【Codeforces Round #389】Codeforces 752E Santa Claus and Tangerines【解法一】

Santa Claus has n tangerines, and the i-th of them consists of exactly

ai slices. Santa Claus came to a school which has k pupils. Santa

decided to treat them with tangerines.

However, there can be too few tangerines to present at least one

tangerine to each pupil. So Santa decided to divide tangerines into

parts so that no one will be offended. In order to do this, he can

divide a tangerine or any existing part into two smaller equal parts.

If the number of slices in the part he wants to split is odd, then one

of the resulting parts will have one slice more than the other. It’s

forbidden to divide a part consisting of only one slice.

Santa Claus wants to present to everyone either a whole tangerine or

exactly one part of it (that also means that everyone must get a

positive number of slices). One or several tangerines or their parts

may stay with Santa.

Let bi be the number of slices the i-th pupil has in the end. Let

Santa’s joy be the minimum among all bi’s.

Your task is to find the maximum possible joy Santa can have after he

treats everyone with tangerines (or their parts). Input

The first line contains two positive integers n and k (1 ≤ n ≤ 106,

1 ≤ k ≤ 2·109) denoting the number of tangerines and the number of

pupils, respectively.

The second line consists of n positive integers a1, a2, …, an

(1 ≤ ai ≤ 107), where ai stands for the number of slices the i-th

tangerine consists of. Output

If there’s no way to present a tangerine or a part of tangerine to

everyone, print -1. Otherwise, print the maximum possible joy that

Santa can have.

解法二见【这里】。

很明显可以二分答案，然后对于给定的数量记忆化搜索求出最大能分成多少块。

先把橘子数量从小到大排序，这样可以减少递归次数。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
int rd()
{
int x=0;
char c=getchar();
while (c<'0'||c>'9') c=getchar();
while (c>='0'&&c<='9')
{
x=x*10+c-'0';
c=getchar();
}
return x;
}
int a[1000010],n,f[10000010],last[10000010],clo,mid,k;
int dfs(int x)
{
if (last[x]==clo) return f[x];
last[x]=clo;
if (x<mid) return f[x]=0;
return f[x]=max(1,dfs(x/2)+dfs(x-x/2));
}
bool ok()
{
int i;
LL tot=0;
clo++;
for (i=1;i<=n;i++)
tot+=dfs(a[i]);
return tot>=k;
}
int main()
{
int i,l,r=0;
LL s=0;
n=rd();
k=rd();
for (i=1;i<=n;i++)
a[i]=rd(),s+=a[i],r=max(r,a[i]);
if (s<k)
{
printf("-1\n");
return 0;
}
sort(a+1,a+n+1);
l=1;
while (l<r)
{
mid=(l+r+1)/2;
if (ok()) l=mid;
else r=mid-1;
}
printf("%d\n",l);
}