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.
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.
Print a single integer, denoting the minimum possible height of the tallest tower.
1 3
9
3 2
8
5 0
10
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.
题意就是n个人用2的倍数,m个人用3的倍数,然后所有人要求数不一样,要使得最大的数最小,输出结果。
二分讨论2的倍数,3的倍数,6的倍数,然后注意一下边界,wa在边界上了。。。
代码:
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m,l,r,mid,ans;
int a,b,c;
while(~scanf("%d%d",&n,&m)){
int l=1,r=1e8;
while(l<=r){
mid=(l+r)/2;
a=mid/2;
b=mid/3;
c=mid/6;
if(a>=n&&b>=m&&a+b-c>=n+m){
ans=mid;
r=mid-1;
}
else
l=mid+1;
}
printf("%d
",ans);
}
return 0;
}
智障萌新,加油(ง •_•)ง