2 = 1
5 % 3 = 2
5 % 4 = 1
5 % 5 = 0
5 % 6 = 5
5 % 7 = 5
5 % 8 = 5
5 % 9 = 5
5 % 10 = 5
5 % 11 = 5
5 % 12 = 5
5 % 13 = 5
5 % 14 = 5
5 % 15 = 5
6 % 0 = -1
6 % 1 = 0
6 % 2 = 0
6 % 3 = 0
6 % 4 = 2
6 % 5 = 1
6 % 6 = 0
6 % 7 = 6
6 % 8 = 6
6 % 9 = 6
6 % 10 = 6
6 % 11 = 6
6 % 12 = 6
6 % 13 = 6
6 % 14 = 6
6 % 15 = 6
7 % 0 = -1
7 % 1 = 0
7 % 2 = 1
7 % 3 = 1
7 % 4 = 3
7 % 5 = 2
7 % 6 = 1
7 % 7 = 0
7 % 8 = 7
7 % 9 = 7
7 % 10 = 7
7 % 11 = 7
7 % 12 = 7
7 % 13 = 7
7 % 14 = 7
7 % 15 = 7
8 % 0 = -1
8 % 1 = 0
8 % 2 = 0
8 % 3 = 2
8 % 4 = 0
8 % 5 = 3
8 % 6 = 2
8 % 7 = 1
8 % 8 = 0
8 % 9 = 8
8 % 10 = 8
8 % 11 = 8
8 % 12 = 8
8 % 13 = 8
8 % 14 = 8
8 % 15 = 8
9 % 0 = -1
9 % 1 = 0
9 % 2 = 1
9 % 3 = 0
9 % 4 = 1
9 % 5 = 4
9 % 6 = 3
9 % 7 = 2
9 % 8 = 1
9 % 9 = 0
9 % 10 = 9
9 % 11 = 9
9 % 12 = 9
9 % 13 = 9
9 % 14 = 9
9 % 15 = 9
10 % 0 = -1
10 % 1 = 0
10 % 2 = 0
10 % 3 = 1
10 % 4 = 2
10 % 5 = 0
10 % 6 = 4
10 % 7 = 3
10 % 8 = 2
10 % 9 = 1
10 % 10 = 0
10 % 11 = 10
10 % 12 = 10
10 % 13 = 10
10 % 14 = 10
10 % 15 = 10
11 % 0 = -1
11 % 1 = 0
11 % 2 = 1
11 % 3 = 2
11 % 4 = 3
11 % 5 = 1
11 % 6 = 5
11 % 7 = 4
11 % 8 = 3
11 % 9 = 2
11 % 10 = 1
11 % 11 = 0
11 % 12 = 11
11 % 13 = 11
11 % 14 = 11
11 % 15 = 11
12 % 0 = -1
12 % 1 = 0
12 % 2 = 0
12 % 3 = 0
12 % 4 = 0
12 % 5 = 2
12 % 6 = 0
12 % 7 = 5
12 % 8 = 4
12 % 9 = 3
12 % 10 = 2
12 % 11 = 1
12 % 12 = 0
12 % 13 = 12
12 % 14 = 12
12 % 15 = 12
13 % 0 = -1
13 % 1 = 0
13 % 2 = 1
13 % 3 = 1
13 % 4 = 1
13 % 5 = 3
13 % 6 = 1
13 % 7 = 6
13 % 8 = 5
13 % 9 = 4
13 % 10 = 3
13 % 11 = 2
13 % 12 = 1
13 % 13 = 0
13 % 14 = 13
13 % 15 = 13
14 % 0 = -1
14 % 1 = 0
14 % 2 = 0
14 % 3 = 2
14 % 4 = 2
14 % 5 = 4
14 % 6 = 2
14 % 7 = 0
14 % 8 = 6
14 % 9 = 5
14 % 10 = 4
14 % 11 = 3
14 % 12 = 2
14 % 13 = 1
14 % 14 = 0
14 % 15 = 14
15 % 0 = -1
15 % 1 = 0
15 % 2 = 1
15 % 3 = 0
15 % 4 = 3
15 % 5 = 0
15 % 6 = 3
15 % 7 = 1
15 % 8 = 7
15 % 9 = 6
15 % 10 = 5
15 % 11 = 4
15 % 12 = 3
15 % 13 = 2
15 % 14 = 1
15 % 15 = 0
From the data above we can see that if A == B the result will be 0. We can also see that if B > A then B == A. Finally we can see that there are patterns between odd and even values of A while B < A. If you can understand these patterns then most of it becomes algebraic manipulation. From here the next step would be to create an algorithm that would take all of this data and convert it to its binary equivalence.
