Box of Bricks 2088

Problem Description

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds

stacks of different height. “Look, I've built a wall!”, he tells his older sister Alice. “Nah, you should

make all stacks the same height. Then you would have a real wall.”, she retorts. After a little

consideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one,

such that all stacks are the same height afterwards. But since Bob is lazy he wants to do

this with the minimum number of bricks moved. Can you help?

Input

The input consists of several data sets. Each set begins with a line containing the number n of

stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may

assume 1≤n≤50 and 1≤hi≤100.

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange

the bricks such that all stacks have the same height.

The input is terminated by a set starting with n = 0. This set should not be processed.

Output

For each set, print the minimum number of bricks that have to be moved in order to make all the stacks the same height.
Output a blank line between each set.

Sample Input

65 2 4 1 7 50

Sample Output

5
#include <cstdio>#include <vector>#include <algorithm>#include <numeric>int main(int argc, const char* argv[]){    int n = 0;    bool bFlag = true;    while (scanf("%d", &n) != EOF && n > 0)    {        std::vector<int> vec;        int nTemp = 0;        while (n--)        {            scanf("%d", &nTemp);            vec.push_back(nTemp);        }                int nSum = accumulate(vec.begin(), vec.end(), 0);        int nAve = nSum / vec.size();        int nMin = 0;        for (int i=0; i<vec.size(); ++i)        {            if (vec[i] > nAve)            {                nMin += vec[i] - nAve;            }        }        if (bFlag)        {            printf("%d\n", nMin);            bFlag = false;        }        else        {            printf("\n%d\n", nMin);        }    }    return 0;}