【CodeForces】 - 626C 枚举

最新推荐文章于 2021-09-17 20:59:02 发布

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最新推荐文章于 2021-09-17 20:59:02 发布

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分类专栏： 2019年寒假训练日记

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本篇博客探讨了CodeForces 626C问题，主要内容是寻找n个2的倍数和m个3的倍数中，不重复的最大数的最小值。提供了两种解法：一是枚举最大值i，确保2或3的倍数不重复且数量足够；二是枚举重复数字，从6开始每次加6，通过调整结尾位置消除重复，直到满足条件。

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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：

给你n，m，如果 n个2的倍数和m个3的倍数，这n+m个数各不相同，那么求最大的数的最小值。

分析：

方法1：枚举最大值为i，直到 i/2+i/3-i/6（不重复的2或3的倍数）≥n+m，并且要i/2（2的倍数）≥n，i/3（3的倍数）≥m。

方法2：枚举重复的数字i，i最小为6，每次增加6，设置两个结尾初始值为2*n，3*m，当两个结尾都比i大时，那就是还有重复。然后加在比较短的结尾。直到不再有重复。

#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
    int n,m,i;
    scanf("%d%d",&n,&m);
    i=max(2*n,3*m);
    while(i/2+i/3-i/6<n+m||i/2<n||i/3<m)
    {
        i++;
    }
    printf("%d",i);
    return 0;
}

#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
    int n,m,end1,end2;
    scanf("%d%d",&n,&m);
    end1=2*n;
    end2=3*m;
    for(int i=6; i<=min(end1,end2); i+=6)
    {
        if(end1<=end2)//如果相等，加到2的倍数，更小
        {
            end1+=2;
        }
        else
        {
            end2+=3;
        }
    }
    printf("%d",max(end1,end2));
    return 0;
}