【算法题】切割木材

问题如图：

注：木头不一定都要用，用也不一定都要用完

1. 问题可以转化成在1和给的所有原木中最长的长度中间找一个合适的长度，使得把每段原木按此分段后得到的小段木头数目至少为k：在1—max之间用折半查找，找到最大的长度

//def woodCut(self, L, k){//  if sum(L) < k ://  return 0////      start, end = 1, max(L)//      while start + 1 < end ://          mid = start + (end - start) / 2//          pieces = sum([l / mid for l in L])//          if pieces >= k ://              start = mid//          else ://          end = mid////          if sum([l / end for l in L]) > k://  return end//      return start//}#include <stdio.h>#include <numeric>using namespace std;int k(0);int array[10000];int len(0), max_wood(0);int WoodCut()//二分查找上界{    int sum = accumulate(array, array+len,0);    if (sum < k)        return 0;    int left(1), right(max_wood);    int mid;    while (left+1<right)    {        mid = left + (right - left) / 2;        int k_tmp(0);        for (auto i = 0; i <= len ;i++)        {            k_tmp += array[i] / mid;        }        if (k_tmp>=k)        {            left = mid;        }        else        {            right = mid;        }    }    int right_k(0);    for (auto i = 0; i <= len; i++)    {        right_k += array[i] / right;    }    if (right_k>=k)    {        return right;    }    else    {        return left;    }    return 0;}int main(){    char c;    scanf("%d", &k);    while (1)    {        scanf("%d", &array[len]);        if (array[len]>max_wood  )        {            max_wood = array[len];        }        c = getchar();        if (c=='\n')        {            break;        }        ++len;    }    printf("%d",WoodCut());    return 1;}