CodeForces - 270C Magical Boxes（思维）

最新推荐文章于 2020-07-02 15:02:28 发布

我好想逃

最新推荐文章于 2020-07-02 15:02:28 发布

阅读量283

点赞数

CC 4.0 BY-SA版权

分类专栏：思路

本文链接：https://blog.youkuaiyun.com/zvenWang/article/details/84199822

思路专栏收录该内容

9 篇文章

订阅专栏

Emuskald is a well-known illusionist. One of his trademark tricks involves a set of magical boxes. The essence of the trick is in packing the boxes inside other boxes.

From the top view each magical box looks like a square with side length equal to 2k (k is an integer, k ≥ 0) units. A magical box v can be put inside a magical box u, if side length of v is strictly less than the side length of u. In particular, Emuskald can put 4 boxes of side length 2k - 1 into one box of side length 2k, or as in the following figure:

Emuskald is about to go on tour performing around the world, and needs to pack his magical boxes for the trip. He has decided that the best way to pack them would be inside another magical box, but magical boxes are quite expensive to make. Help him find the smallest magical box that can fit all his boxes.

Input

The first line of input contains an integer n (1 ≤ n ≤ 105), the number of different sizes of boxes Emuskald has. Each of following n lines contains two integers ki and ai (0 ≤ ki ≤ 109, 1 ≤ ai ≤ 109), which means that Emuskald has ai boxes with side length 2ki. It is guaranteed that all of ki are distinct.

Output

Output a single integer p, such that the smallest magical box that can contain all of Emuskald’s boxes has side length 2p.

Examples

input

Copy
2
0 3
1 5
output

Copy
3
input

Copy
1
0 4
output

Copy
1
input

Copy
2
1 10
2 2
output

Copy
3
Note

Picture explanation. If we have 3 boxes with side length 2 and 5 boxes with side length 1, then we can put all these boxes inside a box with side length 4, for example, as shown in the picture.

In the second test case, we can put all four small boxes into a box with side length 2.

题意：

小箱子可以装在大箱子里，给出各种箱子的边长及个数，求把所有箱子装入一个箱子，那么此箱子最小边长是2^i,输出i

思路：先将箱子的尺寸升序排序，在遍历一遍判断每个箱子能否放入尺寸最大的箱子里边，如果不能那么让最大箱子的尺寸++

若可以，那判断下一个。（在这我看很多博主写的是遍历每个箱子能不能放如比第一个比他大的箱子，我没有这样做，因为你想，如果这些小箱子的数量能放入最大的那个，其实肯定是有办法放进去的）

#include <iostream>
#include <cstdio>
#define LL long long
#include <cstdlib>
#include <map>
#define MAXN 200005
#include <algorithm>
#include <cmath>
using namespace std;
 
struct Edge
{
    long long ki;
    long long ai;
    bool operator<(const Edge& ee)
    {
        return ki > ee.ki;
    }
}e[100005];
 
int main()
{
    long n,i;
    long long index,ans;
    while(cin>> n)
    {
        index = -1;
        for(i=0; i<n; i++)
        {
            cin>>e[i].ki>>e[i].ai;
            if(index < e[i].ki)
                index = e[i].ki;
        }
        sort(e,e+n);
        ans = index + 1;   //ans为当前最大的盒子的边长 
        for(int i=0; i<n; i++)
        {
            index = ans - e[i].ki;
            if(index>30)  //如果最大的和一个小盒子的边长差大于30，那算下来要pow（4，index）超范围 
                continue;
            index = pow(4.0,index);//直接看当前最大的盒子能不能装下当前全部的小盒子，如果能，那么肯定 
            while(e[i].ai>index)   //是有办法可以放进去的 （大小的嵌套），但是从俯视的角度看，都是 
            {					//一样的 
                index*=4;
                ans++;
            }
        }
        cout << ans <<endl;
    }
 
    return 0;
}