C. Rings

先贴上我错误的代码：

#include <bits/stdc++.h>
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define f(i,a,n) for(int i=a;i<n;++i)
#define ff(i,a,n) for(int i=a;i<=n;++i)
#define OP freopen("D:\\Users\\HP ZBook15\\CLionProjects\\untitled\\demo\\in.TXT","r",stdin);
#define CL freopen("D:\\Users\\HP ZBook15\\CLionProjects\\untitled\\demo\\out.TXT","w",stdout);
const int INF=0x3f3f3f3f;
using namespace std;
typedef long long ll;
typedef double dbl;
typedef pair<int, int> pi;
//
int n;
void solve(){
    
	cin>>n;
	string s;
	cin>>s;
	int flag=0,p=0;
	for(int i=n-1;i>=n/2;i--){
    
		if(s[i]=='0'){
    
			flag=i;
			break;
		}
	}
	if(flag){
    
		cout<<1<<" "<<flag+1<<" "<<1<<" "<<flag<<endl;
	}
	else{
    
		cout<<n/2<<" "<<n-1<<" "<<n/2+1<<" "<<n<<endl;
	}
}
int main(){
    
    multiCase solve();
    return 0;
}

错误原因：

在else里面判断失误，错误样例

2
4
1011
5
10111

我的错解：

2 3 3 4
2 4 3 5

正解1：

1 2 2 3 
1 2 2 3 

正解2：

2 4 3 4
2 5 3 5

解释：

当后一半（n为奇数的时候后一半多一个元素）没有0的时候，要用到前面一个数字去构造，如果前面一个是1，那么可以用，如果前面的不是1，那么就不能用这个数字了，下面两个代码看个人 思维习惯 理解吧

ac代码：

一种思维习惯：

#include <bits/stdc++.h>
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define f(i,a,n) for(int i=a;i<n;++i)
#define ff(i,a,n) for(int i=a;i<=n;++i)
#define OP freopen("D:\\Users\\HP ZBook15\\CLionProjects\\untitled\\demo\\in.TXT","r",stdin);
#define CL freopen("D:\\Users\\HP ZBook15\\CLionProjects\\untitled\\demo\\out.TXT","w",stdout);
const int INF=0x3f3f3f3f;
using namespace std;
typedef long long ll;
typedef double dbl;
typedef pair<int, int> pi;
//
int n;
void solve(){
    
	cin>>n;
	string s;
	cin>>s;
	int flag=0,p=0;
	for(int i=n-1;i>=n/2;i--){
    
		if(s[i]=='0'){
    
			flag=i;
			break;
		}
	}
	if(flag) cout<<1<<" "<<flag+1<<" "<<1<<" "<<flag<<endl;
	else cout<<n/2<<" "<<(s[n/2-1]=='1'?n-1:n)<<" "<<n/2+1<<" "<<n<<endl;
}
int main(){
    
    multiCase solve();
    return 0;
}
/*
4
4
1011
5
10111
4
1111
5
11111
*/

另一种思维习惯：

#include <bits/stdc++.h>
using namespace std;
void solve() {
    
    int n;
    cin >> n;

    string s;
    cin >> s;

    for (int i = 0; i < n; i++) {
    
        if (s[i] == '0') {
    
            if (i < n / 2) {
    
                cout << i + 1 << " " << n << " " << i + 2 << " " << n << "\n";
            } else {
    
                cout << 1 << " " << i + 1 << " " << 1 << " " << i << "\n";
            }
            return;
        }
    }

    cout << 1 << " " << n / 2 * 2 << " " << 1 << " " << n / 2 << "\n";
}

int main() {
    
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

    while (t--) {
    
        solve();
    }

    return 0;
}