zoukankan      html  css  js  c++  java
  • Codeforces 626 C. Block Towers-二分 (8VC Venture Cup 2016-Elimination Round)

     
     
    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

     

                    

             

          

  • 相关阅读:
    LOJ 2553 「CTSC2018」暴力写挂——边分治+虚树
    hdu 1028 & hdu 1398 —— 整数划分(生成函数)
    bzoj 4827 [Hnoi2017] 礼物 —— FFT
    bzoj 4503 两个串 —— FFT
    bzoj 3527 [Zjoi2014] 力 —— FFT
    bzoj 3160 万径人踪灭 —— FFT
    bzoj 2194 快速傅立叶之二 —— FFT
    bzoj 2179 FFT快速傅立叶 —— FFT
    洛谷 P3803 多项式乘法(FFT) —— FFT
    CF 1009 F Dominant Indices —— 长链剖分+指针
  • 原文地址:https://www.cnblogs.com/ZERO-/p/9692678.html
Copyright © 2011-2022 走看看