题目链接：
http://codeforces.com/problemset/problem/626/C
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
9
input
Copy
3 2
output
Copy
8
input
Copy
5 0
output
Copy
10
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.
题意：
有n个人用2米高的木板堆塔，有m个人用3米高的木板堆塔。木板可以随便使用，要求任意两人堆成塔的高度不一，问满足该条件下所有人堆成塔的最大高度的最小值。
思路：
最大值的最小化问题，二分找最值。
This is the code
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long //1844674407370955161
#define INT_INF 0x7f7f7f7f //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]= {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]= {-1, 1, 0, 0, -1, 1, -1, 1};
bool check(int num,int n,int m)
{
int o2=num/2;
int o3=num/3;
int o6=num/6;
if(n<=o2 && m<=o3 && n+m<=o2+o3-o6)
return true;
return false;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
int l=2;//注意范围
int r=3000000;
int mid;
while(l<=r)
{
mid=(l+r)>>1;
if(check(mid,n,m))
r=mid-1;
else
l=mid+1;
}
printf("%d\n",r+1);//因为最后一个满足的mid传给r的时候是mid-1，所以要加一输出
}
return 0;
}