标签：技术分享 gif ret ons splay imu spec not pos
个人心得：这题没怎么看，题意难懂。后面比完再看的时候发现很好做但是怎么卡时间是个问题。
题意：就是有m个可以用2层积木的，n个可以用三层积木的，但是他们不允许重复所以可以无限添加。
比如 3 2
一开始是2层的开始2，然后 3，然后 4，此时再添加都一样了，为了保证最小高度所以3+3=6，此时的2层的就要添加2个才不重样
网上大神多，这个代码我服，按照函数关系俩着重复的地点都有规律，所以只要找到此时最大m*2,n*3然后碰到一次相同就让最小的最大值增加就可以了
题目：
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.
Example
1 3
9
3 2
8
5 0
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 9blocks. 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.
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<iomanip> 6 #include<algorithm> 7 using namespace std; 8 const int maxn=1005; 9 int main() 10 { 11 int a,b; 12 int m,n; 13 while(cin>>m>>n){ 14 int a=m*2,b=n*3; 15 for(int i=6;i<=min(a,b);i+=6){ 16 if(a<=b) a+=2; 17 else b+=3; 18 } 19 cout<<max(a,b)<<endl; 20 } 21 return 0; 22 }
标签：技术分享 gif ret ons splay imu spec not pos
原文地址：http://www.cnblogs.com/blvt/p/7788076.html
