Sorting vector (c++) - c++

Question

I have to sort a vector N into M equals parts(M less than N).
But those equal parts have to be in the middle. If I have parts that are not equal i need to put them in the first and/or last element.Difference between first and last has to be minimal. I have managed to make a sort of that function. Function is making new vector(sizeof(M)) and inside stores number of parts from vector N.
ex. 10/7 vector M [1 1 2 2 2 1 1] 1+1+2+2+2+1+1=10 into 7 parts. Meaning I am taking n or n+1 objects for vector N and storing index in vector M. Values inside do not have to be equal.
Now I am having trouble because my deviations can only be first and last element.
And also i am having problems whit for
ex. 12/9 because i get M[1 1 1 2 2 2 1 1 1 ]
but if I can only have first and last as different
then it should be like M[3 1 1 1 1 1 1 1 2]
So my question : Is there any better way of making this?
#include "stdafx.h"
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;
int N = 12; // size of vector
int M = 9; // numbers of divisions
static void subsizes(int vecSize, int subCount, vector<int> &v)
{
int baseSize = vecSize / subCount;
int bumps = vecSize % subCount;
int i,n=0,nPlus,counterPlus=0,counterN,counterLeft;
vector<int> temp(subCount); // for new results
vector<int> proba(subCount);
for (i = 0; i < subCount; i++) //dividing to n, n+1 and placing them in vector
{
temp[i]= baseSize + (i < bumps);
}
for (i=0; i<subCount; i++) // finding what numbers are n, n+1
{
nPlus=temp[i];
if(n==0 && n<nPlus){
n=nPlus;}
}
for(i=0; i<subCount;i++) //counting n, n+1
{
if(nPlus==temp[i])counterPlus++;
}
counterN=subCount-counterPlus;
counterLeft=counterPlus-2;
for(i=0; i<counterPlus/2; i++) //placing them in right order
temp[i]=nPlus;
for(i=counterPlus/2; i<counterPlus/2+counterN; i++)
temp[i]=n;
for(i=counterPlus/2+counterN; i<subCount; i++)
temp[i]=nPlus;
cout<<endl<<"Divided vector is :"<<endl; //printing results
for(i=0; i<subCount;i++)
{
int part = temp[i];
cout<<"At : vector["<<i<<"] nubmer of objects is --- "<<part<<endl;
}
putchar('\n');
}
int _tmain(int argc, _TCHAR* argv[])
{
vector<int> vec(N);
int vecSize = vec.size(); //length of original vector
int subCount=M; //number of sub-vectors parts
generate(vec.begin(), vec.end(), rand); // filling vector with C function rand()
cout<<"Vector is [ ";
for (auto i : vec) // printing out a vector
{
cout<<i<<" ";
}
cout<<"]"<<endl;
subsizes(vecSize,subCount,vec); // calling funciton that divideds and print results
system("PAUSE");
return 0;
}

Answer 1

You are complicating this, a lot.
If I understand you correctly
you want to take a vector of size N and divide it into M parts
Each of the M parts should have the same length
If there is remaining elements in N, you want to distribute the remainder to extra sub-parts at the beginning and end
Your output is a vector of the sizes of sub-parts
edit: it seems my assumption of "you want to distribute the remainder to extra sub-parts at the beginning and end" was wrong
If that is indeed what you want then you can just do this
void subsizes(int vecSize, int subCount, vector<int> &v)
{
int regular_size = (vecSize+2) / subCount;
int remainder = vecSize - regular_size * subCount;
vector<int> sizes(subCount, regular_size);
int front_remainder = remainder - (remainder / 2);
int back_remainder = remainder / 2;
sizes.front() += front_remainder;
sizes.back() += back_remainder;
for (int i = 0; i<sizes.size(); i++)
{
int part = sizes[i];
cout << "At : vector[" << i << "] nubmer of objects is --- " << part << endl;
}
putchar('\n');
}

Answer 2

I'm not sure I have a complete grip on your problem, but it appears that you are allowed to change the elements contained in vec. In which case, this is probably what you are looking for:
void subsizes(vector<int>& vec)
{
if(vec.size() > 1) // Don't do anything if there aren't at least 2 elements
{
// Get the sum of all the elements,
// but we're going to be adding back in 1s in every element but the last
// so subtract (vec.size() - 1) from this total.
// This is done by initializing accumulate to 1 - vec.size()
int last = accumulate(vec.begin(), vec.end(), 1 - vec.size());
// Now put 1s in all the elements
fill(vec.begin(), vec.end(), 1);
// Change the last element to the value accumulated in last
vec.back() = last;
}
}