I chose the value of N above as 15 for a reason. This is due to the binary representation of all the possible combinations of binary digits before they repeat again. We know that a single byte of data is 8 bits; we know that the values from [0,15] will fit into half of that; for example:
binary byte: hex decimal
0000 0000 0x00 0
...
0000 1111 0xFF 15
After these 15 different sequences of 0s and 1s these patterns will repeat. So by taking the table above you can convert these into binary representations. Now once you examine the representations of A & B inputs with their C outputs in binary and understand the 3 properties of the results that I had mentioned above; you should be able to design an algorithm to quickly compute the modulo of any A B combination quite easily. One trick to remember is that there are 3 other things to take into consideration. The first is what user eerokia had stated:
"In particular, modulo with a power of 2 can be replaced by a bitwise operations."
The next beyond that is are the values even or odd as the even and odd cases do present different patters of A mod B when B < A.
I have give you some tools of information to start out on, but the rest I'll leave up to you including the task of converting the A, B, and C values into their binary representations.
Once you know the binary patterns of both the A and B inputs according to their C outputs and you understand the truth tables of the logical gates - operators such as And - &, Or - |, Nand - (!&), Nor - (!|), Xor - ^ Xnor - (!^) and Not - ! as well as the compliment (~). You should be able to design your algorithm with efficiency.
I wrote an implementation of the following tutorial: LINK
Basically, since C/C++ does not have BIG Integer we are storing the factorial decimal values in an array. This is equivalent to writing a multiplication that performs the multiplication kids are taught at schools.
Problem: It works fine for values up to 17! after that (18!, 19!,... ) it does not output correct values.
#include <iostream>
using namespace std;
int main(){
int fact[1000]={1};
int n; scanf("%d", &n); //n are the number of factorials we will calculate
while(n--){
int number; scanf("%d", &number); //scan the number
if(number == 0) printf("%d", 1);
int flag = number;
int index = 0, length = 0;
//following lines we find the length of the entered number
while(flag!=0){
fact[index] = flag%10;
flag /= 10;
index++; length++;
}
//following lines are the multiplication code
while(number>1){
index = 0;
int temp = 0;
number--;
for(index = 0; index<length; index++){
int x = (fact[index] * number) + temp;
fact[index] = x%10;
temp = x/10;
}
//here we append the carry over left from multiplication
while(temp){
fact[index] = temp%10;
temp /= 10;
length++;
}
}
//print the array from most to least significant digit
for(int i = length-1; i>=0; i--){
printf("%d", fact[i]);
}
printf("\n");
}
return 0;
}
For a start, you need to be very careful with:
long long int x = (fact[index] * number) + temp;
Since fact[], number and temp are all int types, the calculation will be done as an int and only widened to a long long when placing the value into x.
You would be better off with:
long long x = fact[index];
x *= number;
x += temp;
That way, it becomes a long long early enough that the calculations will be done with that type.
