栈问题2（四则运算）

来源：互联网发布：嘉合信网络信息技术编辑：程序博客网时间：2024/06/05 16:11
利用栈实现形如 a+b*c+(d*e+f)*g 的运算
先从中缀式转为后缀式然后再计算可输入空格
#include<iostream>#include<stack>#include<cstdio>using namespace std;//a+b*c+(d*e+f)*g  //1+1*1+(2*1+1)*2 = 8//3 + 5 * 1 + ( 2 * 0 + 1 ) * 1 = 9//20+3*6+(1+1)*9 = 56int main(){    char Operator[10] = { '+','-','*','/','(',')' };    stack<char> st;    stack<int> sum;    char s[50],o[50]; //s存放输入 o存放操作符    int num[50];  //存储数字    int nc, oc;    //nc数字数量 oc操作符数量    while(gets_s(s))    {        while (!st.empty())            st.pop();        while (!sum.empty())            sum.pop();        nc = oc = 0;        //中缀转后缀        for (int i = 0; i < strlen(s); i++)        {            if (s[i] == ' ')                continue;            if ( strchr(Operator,s[i]) != NULL )            {                if (st.empty() || s[i] == '(' || st.top() == '(' )                {                    st.push(s[i]);                }                else if (s[i] == '*' || s[i] == '/')                {                    if (st.top() == '+' || st.top() == '-')                    {                        st.push(s[i]);                    }                    else                    {                        num[nc++] = -1;                        o[oc++] = st.top();                        st.pop();                        st.push(s[i]);                    }                }                else if (s[i] == '+' || s[i] == '-')                {                    while (!st.empty() && st.top() != '(' )                    {                        o[oc++] = st.top();                        st.pop();                        num[nc++] = -1;                    }                    st.push(s[i]);                }                else    //s[i] == ')'                {                    while (!st.empty() && st.top() != '(')                    {                        o[oc++] = st.top();                        st.pop();                        num[nc++] = -1;                    }                    st.pop();   // pop ')'                }            }            else            {                num[nc] = s[i] - '0';                while (i + 1 < strlen(s) && strchr(Operator, s[i+1]) == NULL && s[i+1] != ' ')                    num[nc] = num[nc] * 10 + (int)(s[++i] - '0');                nc++;            }        }        while (!st.empty())        {            {                o[oc++] = st.top();                st.pop();                num[nc++] = -1;            }        }        for (int i = 0, j = 0; i < nc; i++)        {            if (num[i] != -1)                cout << num[i];            else                cout << o[j++];        }        cout << endl;        //计算后缀式        bool flag = true;        for (int i = 0, j=0; i < nc; i++)        {            if (num[i] != -1)            {                sum.push(num[i]);            }            else            {                int s1 = sum.top();                sum.pop();                int s2 = sum.top();                sum.pop();                if (o[j] == '+')                    s1 += s2;                else if (o[j] == '-')                    s1 -= s2;                else if (o[j] == '*')                    s1 *= s2;                else                {                    if (s2 == 0)                        cout << "除数为0!\n";                    flag = false;                    break;                    s1 /= s2;                }                j++;                sum.push(s1);            }        }        if(flag)            cout << sum.top() << endl;    }}
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