预处理两个数组:
vis[x]，有几个数能够变成x
num[x]，所有数变成x最少需要变化的步数
ans=min(num[x])，vis[x]==n
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int num[int(1e5)+10],vis[int(1e5)+10];
void work(int val)
{
int t=val*2,step=1;
while(t<=int(1e5))
{
vis[t]++;
num[t]+=step;
step++;
t<<=1;
}
t=val,step=0;
while(t>0)
{
vis[t]++;
num[t]+=step;
if((t&1)&&t>1)
{
int ts=step+2,tt=t/2*2;
while(tt<=int(1e5))
{
vis[tt]++;
num[tt]+=ts;
tt<<=1;
ts++;
}
}
step++;
t>>=1;
}
}
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++)
{
int t;
cin>>t;
work(t);
}
int ans=INT_MAX;
for(int i=1;i<=int(1e5);i++)
if(vis[i]==n)
ans=min(ans,num[i]);
cout<<ans;
}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.
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?
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 one integer the minimum number of operations required to make all the chemicals volumes equal.
3 4 8 2
2
3 3 5 6
5
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.
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codeforces 558 C Amr and Chemistry
原文：http://blog.csdn.net/stl112514/article/details/46889707