E题
题目链接
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;
}
/*
*/