题意:略。

思路:思考一下,最先拿去对折的绳子会参与之后的每次对折,而对一条绳子而言,对折的次数越多剩下的就越短,因此,要让最终的结果尽可能长,应该先让较短的绳子先对折。

代码:

#include <cstdio>
#include <algorithm>
using namespace std;
;
int segment[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    ;i<n;i++)
        scanf("%d",&segment[i]);
    sort(segment,segment+n);
    ];
    ;i<n;i++)
        ans=(ans+segment[i])/;
    printf("%d",ans);
    ;
}

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