L. Binary String

最新推荐文章于 2022-11-24 14:08:45 发布

newhonor

最新推荐文章于 2022-11-24 14:08:45 发布

阅读量467

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/weixin_39132605/article/details/98519922

本文探讨了如何通过删除操作将二进制字符串转化为比给定整数k小的最大数值，详细解析了算法思路和实现过程，包括优先删除1保持首位1的方法，以及通过位运算和字符串操作实现的解决方案。

https://codeforc.es/gym/102001/problem/L

思路：优先删除1，保持首位1；

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <map>
#include <set>
using namespace std;

#define sfi(x) scanf("%d",&x)
#define sfc(x) scanf("%c",x)
#define sfl(x) scanf("%lld",&x)
#define sfs(x) scanf("%s",x)

#define pb push_back
#define fl() printf("flag!\n")
#define INF 0x3f3f3f3f
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int maxn=1e5+9;
const int mod=1e9+7;

ll power(ll x,ll n)
{
    ll ans=1;
    while(n)
    {
        if(n&1) ans=ans*x;
        x=x*x;
        n>>=1;
    }
    return ans;
}

ll get_num(string s)
{
    ll num=0;
    int len=s.length();
    for(int i=0;i<len;i++)
    {
        if(s[i]=='1')
            num+=power(2,len-1-i);
    }
    return num;
}

int main()
{
    //FAST_IO;

    ll k;
    cin>>k;

    string s;
    cin>>s;
    int len=s.length();

    ll tmp=k;
    string t="";
    while(tmp)
    {
        if(tmp&1)
        {
            t+='1';
        }
        else t+='0';
        tmp>>=1;
    }

    reverse(t.begin(),t.end());
    //cout<<t<<endl;

    int lent=t.length();

    int d=len-lent;

    if(d<=0)
    {
        ll num=get_num(s);
        if(num>k) cout<<1<<endl;
        else cout<<0<<endl;
        return 0;
    }

    int ans=0;
    string res="";

    for(int i=1;i<len;i++)
    {
        if(s[i]=='1')
        {
            s[i]='#';
            ans++;
            d--;
            if(d==0)break;
        }

    }


    if(d>0)
    {
        for(int i=1;i<len;i++)
        {
            if(s[i]=='0')
            {
                s[i]='#';
                ans++;
                d--;

                if(d==0) break;
            }
        }
    }

    for(int i=0;i<len;i++)
    {
        //cout<<s[i];
        if(s[i]!='#') res+=s[i];
    }
    //cout<<endl;

    ll num=get_num(res);
    if(num>k) ans++;

    //cout<<res<<endl;

    cout<<ans<<endl;


    return 0;
}

#include<cstdio>
#include<queue>
#include<map>
#include<cmath>
#include<iomanip>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#define IT set<node>::iterator
#define endl '\n'
#define FAST_IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
typedef long long LL;
typedef long long ll;
const ll inf = 1e18;
typedef unsigned long long ull;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
ll get_num(string s)
{
	ll num = 0;
	int len = s.length();
	for (ll i = 0;i<len;i++)
	{
		//vif (s[i] == '1')
			num = num * 2 + (s[i] - '0');
	}
	return num;
}

int main()
{
	//FAST_IO;

	ll k;
	cin >> k;

	string s;
	cin >> s;
	int ans = 0, flag;
	while (get_num(s) > k) {
		//cout << get_num(s) << endl;
		ans++;
		flag = 0;
		for (int i = 1;i < s.size();i++)
		{
			if (s[i] == '1')
			{
				s.erase(s.begin()+i);
				//cout << s << endl;
				flag = 1;
				break;
			}

		}
		if (!flag) {
			s.erase(s.begin()+s.size()-1);
			//cout << s << endl;
		}
	}
	//cout << ans << endl;
	/*
	if (d>0)
	{
		for (int i = 1;i<len;i++)
		{
			if (s[i] == '0')
			{
				s[i] = '#';
				ans++;
				d--;

				if (d == 0) break;
			}
		}
	}

	for (int i = 0;i<len;i++)
	{
		if (s[i] != '#') res += s[i];
	}

	ll num = get_num(res);
	if (num>k) ans++;
	*/

	cout << ans << endl;


	return 0;
}