【模板】表达式运算

这些都是emmmm
http://www.tyvj.cn/p/1041
http://www.tyvj.cn/p/1042
http://codevs.cn/problem/2178/

描述

给出一个表达式,其中运算符仅包含+,-,*,/,^要求求出表达式的最终值数据可能会出现括号情况 还有可能出现多余括号情况数据保证不会出现>maxlongint的数据。数据可能回出现负数情况

样例输入 Sample Input

(2+2)^(1+1)

样例输出 Sample Output

16

分析

真的是超级麻烦emmmm，今天考试的时候看了看接着就弃了。。。说实话这个真不应该弃掉，但是真的不会了w
需要开long long，不开的话炸了一个点，真是麻烦

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <climits>
using namespace std;
typedef long long LL;
char ch[10005];
LL num[10005], top1;
LL opt[10005], top2;
void get_num()
{
    if (!top2)
        return;
    LL y = num[top1--], x = num[top1--];
    LL pos = opt[top2--];
    if (pos == 5) x = pow(x, y);
    if (pos == 4) x = x / y;
    if (pos == 3) x = x * y;
    if (pos == 2) x = x - y;
    if (pos == 1) x = x + y;
    num[++top1] = x;
}
int main()
{
    scanf("%s", ch);
    LL n = strlen(ch);
    for (int i = 0; i < n;) 
    {
        if (ch[i] >= '0' && ch[i] <= '9')
        {
            LL x = 0;
            while(ch[i] >= '0' && ch[i] <= '9')
            {
                x = x * 10 + ch[i] - '0'; 
                i++;
            }
            num[++top1] = x;
        }
        if (ch[i] == '(') 
        {
            opt[++top2] = 0;
                i++;
        }
        if (ch[i] == '+') 
        {
            while(top2 && opt[top2] != 0) get_num();
            opt[++top2] = 1; i++;
        }
        if (ch[i] == '-') 
        {
            if (i == 0 || ch[i - 1] == '(')
            {
                int x = 0; i++;
                while(ch[i] >= '0' && ch[i] <= '9')
                {
                    x = x * 10 - (ch[i] - '0'); 
                    i++;
                }
                num[++top1] = x;
            }
            else 
            {
                while(top2 && opt[top2] != 0)
                    get_num();
                opt[++top2] = 2;
                i++;
            }
        }
        if (ch[i] == '*')
         {
            while(top2 && opt[top2] > 2)
                get_num();
            opt[++top2] = 3;
            i++;
        }
        if (ch[i] == '/')
        {
            while(top2 && opt[top2] > 2)
                get_num();
            opt[++top2] = 4;
            i++;
        }
        if (ch[i] == '^')
        {
            while(top2 && opt[top2] > 4)
                get_num();
            opt[++top2] = 5;
            i++;
        }
        if (ch[i] == ')')
        {
            while(top2 && opt[top2] != 0)
                get_num();
            top2--;
            i++;
        }
    }
    while(top2)
    {
        if (opt[top2] == 0)
            top2--;
        else
            get_num();
    }
    printf("%lld\n", num[top1]);
    return 0;
}

然后是std：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int MAXN=100000+5;
typedef long long ll;
char opt[MAXN],ch[MAXN];
ll sta[MAXN];
int tpo,tp,len;

ll pow(ll x,ll a)
{
    if(a==0) return 1;
    if(a==1) return x;
    if(a<0) return 0;
    ll t=pow(x,a>>1);
    t*=t;
    if(a&1) t*=x;
    return t;
}

inline int prty(char c)//priority
{
    switch(c)
    {
        case '+':
        case '-':return 1;
        case '*':
        case '/':return 2;
        case '^':return 3;
        case '(':return 0;
        default :return -1;
    }
}

inline ll calc(ll b,char c,ll a)
{
    switch(c)
    {
        case '+':return a+b;
        case '-':return a-b;
        case '*':return a*b;
        case '/':return a/b;
        case '^':return pow(a,b);
    }
}

int main()
{
    freopen("formula.in","r",stdin);
    freopen("formula.out","w",stdout);
    scanf("%s",ch+1);
    len=strlen(ch+1);
    ch[0]='(',ch[++len]=')';
    for(int i=0;i<=len;++i)
    {
        if((ch[i]=='-'&&ch[i-1]=='(')||ch[i]>='0'&&ch[i]<='9')
        {
            bool f=0;
            if(ch[i]=='-')
                f=1,++i;
            ll t;
            for(t=0;ch[i]>='0'&&ch[i]<='9';++i)
                t=(t<<1)+(t<<3)+ch[i]-'0';
            --i;
            if(f) t*=-1;
            sta[++tp]=t;
        }
        else if(ch[i]==')')
        {
            while(opt[tpo]!='(')
            {
                ll t=calc(sta[tp],opt[tpo],sta[tp-1]);
                --tp,--tpo;
                sta[tp]=t;
            }
            --tpo;
        }
        else
        {
            while(ch[i]!='('&&prty(opt[tpo])>=prty(ch[i]))
            {
                ll t=calc(sta[tp],opt[tpo],sta[tp-1]);
                --tp,--tpo;
                sta[tp]=t;
            }
            opt[++tpo]=ch[i];
        }
    }
    printf("%lld",sta[1]);
    fclose(stdin);
    fclose(stdout);
    return 0;
}