全网最高端?中缀表达式转为后缀表达式以及求值(可用于负数,阶乘)
Posted nightraven
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了全网最高端?中缀表达式转为后缀表达式以及求值(可用于负数,阶乘)相关的知识,希望对你有一定的参考价值。
代码里有注释。。。直接上代码。。。
#include<bits/stdc++.h>
#define rep(i,k,n) for(int i=k;i<=n;i++)
#define per(i,n,k) for(int i=n;i>=k;i--)
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define re return
#define se second
#define fi first
using namespace std;
//---------------------------------------------------------------head----------------------------------------------------------------------
struct BigInteger {
typedef unsigned long long LL;
static const int BASE = 100000000;
static const int WIDTH = 8;
vector<int> s;
BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
BigInteger(LL num = 0) {*this = num;}
BigInteger(string s) {*this = s;}
BigInteger& operator = (long long num) {
s.clear();
do {
s.push_back(num % BASE);
num /= BASE;
} while (num > 0);
return *this;
}
BigInteger& operator = (const string& str) {
s.clear();
int x, len = (str.length() - 1) / WIDTH + 1;
for (int i = 0; i < len; i++) {
int end = str.length() - i*WIDTH;
int start = max(0, end - WIDTH);
sscanf(str.substr(start,end-start).c_str(), "%d", &x);
s.push_back(x);
}
return (*this).clean();
}
BigInteger operator + (const BigInteger& b) const {
BigInteger c; c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = g;
if (i < s.size()) x += s[i];
if (i < b.s.size()) x += b.s[i];
c.s.push_back(x % BASE);
g = x / BASE;
}
return c;
}
BigInteger operator - (const BigInteger& b) const {
assert(b <= *this); // 减数不能大于被减数
BigInteger c; c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = s[i] + g;
if (i < b.s.size()) x -= b.s[i];
if (x < 0) {g = -1; x += BASE;} else g = 0;
c.s.push_back(x);
}
return c.clean();
}
BigInteger operator * (const BigInteger& b) const {
int i, j; LL g;
vector<LL> v(s.size()+b.s.size(), 0);
BigInteger c; c.s.clear();
for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
for (i = 0, g = 0; ; i++) {
if (g ==0 && i >= v.size()) break;
LL x = v[i] + g;
c.s.push_back(x % BASE);
g = x / BASE;
}
return c.clean();
}
BigInteger operator / (const BigInteger& b) const {
assert(b > 0); // 除数必须大于0
BigInteger c = *this; // 商:主要是让c.s和(*this).s的vector一样大
BigInteger m; // 余数:初始化为0
for (int i = s.size()-1; i >= 0; i--) {
m = m*BASE + s[i];
c.s[i] = bsearch(b, m);
m -= b*c.s[i];
}
return c.clean();
}
BigInteger operator % (const BigInteger& b) const { //方法与除法相同
BigInteger c = *this;
BigInteger m;
for (int i = s.size()-1; i >= 0; i--) {
m = m*BASE + s[i];
c.s[i] = bsearch(b, m);
m -= b*c.s[i];
}
return m;
}
int bsearch(const BigInteger& b, const BigInteger& m) const{
int L = 0, R = BASE-1, x;
while (1) {
x = (L+R)>>1;
if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
else R = x;
}
}
BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}
bool operator < (const BigInteger& b) const {
if (s.size() != b.s.size()) return s.size() < b.s.size();
for (int i = s.size()-1; i >= 0; i--)
if (s[i] != b.s[i]) return s[i] < b.s[i];
return false;
}
bool operator >(const BigInteger& b) const{return b < *this;}
bool operator<=(const BigInteger& b) const{return !(b < *this);}
bool operator>=(const BigInteger& b) const{return !(*this < b);}
bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
};
ostream& operator << (ostream& out, const BigInteger& x) {
out << x.s.back();
for (int i = x.s.size()-2; i >= 0; i--) {
char buf[20];
sprintf(buf, "%08d", x.s[i]);
for (int j = 0; j < strlen(buf); j++) out << buf[j];
}
return out;
}
istream& operator >> (istream& in, BigInteger& x) {
string s;
if (!(in >> s)) return in;
x = s;
return in;
}
stack<BigInteger> si;
stack<char> sc;
vector<BigInteger> vc;
bool isnum[1005];
bool flag=0;
BigInteger pow(BigInteger a,BigInteger b){BigInteger base=a,ans=1;while(b!=0){if(b%2==1)ans*=base;base*=base;b/=2;}return ans;}
BigInteger clas(char c)//符号优先级
{
if(c=='(' || c==')') re 0;
if(c=='+' || c=='-') re 1;
if(c=='*' || c=='/') re 2;
if(c=='^') re 3;
re 4;
}
BigInteger cal(BigInteger a,BigInteger b,BigInteger c)//符号求值
{
BigInteger ans=1;
if(c=='+') re a+b;
if(c=='-') re b-a;
if(c=='*') re a*b;
if(c=='/') re b/a;
re pow(b,a);
}
int main()
{
string s;cin>>s;
s="("+s+")";//可以用来防止栈为空找top时的溢出情况
int n=(int)s.size();
rep(i,0,n-1)
{
if(s[i]=='(')//左括号直接压入
{
sc.push(s[i]);
continue;
}
if(s[i]==')')//找到右括号将栈清空到左括号为止
{
char c=sc.top();sc.pop();
while(c!='(')
{
vc.pb(c);
c=sc.top();sc.pop();
}
continue;
}
if(s[i]>='0' && s[i]<='9')//数字
{
BigInteger sum=0;
while(s[i]>='0' && s[i]<='9')
sum=sum*10+s[i++]-'0';
if(flag)
sum=BigInteger(-1)*sum,flag=0;
vc.pb(sum);
isnum[vc.size()-1]=1;
i--;continue;
}
if(s[i]=='-' && (clas(s[i-1])!=4 && s[i-1]!=')'))
{
flag=1;
continue;
}
while(!sc.empty() && clas(s[i])<=clas(sc.top()))//如果是符号就找到第一个优先级比他小的
vc.pb(sc.top()),sc.pop();
sc.push(s[i]);
}
while(!sc.empty())//清空栈
vc.pb(sc.top()),sc.pop();
rep(i,0,vc.size()-1)//根据后缀表达式求值
{
if(!isnum[i])
{
BigInteger a=si.top();si.pop();
BigInteger b=si.top();si.pop();
BigInteger ans=cal(a,b,vc[i]);
si.push(ans);
}
else
si.push(vc[i]);
}
cout<<si.top();
return 0;
}
//----------------------------------------------------------------end----------------------------------------------------------------------
以上是关于全网最高端?中缀表达式转为后缀表达式以及求值(可用于负数,阶乘)的主要内容,如果未能解决你的问题,请参考以下文章