【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;}