1125 Chain the Ropes (25 分)

1125 Chain the Ropes (25 分)

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.

Your job is to make the longest possible rope out of N given segments.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (2≤N≤10​4​​). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10​4​​.

Output Specification:

For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

Sample Input:

8
10 15 12 3 4 13 1 15

Sample Output:

14

题解

• 仍然是非常不严谨的题目：“The result must be rounded to the nearest integer that is no greater than the maximum length.” 看到这里我以为每次chain之后绳子长度是一个float数，而实际上却是整数，比如3和10chain之后是6二不是6
• 这样的话题目最后一句话不仅是多余的而且是给人以错误的误导
• 用浮点型存答案再用round四舍五入输出会导致最后一个test WA
• 至于思路：随手测试发现按从小到大顺序chain能得到最大值，没有深思直接写了

code

#include <vector>
#include <bits/stdc++.h>
using namespace std;

const int N = 1e4 + 1;
int a[N];

int main()
{
    int n, ans;
    scanf("%d",&n);
    for(int i = 0; i < n; ++i){
        scanf("%d",&a[i]);
    }
    sort(a, a + n);
    ans = a[0];
    for(int i = 1; i < n; ++i){
        ans = (a[i] + ans) / 2;
    }
    printf("%d",ans);
    return 0;
}