However, that doesn't actually fix your problem, so let's modify your code a little to see where the problem lies:
#include <iostream>
using namespace std;
int main(){
int fact[1000]={1};
int n = 18, numberx = 0;
while(n-- > 0){
int number = ++numberx;
if(number == 0) { printf("%d", 1); continue; }
int flag = number;
int index = 0, length = 0;
//following lines we find the length of the entered number
while(flag!=0){
fact[index] = flag%10;
flag /= 10;
index++; length++;
}
//following lines are the multiplication code
while(number>1){
index = 0;
int temp = 0;
number--;
for(index = 0; index<length; index++){
long long int x = fact[index];
x *= number;
x += temp;
fact[index] = x%10;
temp = x/10;
}
//here we append the carry over left from multiplication
while(temp){
fact[index] = temp%10;
temp /= 10;
length++;
}
}
//print the array from most to least significant digit
printf("%d! = ", number);
for(int i = length-1; i>=0; i--){
printf("%d ", fact[i]);
}
printf("\n");
}
return 0;
}
Running this gives you:
1! = 1
2! = 2
3! = 6
4! = 2 4
5! = 1 2 0
6! = 7 2 0
7! = 5 0 4 0
8! = 4 0 3 2 0
9! = 3 6 2 8 8 0
10! = 3 6 2 8 8 0 0
11! = 3 9 9 1 6 8 0 0
12! = 4 7 9 0 0 1 6 0 0
13! = 6 2 2 7 0 2 0 8 0 0
14! = 8 7 1 7 8 2 9 1 2 0 0
15! = 1 3 0 7 6 7 4 3 6 8 0 0 0
16! = 2 0 9 2 2 7 8 9 8 8 8 0 0 0
17! = 3 5 5 6 8 7 4 2 8 0 9 6 0 0 0
18! = 1 9 9 1 0 4 7 1 7 3 8 5 7 2 8 0 0 0
which is, as you state okay up until 18!, where if fails. And, in fact, you can see the ratio between 17! and 18! is about 500 rather than 18 so that's where we should look.
Let's first strip out the extraneous stuff by starting at 17!. That can be done simply by changing a couple of starting values:
int n = 2, numberx = 16;
and that gives:
17! = 3 5 5 6 8 7 4 2 8 0 9 6 0 0 0
18! = 1 9 9 1 0 4 7 1 7 3 8 5 7 2 8 0 0 0
Then we can add debug code to see what's happening, outputting temporary results along the way. The main loop can become:
while(number>1){
index = 0;
int temp = 0;
number--;
if (numberx > 17) printf("\n");
for(index = 0; index<length; index++){
if (numberx > 17) printf("index %d fact[] %d number %d temp %d", index, fact[index], number, temp);
long long int x = fact[index];
x *= number;
x += temp;
fact[index] = x%10;
temp = x/10;
if (numberx > 17) printf(" -> fact[] %d temp %d\n", fact[index], temp);
}
//here we append the carry over left from multiplication
while(temp){
fact[index] = temp%10;
temp /= 10;
length++;
}
if (numberx > 17) {
printf("temp: ");
for(int i = length-1; i>=0; i--){
printf("%d ", fact[i]);
}
printf("\n");
}
}
This shows you *exactly where things start to go wrong (// bits are added by me):
17! = 3 5 5 6 8 7 4 2 8 0 9 6 0 0 0
index 0 fact[] 8 number 17 temp 0 -> fact[] 6 temp 13
index 1 fact[] 1 number 17 temp 13 -> fact[] 0 temp 3
temp: 3 0 6 // okay: 18 * 17 = 306
index 0 fact[] 6 number 16 temp 0 -> fact[] 6 temp 9
index 1 fact[] 0 number 16 temp 9 -> fact[] 9 temp 0
index 2 fact[] 3 number 16 temp 0 -> fact[] 8 temp 4
temp: 4 8 9 6 // okay 306 * 16 = 4896
index 0 fact[] 6 number 15 temp 0 -> fact[] 0 temp 9
index 1 fact[] 9 number 15 temp 9 -> fact[] 4 temp 14
index 2 fact[] 8 number 15 temp 14 -> fact[] 4 temp 13
index 3 fact[] 4 number 15 temp 13 -> fact[] 3 temp 7
temp: 7 3 4 4 0 // okay 4896 * 15 = 73440
index 0 fact[] 0 number 14 temp 0 -> fact[] 0 temp 0
index 1 fact[] 4 number 14 temp 0 -> fact[] 6 temp 5
index 2 fact[] 4 number 14 temp 5 -> fact[] 1 temp 6
index 3 fact[] 3 number 14 temp 6 -> fact[] 8 temp 4
index 4 fact[] 7 number 14 temp 4 -> fact[] 2 temp 10
temp: 8 1 2 8 1 6 0 // no good: 73440 * 14 = 10128160 !!!
1 0 2 8 1 6 0 // is what it should be
With a bit of thought, it appears to be the point where the final "carry" from the multiplication is greater than nine, meaning it's almost certainly in the code for handling that:
while(temp){
fact[index] = temp%10;
temp /= 10;
length++;
}
Thinking about that (and comparing it to other code that changes index and length together), it becomes obvious - even though you increase the length of the array, you're not increasing the index. That means, for a final carry of ten or more, the subsequent carry will not populate the correct index, it will simply overwrite the same index each time.
