CodeForces 626C - Block Towers(二分)

最新推荐文章于 2020-05-19 21:20:18 发布

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本文介绍了一个编程问题，学生们使用两种不同长度的积木构建塔，要求每个塔的高度不同且尽可能低。文章通过二分查找算法给出了最优解，并提供了AC代码。

C. Block Towers
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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:
分为两个阵营,一个是2的倍数,一个是3的倍数,他们都不相同,给出这两个阵营人数的个数,求这个集合中的最大值的最小值.

解题思路:
他们相互独立,那么要考虑他们的公倍数,也就是6的倍数,即对于同一个6的倍数,他只能在两个阵营中的一个.

AC代码

#include<stdio.h>
int main()
{
    int n;
    int m;
    scanf("%d%d",&n,&m);
    int left = 1;
    int right = 1000000000;
    int mid;
    int ans;
    while(right >= left)
    {
        mid = (right + left) >> 1;
        int tmp_n = mid/2;
        int tmp_m = mid/3;
        int tmp_s = mid/6;
        if(tmp_n>=n&&tmp_m>=m&&tmp_n+tmp_m-tmp_s>=n+m)  
        {
            ans = mid;
            right = mid-1;
        }
        else    left = mid+1;
    }
    printf("%d\n",ans);
    return 0;
}