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/*
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测试用例1:
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0&(1|0)|(1|1|1&0)
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答案:
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1
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1 2
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测试用例2:
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(0|1&0|1|1|(1|1))&(0&1&(1|0)|0|1|0)&0
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答案:
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0
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2 3
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*/
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#include <bits/stdc++.h>
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using namespace std;
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struct Node {
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int v, a, b; // v:代表当前的结果值,a: &短路的次数 b:|短路的次数
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};
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stack<Node> num;
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stack<char> stk;
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unordered_map<char, int> h{{'|', 1}, {'&', 2}};
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void eval() {
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// 这里要注意从栈中弹出元素的顺序,先出来的是y,后出来的是x
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Node y = num.top();
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num.pop();
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Node x = num.top();
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num.pop();
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char p = stk.top();
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stk.pop();
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Node r;
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if (p == '|') {
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if (x.v == 1)
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r = {1, x.a, x.b + 1};
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else
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r = {y.v, x.a + y.a, x.b + y.b};
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} else if (p == '&') {
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if (x.v == 1)
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r = {y.v, x.a + y.a, x.b + y.b};
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else
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r = {0, x.a + 1, x.b};
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}
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num.push(r);
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}
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int main() {
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string s;
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cin >> s;
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for (int i = 0; i < s.size(); i++) {
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if (isdigit(s[i])) {
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int x = 0;
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while (i < s.size() && isdigit(s[i])) {
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x = x * 10 + s[i] - '0';
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i++;
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}
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i--;
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num.push({x, 0, 0});
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} else if (s[i] == '(')
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stk.push(s[i]);
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else if (s[i] == ')') {
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while (stk.top() != '(') eval();
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stk.pop();
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} else {
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while (stk.size() && h[stk.top()] >= h[s[i]]) eval();
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stk.push(s[i]);
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}
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}
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while (stk.size()) eval();
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printf("%d\n", num.top().v);
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printf("%d %d\n", num.top().a, num.top().b);
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return 0;
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}
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