思路是先把中缀表达式转成后缀表达式，然后建树求解。代码应该是很好理解的，但是想了很久也不明白问题出在哪。

#include<bits/stdc++.h>
#define N 1000005
using namespace std;

int n;
char s[N],t[N];
int tot;
char S[N],top;
int S2[N],l[N],r[N],root;
int ans1,ans2;

int dfs(int u)
{
	if(t[u] == '0' || t[u] == '1')return t[u]-'0';
	int L = dfs(l[u]);
	if(L == 0 && t[u] == '&')
	{
		ans1++;
		return 0;
	}
	if(L == 1 && t[u] == '|')
	{
		ans2++;
		return 1;
	}
	int R = dfs(r[u]);
	return R;
}

int main()
{
	scanf("%s",s+1),n = strlen(s+1);
	for(int i = 1;i <= n;i++)
	{
		if(s[i] == '0' || s[i] == '1')t[++tot] = s[i];
		else
		{
			if(s[i] == '&'){while(S[top]=='&'&&top)t[++tot] = S[top--];S[++top]='&';}
			if(s[i] == '|'){while(S[top]!='('&&top)t[++tot] = S[top--];S[++top]='|';}
			if(s[i] == '(')S[++top] = s[i];
			if(s[i] == ')'){while(S[top]!='('&&top)t[++tot] = S[top--];top--;}
		}
	} 
	while(top)t[++tot] = S[top--];
	//for(int i = 1;i <= tot;i++)printf("%c",t[i]);
	//printf("\n");
	for(int i = 1;i <= tot;i++)
	{
		if(t[i] == '0' || t[i] == '1')S2[++top] = i;
		else
		{
			r[i] = S2[top--],l[i] = S2[top--];
			S2[++top] = root = i;
		}
	}
	//for(int i = 1;i <= tot;i++)printf("(%d)-%d-%d\n",i,l[i],r[i]);
	printf("%d\n",dfs(tot));
	printf("%d %d\n",ans1,ans2);
	return 0;
}