思路
先考虑到min(xdy)=x,max(xdy)=x*y
接着就是把d的优先级设置为大于*
最后就是中缀表达式转后缀表达式求值即可。
模板
#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define x first
#define y second
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=a-1;i>=(b);--i)
#define fuck(x) cout<<'['<<#x<<' '<<(x)<<']'
#define sub(x,y) x=((x)-(y)<0)?(x)-(y)+mod:(x)-(y)
#define clr(a,b) memset(a,b,sizeof(a))
#define eps 1e-10
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef unsigned int ui;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 1e9 + 7;
const int MX = 2e6 + 5;
vector<string>pre, s;
string str;
bool isoperator(string op) {
if(op == "+" || op == "-" || op == "*" || op == "d") return 1;
return 0;
}
int priority(string op) {
if(op == "#") return -1;
if(op == "(") return 0;
if(op == "+" || op == "-") return 1;
if(op == "*") return 2;
if(op == "d") return 3;
return -1;
}
void postfix() {
stack<string> OPTR; //运算符栈
stack<string> OPND; //数据栈
OPTR.push("#");
rep(i, 0, pre.size()) {
if (pre[i] == "(") OPTR.push(pre[i]);
else if(pre[i] == ")") {
while(OPTR.top() != "(") {
OPND.push(OPTR.top());
OPTR.pop();
}
OPTR.pop();
} else if (isoperator(pre[i])) {
while(!OPTR.empty() && (priority(pre[i]) < priority(OPTR.top()) || priority(pre[i]) == priority(OPTR.top()) && pre[i] != "d")) {
OPND.push(OPTR.top());
OPTR.pop();
}
OPTR.push(pre[i]);
} else OPND.push(pre[i]);
}
while(OPTR.top() != "#") {
OPND.push(OPTR.top());
OPTR.pop();
}
OPTR.pop();
//利用操作符栈逆序即可得到后缀表达式
while(!OPND.empty()) {
OPTR.push(OPND.top());
OPND.pop();
}
s.clear();
while(!OPTR.empty()) {
s.push_back(OPTR.top());
OPTR.pop();
}
}
bool is_dig(char ch) {return ch >= '0' && ch <= '9';}
void pre_solve() {
pre.clear();
rep(i, 0, str.length()) {
if(is_dig(str[i])) {
rep(j, i, str.length()) {
if(!is_dig(str[j])) {
pre.push_back(str.substr(i, j - i));
i = j - 1;
break;
}
if(j == str.length() - 1) {
pre.push_back(str.substr(i, j - i + 1));
i = j;
break;
}
}
} else pre.push_back(str.substr(i, 1));
}
}
ll str_to_int(string st) {
ll ret = 0;
rep(i, 0, st.length()) ret = ret * 10 + st[i] - '0';
return ret;
}
struct node {
ll l, r;//重载运算付即可
node(ll l = 0, ll r = 0): l(l), r(r) {}
node(string st) {l = r = str_to_int(st);}
node operator-(const node& _A)const {
return node(l - _A.r, r - _A.l);
}
node operator+(const node& _A)const {
return node(l + _A.l, r + _A.r);
}
node operator*(const node& _A)const {
node ret;
ll a = l * _A.l;
ll b = l * _A.r;
ll c = r * _A.l;
ll d = r * _A.r;
ret.l = min(min(a, b), min(c, d));
ret.r = max(max(a, b), max(c, d));
return ret;
}
node operator/(const node& _A)const {
node ret;
ll l1 = max(l, 0ll), r1 = r;
ret.l = l1, ret.r = r1 * _A.r;
return ret;
}
};
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
while(cin >> str)
{
pre_solve();
postfix();
stack<node> stk;
node a, b;
for(int i=0;i<s.size();i++)
{
if(isoperator(s[i]))
{
b = stk.top(); stk.pop();
a = stk.top(); stk.pop();
if(s[i] == "-") a = a - b;
if(s[i] == "+") a = a + b;
if(s[i] == "*") a = a * b;
if(s[i] == "d") a = a / b;
stk.push(a);
}
else stk.push(node(s[i]));
}
a = stk.top();
printf("%lld %lld\n", a.l, a.r);
}
return 0;
}