Codeforces Round 988 (Div. 3) E题解析

E题

题目链接

Codeforces Round 988 (Div. 3)

题目描述

题目的思路

根据题目的意思，我们可以推断出算法时间复杂度应该在O（N）

对于这道题而言，我们可以分析下思路

首先我们先从1~n的范围里面询问答案

如果出现0就跳过，因为无序操作

如果出现非0答案temp就记录下1~i的01序列的个数

如果我询问出来的答案都是0的话，说明我的序列肯定全是0，或者全是1，这样是无法确定的，所以我们可以输出“IMPOSSIBLE”

否则我们就可以构造满足非零答案temp的目标子串

我们可以先放i-temp-1个1，再放temp个0，再放1个1，这个1可以和前面的temp个0组成temp个01序列

然后再对i+1~n的范围进行询问答案，询问出来的答案标记为now，用temp来表示上一次询问的答案

这里以正序询问出来的答案为例

如果询问出来的答案大于上一次询问出来的答案，就说明需要增加，就向ans加“1",这个操作会和前面的0结合成01序列。

如果询问出来的答案不变，那么就说明不需要增加，向ans后面加“0”,这个1无论和前面的“0”或者“1”都不会组成01序列。

最后输出即可

AC代码

逆序查询（判断答案的逻辑和正序相反，因为构造出来的答案是反的）

点击查看代码

// Problem: E. Kachina's Favorite Binary String
// Contest: Codeforces - Codeforces Round 988 (Div. 3)
// URL: https://codeforces.com/contest/2037/problem/E#
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define debug1(a) cout << #a << '=' << a << endl;
#define debug2(a, b) cout << #a << " = " << a << "  " << #b << " = " << b << endl;
#define debug3(a, b, c) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << endl;
#define debug4(a, b, c, d) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << endl;
#define debug5(a, b, c, d, e) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << "  " << #e << " = " << e << endl;
#define vec(a)                         \
    for (int i = 0; i < a.size(); i++) \
        cout << a[i] << ' ';           \
    cout << endl;
#define darr(a, _i, _n)               \
    cout << #a << ':';                \
    for (int ij = _i; ij <= _n; ij++) \
        cout << a[ij] << ' ';         \
    cout << endl;

#define endl "\n"

#define fi first
#define se second
#define caseT \
    int T;    \
    cin >> T; \
    while (T--)
// #define int long long
// #define int __int128

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 99999999;

// const int N = ;

void solve()
{
	int n;
	cin>>n;
    int s1=0,s0=0;
    int temp;
    string s;
    for(int i=n-1;i>=1;i--)
    {
    	cout<<"? "<<i<<" "<<n<<endl;
    	cin>>temp;
    	if(!temp)continue;
    	else{
    		s1=temp;
    		s0=n-i-temp;
    		break;
    	}
    }
    if(!s1)
    {
    	 cout<<"! IMPOSSIBLE"<<endl;
    	 return;
    }
    for(int i=1;i<=s0;i++)s+="0";
    for(int i=1;i<=s1;i++)s+="1";
    s+="0";
    //这样构造就保证了第一个询问出来不为0的答案是temp
    for(int i=n-s1-s0-1;i>=1;i--)
    {
    	cout<<"? "<<i<<" "<<n<<endl;
    	int now;
    	cin>>now;
    	if(now>temp)s+="0";
    	else s+="1";
    	temp=now;
    }
    reverse(s.begin(),s.end());
    cout<<"! "<<s<<endl;
}

signed main()
{

    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
	//caseT
    int t;
    cin>>t;
    while(t--)
    {
       solve();
    }
    return 0;
}
/*

*/

正序查询

点击查看代码

// Problem: E. Kachina's Favorite Binary String
// Contest: Codeforces - Codeforces Round 988 (Div. 3)
// URL: https://codeforces.com/contest/2037/problem/E#
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define debug1(a) cout << #a << '=' << a << endl;
#define debug2(a, b) cout << #a << " = " << a << "  " << #b << " = " << b << endl;
#define debug3(a, b, c) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << endl;
#define debug4(a, b, c, d) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << endl;
#define debug5(a, b, c, d, e) cout << #a << " = " << a << "  " << #b << " = " << b << "  " << #c << " = " << c << "  " << #d << " = " << d << "  " << #e << " = " << e << endl;
#define vec(a)                         \
    for (int i = 0; i < a.size(); i++) \
        cout << a[i] << ' ';           \
    cout << endl;
#define darr(a, _i, _n)               \
    cout << #a << ':';                \
    for (int ij = _i; ij <= _n; ij++) \
        cout << a[ij] << ' ';         \
    cout << endl;

#define endl "\n"

#define fi first
#define se second
#define caseT \
    int T;    \
    cin >> T; \
    while (T--)
// #define int long long
// #define int __int128

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 99999999;

// const int N = ;

void solve()
{
	int n;
	cin>>n;
    int temp=0;//记录的是上一次的查询结果
    int p=0;//标记的是否发生改变
    vector<int>ans(n+1);
    for(int i=2;i<=n;i++)
    {
    	cout<<"? "<<1<<" "<<i<<endl;
    	int x;
    	cin>>x;
    	if(x&&!p)//如果查询出来的答案大于0且没有改变过
    	{
    		for(int j=1;j<i-x;j++)ans[j]=1;
    		for(int j=i-x;j<i;j++)ans[j]=0;
    		ans[i]=1;
    		p=1;
    	}else{
    		if(x==temp)ans[i]=0;
    		else ans[i]=1;
    	}
    	temp=x;
    }
    cout<<"! ";
    if(!p)cout<<"IMPOSSIBLE"<<endl;
    else{
    	for(int i=1;i<=n;i++)cout<<ans[i];
    	cout<<endl;
    }
}

signed main()
{

    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
	//caseT
    int t;
    cin>>t;
    while(t--)
    {
       solve();
    }
    return 0;
}
/*

*/