Codeforces1560 D. Make a Power of Two(思维+暴力)
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题意:
解法:
2的倍数只有log种,
因此枚举2的倍数t,计算出n变成t至少需要多少次操作即可,
计算过程很简单,从高到低遍历n的数位,用来匹配t的从高到低的数位,
能匹配则匹配,不能用就必须删掉,
如果最后还是不能匹配玩,那么需要在使用加数操作.
对计算出来的数取min就是答案.
code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
vector<int>temp;
int n;
int check(int x){
stack<int>s;
while(x){
s.push(x%10);
x/=10;
}
int ans=0;
for(auto i:temp){
if(s.size()&&i==s.top()){
s.pop();
}else{
ans++;
}
}
ans+=s.size();
return ans;
}
void solve(){
cin>>n;
map<int,int>mp;
int x=n;
temp.clear();
while(x){
temp.push_back(x%10);
x/=10;
}
reverse(temp.begin(),temp.end());
int ans=1e9;
for(int i=1;i<=1e18;i*=2){
ans=min(ans,check(i));
}
cout<<ans<<endl;
}
signed main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
ios::sync_with_stdio(0);cin.tie(0);
int T;cin>>T;while(T--)
solve();
return 0;
}
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