题目

链接

https://www.luogu.com.cn/problem/CF1560D

字面描述

题面翻译

给定一个整数 $n$。每次操作你可以做两件事情中的一件：

• 删去这个数中的一个数位（如果这个数只剩下一位，则可以把它删空）。
• 在这个数的右边添加一个数位。

你可以以任意顺序执行无限次操作。但请注意，在删去一个数位之后，这个数可能包含前导零（例如在删去 $301$ 中的 $3$ 这一位之后，这个数就会变成 $01$ 而不是 $1$）。

你需要执行若干次操作，使得这个数最终变成一个 $2$ 的次幂，或者说存在一个非负整数 $k$ 使得这个数最终是 $2^k$。最终答案不能包含前导零。请求出需要执行的操作的最小次数。

$t$ 组数据，$1⩽t⩽10^4$，$1⩽n⩽10^9$。

题目描述

You are given an integer $ n $ . In $ 1 $ move, you can do one of the following actions:

• erase any digit of the number (it’s acceptable that the number before the operation has exactly one digit and after the operation, it is “empty”);
• add one digit to the right.

The actions may be performed in any order any number of times.

Note that if, after deleting some digit from a number, it will contain leading zeroes, they will not be deleted. E.g. if you delete from the number $ 301 $ the digit $ 3 $ , the result is the number $ 01 $ (not $ 1 $ ).

You need to perform the minimum number of actions to make the number any power of $ 2 $ (i.e. there’s an integer $ k $ ( $ k \ge 0 $ ) such that the resulting number is equal to $ 2^k $ ). The resulting number must not have leading zeroes.

E.g. consider $ n=1052 $ . The answer is equal to $ 2 $ . First, let’s add to the right one digit $ 4 $ (the result will be $ 10524 $ ). Then let’s erase the digit $ 5 $ , so the result will be $ 1024 $ which is a power of $ 2 $ .

E.g. consider $ n=8888 $ . The answer is equal to $ 3 $ . Let’s erase any of the digits $ 8 $ three times. The result will be $ 8 $ which is a power of $ 2 $ .

输入格式

The first line contains one integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Then $ t $ test cases follow.

Each test case consists of one line containing one integer $ n $ ( $ 1 \le n \le 10^9 $ ).

输出格式

For each test case, output in a separate line one integer $ m $ — the minimum number of moves to transform the number into any power of $ 2 $ .

样例 #1

样例输入 #1

12
1052
8888
6
75
128
1
301
12048
1504
6656
1000000000
687194767

样例输出 #1

2
3
1
3
0
0
2
1
3
4
9
2

提示

The answer for the first test case was considered above.

The answer for the second test case was considered above.

In the third test case, it’s enough to add to the right the digit $ 4 $ — the number $ 6 $ will turn into $ 64 $ .

In the fourth test case, let’s add to the right the digit $ 8 $ and then erase $ 7 $ and $ 5 $ — the taken number will turn into $ 8 $ .

The numbers of the fifth and the sixth test cases are already powers of two so there’s no need to make any move.

In the seventh test case, you can delete first of all the digit $ 3 $ (the result is $ 01 $ ) and then the digit $ 0 $ (the result is $ 1 $ ).

思路

本题最终的目的是将一个[1,1e9]的整数通过2种操作变为2的非负整数次幂。

将2⁰ - 2⁵⁵ 每一个数位上的数打表预处理，用原数列一一与表中元素比较，计算操作次数，记录最小值。

时间复杂度: $O(10t)\approx 1e5$

代码实现

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxn=100;
const int inf=2e9;
int t,ans=inf,tot;
ll op=1,x;
ll a[maxn];
int k[maxn];//记录表中每一元素的长度
ll cnt[maxn][maxn];
int main(){//预处理++k[0];cnt[0][1]=1;for(int i=1;i<=55;i++){op=(ll)op*2;ll op1=op;while(op1){cnt[i][++k[i]]=(ll)op1%10;op1=(ll)op1/10;}}/*for(int i=1;i<=40;i++){for(int j=k[i];j>=1;j--)printf("%lld",cnt[i][j]);printf("\n");}*/scanf("%d",&t);while(t--){ans=inf;scanf("%lld",&x);tot=0;while(x){a[++tot]=(ll)x%10;x=(ll)x/10;}//for(int i=tot;i>=1;i--)printf("%d",a[i]);//计算比较for(int i=0;i<=55;i++){int st1=tot,st2=k[i];while(st1>0&&st2>0){if(a[st1]==cnt[i][st2])--st2;--st1;}//if(i==41)printf("41 %d %d\n",k[i],st2);ans=min(ans,tot-k[i]+2*st2);// tot-(k[i]-st2) 删除的操作次数操作次数 & st2 添加的操作次数}printf("%d\n",ans);}return 0;
}

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