• Codeforces Round #312 (Div. 2) C.Amr and Chemistry


    Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.

    Amr has n different types of chemicals. Each chemical i has an initial volume of ai liters. For this experiment, Amr has to mix all the chemicals together, but all the chemicals volumes must be equal first. So his task is to make all the chemicals volumes equal.

    To do this, Amr can do two different kind of operations.

    • Choose some chemical i and double its current volume so the new volume will be 2ai
    • Choose some chemical i and divide its volume by two (integer division) so the new volume will be 

    Suppose that each chemical is contained in a vessel of infinite volume. Now Amr wonders what is the minimum number of operations required to make all the chemicals volumes equal?

    Input

    The first line contains one number n (1 ≤ n ≤ 105), the number of chemicals.

    The second line contains n space separated integers ai (1 ≤ ai ≤ 105), representing the initial volume of the i-th chemical in liters.

    Output

    Output one integer the minimum number of operations required to make all the chemicals volumes equal.

    Sample test(s)
    input
    3
    4 8 2
    output
    2
    input
    3
    3 5 6
    output
    5
    Note

    In the first sample test, the optimal solution is to divide the second chemical volume by two, and multiply the third chemical volume by two to make all the volumes equal 4.

    In the second sample test, the optimal solution is to divide the first chemical volume by two, and divide the second and the third chemical volumes by two twice to make all the volumes equal 1.

    题解:按值爆搜交上去就满了。。。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cmath>
     4 #include<algorithm>
     5 #include<queue>
     6 #include<cstring>
     7 #define PAU putchar(' ')
     8 #define ENT putchar('
    ')
     9 using namespace std;
    10 const int maxn=100000+10,maxv=100000,inf=-1u>>1;
    11 int vis[maxn],cnt[maxn],stp[maxn],n;
    12 inline int read(){
    13     int x=0,sig=1;char ch=getchar();
    14     while(!isdigit(ch)){if(ch=='-')sig=-1;ch=getchar();}
    15     while(isdigit(ch))x=10*x+ch-'0',ch=getchar();
    16     return x*=sig;
    17 }
    18 inline void write(int x){
    19     if(x==0){putchar('0');return;}if(x<0)putchar('-'),x=-x;
    20     int len=0,buf[15];while(x)buf[len++]=x%10,x/=10;
    21     for(int i=len-1;i>=0;i--)putchar(buf[i]+'0');return;
    22 }
    23 void init(){
    24     n=read();
    25     queue<pair<int,int> >Q;
    26     for(int i=1;i<=n;i++){
    27         int num=read();Q.push(make_pair(num,0));
    28         while(!Q.empty()){
    29             int x=Q.front().first,y=Q.front().second;Q.pop();
    30             if(x>maxv||vis[x]==i)continue;
    31             vis[x]=i;cnt[x]++;stp[x]+=y;
    32             Q.push(make_pair(x<<1,y+1));
    33             Q.push(make_pair(x>>1,y+1));
    34         }
    35     }
    36     int mi=inf;
    37     for(int i=0;i<=maxv;i++)if(cnt[i]==n&&stp[i]<mi)mi=stp[i];
    38     write(mi);
    39     return;
    40 }
    41 void work(){
    42     return;
    43 }
    44 void print(){
    45     return;
    46 }
    47 int main(){init();work();print();return 0;}
  • 相关阅读:
    Using PL/SQL Control Structures
    Oracle JDBC
    PL/SQL 命名规则
    PL/SQL User's Guide and Reference:PL/SQL Architecture
    JAVA中调用存储过程和函数
    Oracle正则表达式
    java根据网卡名称获取IP
    开放通知接口,执行文件同步操作
    Jquery UI Dialog Demo
    esapi2.0GA.jar的license
  • 原文地址:https://www.cnblogs.com/chxer/p/4676798.html
Copyright © 2020-2023  润新知