二分&&Block Towers

r囧r小猫 2022-05-27 10:07 89阅读 0赞

题目链接：[点击打开链接][Link 1]

C. Block Towers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Students in a class are making towers of blocks. Each student makes a (non-zero) tower by stacking pieces lengthwise on top of each other. *n* of the students use pieces made of two blocks and *m* of the students use pieces made of three blocks.

The students don’t want to use too many blocks, but they also want to be unique, so no two students’ towers may contain the same number of blocks. Find the minimum height necessary for the tallest of the students' towers.

Input

The first line of the input contains two space-separated integers *n* and *m* (0 ≤ *n*, *m* ≤ 1 000 000, *n* + *m* > 0) — the number of students using two-block pieces and the number of students using three-block pieces, respectively.

Output

Print a single integer, denoting the minimum possible height of the tallest tower.

Examples

input

Copy

1 3

output

Copy

input

Copy

3 2

output

Copy

input

Copy

5 0

output

Copy

Note

In the first case, the student using two-block pieces can make a tower of height 4, and the students using three-block pieces can make towers of height 3, 6, and 9 blocks. The tallest tower has a height of 9 blocks.

In the second case, the students can make towers of heights 2, 4, and 8 with two-block pieces and towers of heights 3 and 6 with three-block pieces, for a maximum height of 8 blocks.

二分实现：

#include<bits/stdc++.h>
    using namespace std;
    int n,m;
    int f(int x)
    {
        int num1=x/2;//2的数量
        int num2=x/3;//3的数量
        int num3=x/6;//2&&3的数量
        /*
        要同时满足
        除以2的数量大于等于n 
        除以3的数量大于m 
        除以2的数量+除以3的数量-除以6的数量大于等于m+n
        这三个条件时才成立*/
        if(num1<n||num2<m||num1+num2-num3<m+n) return 0;
        else return 1;
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            int l=0,r=1e7;
            int ans;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(f(mid)) ans=mid,r=mid-1;
                else l=mid+1;
            }
            printf("%d\n",ans);
        }
    }

直接实现：

#include<bits/stdc++.h>
    using namespace std;
    int n,m;
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
          for(int i=0;;i++)
          {
              if(i/2>=n&&i/3>=m&&i/2+i/3-i/6>=m+n)
              {
                  printf("%d\n",i);
                  break;
              }
          }
        }
        return 0;
    }

[Link 1]: http://codeforces.com/problemset/problem/626/C