rt

思路是先建立后缀表达式，然后扫一遍，如果出现0&或者1|就ans++;

#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;

int n, ans1, ans2;
char res;
string s;
stack<char> op, num;
vector<char> v;

void build(){
	for(int i=0; i<=n; i++){
		if(s[i]=='0'||s[i]=='1'){
			v.push_back(s[i]);
		}
		else if(s[i]=='('){
			op.push(s[i]);
		}
		else if(s[i]==')'){
			while(!op.empty() && op.top()!='('){
				v.push_back(op.top());
				op.pop();
			}
			op.pop();
		}
		else if(s[i]=='&'){
			while(!op.empty() && op.top()=='&'){
				v.push_back(op.top());
				op.pop();
			}
			op.push(s[i]);
		}
		else if(s[i]=='|'){
			while(!op.empty() && op.top()!='('){
				v.push_back(op.top());
				op.pop();
			}
			op.push(s[i]);
		}
	}
	while(!op.empty()){
		v.push_back(op.top());
		op.pop();
	}
}

void solve(){
	for(int i=0; i<v.size(); i++){
		if(v[i]=='0'||v[i]=='1'){
			num.push(v[i]);
		}
		else if(v[i]=='&'){
			char a=num.top();num.pop();
			char b=num.top();num.pop();
			char x='0';
			if(a=='0')ans1++;
			if(a=='1' && b=='1') x='1';
			num.push(x);
		}
		else if(v[i]=='|'){
			char a=num.top();num.pop();
			char b=num.top();num.pop();
			char x;
			if(a=='1')ans2++;
			if(a=='1' || b=='1') x='1';
			num.push(x);
		}
	}
	res=num.top();
}

int main()
{
	cin>>s;
	n=s.size()-1;
	build();
	solve();
	cout<<res<<endl;
	cout<<ans1<<" "<<ans2;
	return 0;
}

p.s. 样例2过不了

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