A1125. Chain the Ropes

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

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
double rope[100000];
int N;
int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; i++){
        scanf("%lf", &rope[i]);
    }
    sort(rope, rope + N);
    double ans = rope[0];
    for(int i = 0; i < N; i++){
        ans += rope[i];
        ans = ans / 2.0;
    }
    int prt = ans;
    printf("%d", prt);
    cin >> N;
    return 0;
}

 

题意： 两段绳子可以合并成一段，但它们一合并，新绳子的长度就会变为原来两绳之长的和的一半，求能得到的最长的绳子。显然越早合并的绳子被1/2的次数就越多，所以要让越长的绳子尽量最后合并。