PAT甲级 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:

810 15 12 3 4 13 1 15

Sample Output:

14

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题目的意思是将n个绳子两两合并，每次长度变为和的一半，问最后变成1根的长度最大是多少

思路，每次合并每一根绳子会变为原来的1/2，所以最长的肯定合的越少越好，贪心妹子选择最短的合并

include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<queue>#include<stack>#include<map>#include<set>using namespace std;#define LL long longconst int inf=0x3f3f3f3f;int main(){   int n;   int a[100005];   scanf("%d",&n);   for(int i=0;i<n;i++)    scanf("%d",&a[i]);   sort(a,a+n);   for(int i=1;i<n;i++)   {       a[i]=(a[i]+a[i-1])/2;   }   printf("%d\n",a[n-1]);    return 0;}