This can be seen here:
temp: 8 1 2 8 1 6 0 // no good: 73440 * 14 = 10128160 !!!
1 0 2 8 1 6 0 // is what it should be
where it will have placed the zero (10 % 10) at that second location (increasing the length) but then placed the one (10 / 10) at the same index, leaving the 8 at whatever value it had before.
So, if we increment index as well, what do we see (going back to the less verbose code)?
1! = 1
2! = 2
3! = 6
4! = 2 4
5! = 1 2 0
6! = 7 2 0
7! = 5 0 4 0
8! = 4 0 3 2 0
9! = 3 6 2 8 8 0
10! = 3 6 2 8 8 0 0
11! = 3 9 9 1 6 8 0 0
12! = 4 7 9 0 0 1 6 0 0
13! = 6 2 2 7 0 2 0 8 0 0
14! = 8 7 1 7 8 2 9 1 2 0 0
15! = 1 3 0 7 6 7 4 3 6 8 0 0 0
16! = 2 0 9 2 2 7 8 9 8 8 8 0 0 0
17! = 3 5 5 6 8 7 4 2 8 0 9 6 0 0 0
18! = 6 4 0 2 3 7 3 7 0 5 7 2 8 0 0 0
19! = 1 2 1 6 4 5 1 0 0 4 0 8 8 3 2 0 0 0
20! = 2 4 3 2 9 0 2 0 0 8 1 7 6 6 4 0 0 0 0
That solves your specific problem and hopefully provides some education on debugging as well :-)
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I want to sum digits of an integer. If I have 123 I want to sum 1+2+3. I found a program that works well but I don't understand how it works. Can you explain how it works?
#include <iostream>
using namespace std;
int main()
{
int n, sum;
cout << "Enter integer\n";
cin >> n;
sum = n/100 + (n/10)%10 + n%10;
cout << "sum = " << sum << endl;
}
How n/100 generates 1 from 123, (n/10)%10 2 from 123 and n%10 3 from 123
sum = n/100 + (n/10)%10 + n%10;
1) n/100
(n=123)
in this statement 123/100 means ans is = 1
2)(n/10)%10
here (123/10) firstly evaluate and ans is = 12, and then 12%10 gets evaluate and ans is = 2
3)n%10 again 123%10 evaluate ans is 3
then statement becomes
sum = 1 + 2 + 3
Note: % symbol gives remainder
These are very simple maths.
Here, n = 123
100 | 123 | 1
100
_________
23
Therefore the quotient is 1.
The same way, for 123/10,
10 | 123 | 12
10
___________
23
20
___________
3
So, 123/10 = 12. Now for (123/10)%10 = 12%10,
10 | 12 | 1
10
______
2
Therefore, (123/10)%10 = 12%10 = 2.
The same way, 123%10 = 3
Therefore, the answer is: 123/100 + (123/10)%10 + 123%10 = 1 + 2 + 3 = 6
(Note: a % b, here b divides a and returns the remainder)
I am newbie on codechef and i was trying to solve the following question however my code runs fine on my machine, i also tested it with some cases.
Question is as follows :-
In Byteland it is always the military officer's main worry to order his soldiers on parade correctly. Luckily, ordering soldiers is not really such a problem. If a platoon consists of n men, all of them have different rank (from 1 - lowest to n - highest) and on parade they should be lined up from left to right in increasing order of rank.
Sounds simple, doesn't it? Well, Sgt Johnny thought the same, until one day he was faced with a new command. He soon discovered that his elite commandos preferred to do the fighting, and leave the thinking to their superiors. So, when at the first rollcall the soldiers lined up in fairly random order it was not because of their lack of discipline, but simply because they couldn't work out how to form a line in correct order of ranks. Sgt Johnny was not at all amused, particularly as he soon found that none of the soldiers even remembered his own rank. Over the years of service every soldier had only learned which of the other soldiers were his superiors. But Sgt Johnny was not a man to give up easily when faced with a true military challenge. After a moment's thought a solution of brilliant simplicity struck him and he issued the following order: "men, starting from the left, one by one, do: (step forward; go left until there is no superior to the left of you; get back in line).". This did indeed get the men sorted in a few minutes. The problem was solved... for the time being.
