PAT-1125. Chain the Ropes (25)

1125. Chain the Ropes (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 <= 10⁴). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10⁴.

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

---

提交代码

 

这道题首先是对题意的不懂，两条绳子合成一条链，这条链又会和下一条绳子合成新链，直到绳子用完。

然后题意懂了，就很简单了，先合成链的绳子折叠次数会较多，所以链要长久必须把长的绳子放在后面合成，所以排序然后依次合成。

值得注意的一个小错误就是，链初试为a[0]，但这是在排序后的，排序前不能初始化。

 

#include <bits/stdc++.h>

using namespace std;

int N;
int a[10004];

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