写在前面
- 思路分析
- 越早加入段，对折次数越多。所以希望得到结绳最大长度，必须让长的段对折次数尽可能的少
- 将所有段从小到大排序，从小到大分别将每段依次计入结绳
- 打印最后结果
-
The result must be rounded to the nearest integer that is no greater than the maximum length.
- 英文单词
round :- 题目简单，5分钟a题
测试用例
input:
8
10 15 12 3 4 13 1 15
output:
14
ac代码
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
int n;
scanf("%d", &n);
vector<int> v(n);
for (int i = 0; i < n; i++)
scanf("%d", &v[i]);
sort(v.begin(), v.end(), less<int>());
double result = v[0];
for (int i = 1; i < n; i++)
result = (result + v[i]) / 2.0;
printf("%d", (int)result);
return 0;
}