The next day, the soldiers came in exactly the same order as the day before, and had to be rearranged using the same method. History repeated. After some weeks, Sgt Johnny managed to force each of his soldiers to remember how many men he passed when going left, and thus make the sorting process even faster.
If you know how many positions each man has to walk to the left, can you try to find out what order of ranks the soldiers initially line up in?
Input
The first line of input contains an integer t<=50, the number of test cases. It is followed by t test cases, each consisting of 2 lines. The first line contains a single integer n (1<=n<=200000). The second line contains n space separated integers wi, denoting how far the i-th soldier in line must walk to the left when applying Sgt Johnny's algorithm.
Output
For each test case, output a single line consisting of n space separated integers - the ranks of the soldiers, given from left to right in their initial arrangement.
Example
Input:
2
3
0 1 0
5
0 1 2 0 1
Output:
2 1 3
3 2 1 5 4
Warning: large Input/Output data, be careful with certain languages
#include <iostream>
#include <string.h>
using namespace std;
int main ()
{
int t,n;
cin >> t;
while(t>0){
cin >> n;
int array[n+1];
int stepsmoved,i;
for(i = 1; i <= n; i++){
array[i] = i;
}
for(i = 1; i <=n; i++){
cin >> stepsmoved;
if(stepsmoved == 0){}
else{
int x;
x = array[i];
for (int j = i; j> i- stepsmoved; j--){
array[j] = array[j-1];
}
array[i-stepsmoved] = x;
}
}
for(i = 1; i <= n; i++){
cout<<array[i]<<" ";
}
cout<<endl;
t--;
}
return 0;
}
So is there something logically or syntactically wrong?
The order of 'unwinding' the sorting is relevant.
Here is the code that demonstrates the statement above (the ranks are 1-based, the 1 - is highest, 10 - is lowest, array indices are 0-based):
#include <stdio.h>
void dump(int *a) {
int i;
for (i = 0; i < 10; i++)
printf("%d ", a[i]);
printf("\n");
}
int main() {
int array[10] = {0}, steps[10] = {0};
int i,j;
srand(0);
// Assign ranks in random order
for (i = 0; i < 10;) {
j = rand() % 10;
if (!array[j])
array[j] = ++i;
}
dump(array);
// Sort according to the Sgt Johnny's initial idea
for (i = 1; i < 10; i++) {
for (j = 0; array[j] < array[i]; j++);
if (j < i) {
int k, temp = array[i];
for (k = i; k > j; k--) {
array[k] = array[k-1];
steps[temp-1]++;
}
array[j] = temp;
dump(array);
}
}
printf("Steps:\n");
dump(steps);
printf("\n");
// reconstruct the origina order
#if 1
for (i = 10-1; i >= 0; i--)
#else
for (i = 0; i < 10; i++)
#endif
{
int s = steps[array[i]-1];
for (j = i; s; s--, j++) {
int temp = array[j];
array[j] = array[j+1];
array[j+1] = temp;
}
dump(array);
}
}
If the reconstruction is done in reverse order, then we get a sequence that matches original:
8 7 5 1 10 4 2 3 9 6
7 8 5 1 10 4 2 3 9 6
5 7 8 1 10 4 2 3 9 6
1 5 7 8 10 4 2 3 9 6
1 4 5 7 8 10 2 3 9 6
1 2 4 5 7 8 10 3 9 6
1 2 3 4 5 7 8 10 9 6
1 2 3 4 5 7 8 9 10 6
1 2 3 4 5 6 7 8 9 10
Steps:
3 5 5 4 2 4 1 0 1 0
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 10 9
1 2 3 4 5 6 7 8 10 9
1 2 3 4 5 6 8 7 10 9
1 2 3 4 5 8 7 10 9 6
1 2 3 4 8 7 5 10 9 6
1 2 3 8 7 5 10 4 9 6
