Codeforces 626C Block Towers 【二分】

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

1 3

output

9

input

3 2

output

8

input

5 0

output

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...的最...，二分就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF (2000000000+10)
#define eps 1e-8#define MAXN (200000+10)
#define MAXM (600000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
bool judge(int o, int n, int m)
{
int num1 = o / 2;
int num2 = o / 3;
int num3 = o / 6;
return n <= num1 && m <= num2 && n + m <= num1 + num2 - num3;
}
int main()
{
int n, m; Ri(n); Ri(m);
int l = 0, r = 8000000, ans;
while(r >= l)
{
int mid = (l + r) >> 1;
if(judge(mid, n, m))
{
ans = mid;
r = mid-1;
}
else
l = mid+1;
}
Pi(ans);
return 0;
}