Codeforces 626C Block Towers

C. Block Towerstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputStudents 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.InputThe 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.OutputPrint a single integer, denoting the minimum possible height of the tallest tower.ExamplesinputCopy

1 3

output

9

inputCopy

3 2

output

8

inputCopy

5 0

output

10

NoteIn 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.
题意：给了两种叠塔的同学，一种只能一次用两块砖，另一种用三块砖，而且不同同学叠塔的高度不一样，问叠塔最高的高度的
最小值是多少
这里我们用二分寻找最小值，满足的条件为：假设高度为x, 两块的高度至少要2 * n，三块的高度至少要3 * m,但是有可能有公约数 ，所以n+m <= x / 2 + x / 3 - x / 6
代码： #include<cstdio>
#include<iostream>
using namespace std;
int n,m;
bool judge(int mid)
{
if(n <= mid / 2 && m <= mid / 3 && (mid / 2 + mid / 3 - mid / 6 >= n + m))
return 1;
else
return 0;
}
int main()
{
cin>>n>>m;
int l = 1,r = 3000000,pos;
while(l <= r)
{
int mid = (l + r) / 2;
if(judge(mid))
{
pos = mid;
r = mid - 1;
}
else
{
l = mid + 1;
}
}
cout<<pos<<endl;
return 0;
}