1 2 8 7 5 10 4 3 9 6
1 8 7 5 10 4 2 3 9 6
8 7 5 1 10 4 2 3 9 6
Otherwise, the reconstructed order does not match the original:
8 7 5 1 10 4 2 3 9 6
7 8 5 1 10 4 2 3 9 6
5 7 8 1 10 4 2 3 9 6
1 5 7 8 10 4 2 3 9 6
1 4 5 7 8 10 2 3 9 6
1 2 4 5 7 8 10 3 9 6
1 2 3 4 5 7 8 10 9 6
1 2 3 4 5 7 8 9 10 6
1 2 3 4 5 6 7 8 9 10
Steps:
3 5 5 4 2 4 1 0 1 0
2 3 4 1 5 6 7 8 9 10
2 4 1 5 6 7 3 8 9 10
2 4 5 6 7 1 3 8 9 10
2 4 5 7 1 3 8 6 9 10
2 4 5 7 3 8 6 1 9 10
2 4 5 7 3 8 6 1 9 10
2 4 5 7 3 8 1 9 10 0
2 4 5 7 3 8 1 10 9 0
2 4 5 7 3 8 1 10 0 9
2 4 5 7 3 8 1 10 0 6
I'm trying to read an array in c++, filled with values from 0 to 5, For an unimportant reason, I need to count how many numbers 1, numbers 2, numbers 3, numbers 4 and numbers 5 do stand on the 'iii*DAYS'th position, so when iii = 0 and DAYS is 5, I need to know how many numbers 1, numbers 2, numbers 3, numbers 4 and numbers 5 are located on the 0th, 4th, 9th, 14th position. The code I posted does this quite well, but sometimes, gives a very big unlogical value, -36589245 or 99653256, can somebody tell me why this happens ( it does happen +- one in a hunderd times )
DAYS = 28
NURSES = 33
SHIFTS =5
int calculate_penalty_coverage_offspring(int offspring[NURSES*DAYS])
{
int temporary[DAYS];
int count[DAYS][SHIFTS];
penalty_score_offspring_coverage =0;
for (int ii = 0; ii<DAYS; ii++)
{
int een = 0;
int twee = 0;
int drie = 0;
int vier = 0;
int vijf = 0;
for (int iii = 0; iii<NURSES; iii++)
{
temporary[iii] = offspring[(ii+(iii*DAYS))];
}
for(int a = 0 ; a<DAYS ; a++)
{
if(temporary[a]== 1)
{
een++;
count[ii][0] = een;
}
else if(temporary[a] == 2)
{
twee++;
count[ii][1] = twee;
}
else if(temporary[a]== 3)
{
drie++;
count[ii][2] = drie;
}
else if(temporary[a]== 4)
{
vier++;
count[ii][3] = vier;
}
else if(temporary[a] == 5)
{
vijf++;
count[ii][4] = vijf;
}
}
}
for(int ii = 0; ii<DAYS ; ii++)
{
for (int iii =0 ; iii<SHIFTS ; iii++)
{
cout << count[ii][iii] << '\t';
}
cout << '\n';
}
this is the exeptional output where I talked about, as you can see, there is an onlogical value in the output of -31427696 ... I can't see why the function is working good, except for this one time.
1 2 2 4 4
5 2 2 9 5
9 6 3 5 2
8 3 4 3 8
9 3 3 4 6
5 5 6 8 1
6 8 2 2 5
3 5 8 -31427696 7
5 5 2 5 8
5 7 8 2 3
2 7 1 2 10
5 6 3 5 5
4 4 4 6 7
7 4 6 4 6
6 5 6 4 3
5 3 7 4 6
5 5 6 1 7
5 5 1 6 2
4 6 6 4 5
3 3 4 5 9
6 6 5 4 4
5 5 4 4 5
8 4 4 5 3
5 5 4 7 5
4 8 6 3 3
9 1 5 7 3
3 7 5 2 5
2 6 5 7 5
First you say
int temporary[DAYS];
Where
DAYS = 28
Then you do:
for (int iii = 0; iii<NURSES; iii++)
{
temporary[iii] = offspring[(ii+(iii*DAYS))];
}
Where
NURSES = 33
You're trying to access indices that are out of bounds in temporary.
EDIT: Following our discussion in the comments,
You're additionally not initializing your arrays, specifically count:
int count[DAYS][SHIFTS];
Which you then conditionally fill in (partially) later:
if(temporary[a]== 1)
{
een++;
count[ii][0] = een;
}
// ...
Accesses to count afterwards to indices that were not assigned to will result in the garbage numbers you're seeing. You should probably just default the matrix to all zeros like so:
int count[DAYS][SHIFTS] = {0};