问题如图：
注：木头不一定都要用，用也不一定都要用完
- 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;
}