将矢量拆分为n个子矢量(反弹)

时间:2013-11-06 22:11:31

标签: c++ arrays vector split

我正在尝试将矢量分成n个部分。 我检查了以下解决方案How to split a vector into n "almost equal" parts

我根据以下评论推出了以下代码: 要获得每个零件尺寸的基数,只需将总数除以零件数量:11/3 = 3.显然,某些零件需要大于该零件才能得到合适的零件,但是那就是余数:11%3 = 2.所以现在你知道其中2个部分的大小是3 + 1,剩下的就是3.(Mark Ransom)

int main()
{
std::vector<int> lines;
        int size = 200;
        for(int i = 0; i < size;i++)
        {
            lines.push_back(i);
        }
        int p = 6;
        int right = round((double)size/(double)p);
        for(int i = 0; i < p;i++)
        {
            if( i < size - left)
            {
                vector<int> v;
                for(int j = 0; j < right; j++)
                {
                    v.push_back(lines[j]);
                }
                cout << v.size() << endl;

            }
            else if (i > size - left)
            {
                vector<int> v;
                for(int k = 0; k < right; k++)
                {
                    v.push_back(lines[k]);
                }
                cout << v.size() << endl;
            }
        }
   return 0;
}

p = 6且尺寸= 200的输出为:33,33,33,33,33,33 = 198

p = 6且size = 1000的输出为:167,167,167,167,167,167 = 1002

两个输出都是错误的。我错过了什么?


编辑后:

所以让我明白一下。 我们通过增加 i ,它表示子矢量的大小。 虽然小于尺寸 - 右但我们什么都不做。当我变得更大时,我们必须处理剩余物,我们通过 right = size - i 来改变块的大小。

int main()
{
    std::vector<int> lines;
            int size = 1000;
            for(int i = 0; i < size;i++)
            {
                lines.push_back(i);
            }
            int p = 6;
            int right = round((double)size/(double)p);
            int left = size % p;
            for(int i = 0; i < size; i+= right)
            {
                if(i < size - right)
                {
                    vector<int> v;
                    //MAJOR CORRECTION
                    for(int j = i; j < (i+right); j++)
                    {
                        v.push_back(lines[j]);
                    }
                    cout << v.size() << endl;

                }
                else
                {
                    right = size - i;
                    vector<int> v;
                    //Major Correction
                    for(int k =i; k < size; k++)
                    {
                        v.push_back(lines[k]);
                    }
                    cout << v.size() << endl;

                }

return 0;
}

谢谢。

输出:33 33 33 33 33 33 2 = 200

2 个答案:

答案 0 :(得分:1)

int right = size/p; // don't round! this floors.
int left = size % p; // this one is correct.
for(int i = 0; i < p;i++)
        {
            if( i < size - left)
            {
                vector<int> v;
                for(int j = 0; j < right; j++) // counters, you used i here.
                {
                    v.push_back(lines[j]); // and here.
                }
                cout << v.size() << endl;

            }
            else if (i >= size - left)// sorry equal is here. try >= not > , comment with results.
            {
                vector<int> v;
                for(int j = 0; j < right+1; j++) // and here
                {
                    v.push_back(lines[j]); // and here
                }
                cout << v.size() << endl;
            }
        }

答案 1 :(得分:0)

以其他方式思考你的想法:
p (parts) = 3, size = 11, ceil(11/3) = 4所以4 + 4 + 3 = 11

其他相同 p = 6, size = 200 ceil(200/6) = 34所以,34 + 34 + 34 + 34 + 34 + 30 = 200

int p = 6;
size_t nLimit = ceil((double)lines.size()/p);
// if you don't want to contain the leftover element within p elements, use floor
vector<int>::iterator start = lines.begin();
for(size_t i = 0; i < lines.size(); i+=nLimit){
    // Just use the constructor/insert function
    vector<int> v(start+i, start+std::min<size_t>(i+nLimit, lines.size()));
    cout<<v.size()<<endl;

}

这里的工作代码: http://ideone.com/6V7rSX