中后缀表达式的相互转换和四则运算表达式求值

这篇博客讲的超级好
//所以原理和定义就不多bb了. 不懂的自己看博客吧.
下面附上代码实现:

1：中缀表达式转后缀表达式Code

/** @Cain*/const int maxn = 1e3+5;char zhong[maxn];char hou[maxn];stack<char>S;void solve(){//这个是只有单个整数和运算符,只是直观给出后缀表达式,计算的当然不一样.    scanf("%s",zhong);    int len = strlen(zhong);    int cnt = 0;    for(int i=0;i<len;i++){   //注意一个优先级问题!!!        if(zhong[i]>='0' && zhong[i]<='9') hou[cnt++] = zhong[i];        else{   //也要特别注意判断运算符优先级的问题.            if(zhong[i] == '(' ) S.push(zhong[i]);            else if(zhong[i] == ')' ){                while(!S.empty()){                    char tmp = S.top();                    S.pop();                    if(tmp == '(' ) break;                    hou[cnt++] = tmp;                }            }            else if(zhong[i] == '*' || zhong[i] == '/'){                while(!S.empty()){                    char tmp = S.top();                    if(tmp == '+' || tmp == '-' || tmp == '(') break;                    S.pop();                    hou[cnt++] = tmp;                }                S.push(zhong[i]);            }            else if(zhong[i] == '+' || zhong[i] == '-'){                while(!S.empty()){                    char tmp = S.top();                    if(tmp == '(') break;                    S.pop();                    hou[cnt++] = tmp;                }                S.push(zhong[i]);            }        }    }    while(!S.empty()){  //有可能运算符还有.        char tmp = S.top();        S.pop();        hou[cnt++] = tmp;    }    hou[cnt] = 0;    printf("%s\n",hou);}

模板题

2 : 四则运算表达式求值(有了以上的基础,这个就比较简单)

Code

/** @Cain*/const int maxn = 1e3+5;char zhong[maxn];char hou[maxn];char num[maxn];db res[maxn];stack<char>S;stack<db>cal;void solve(){    while(~scanf("%s",zhong)){        Fill(hou,0); Fill(num,0); Fill(res,0);        int len = strlen(zhong);        int cnt = 0;        for(int i=0;i<len;i++){            if(zhong[i]>='0' && zhong[i]<='9'){                int ans = 0;                Fill(num,0);                while((zhong[i]>='0' && zhong[i]<='9') || zhong[i] == '.'){                    num[ans++] = zhong[i];                    i++;                }                i--;                res[cnt++] = atof(num);            }            else{                if(zhong[i] == '(' ) S.push(zhong[i]);                else if(zhong[i] == ')' ){                    while(!S.empty()){                        char tmp = S.top();                        S.pop();                        if(tmp == '(' ) break;                        hou[cnt++] = tmp;                    }                }                else if(zhong[i] == '*' || zhong[i] == '/'){                    while(!S.empty()){                        char tmp = S.top();                        if(tmp == '+' || tmp == '-' || tmp == '(') break;                        S.pop();                        hou[cnt++] = tmp;                    }                    S.push(zhong[i]);                }                else if(zhong[i] == '+' || zhong[i] == '-'){                    while(!S.empty()){                        char tmp = S.top();                        if(tmp == '(') break;                        S.pop();                        hou[cnt++] = tmp;                    }                    S.push(zhong[i]);                }            }        }        while(!S.empty()){  //有可能运算符还有.            char tmp = S.top();            S.pop();            hou[cnt++] = tmp;        }        for(int i=0;i<cnt;i++){            if(hou[i] == 0) cal.push(res[i]);            else{   //注意这些的写法与之前的处理的关系.                db t1 = cal.top();                cal.pop();                db t2 = cal.top();                cal.pop();                if(hou[i] == '+' ) cal.push(t1+t2);                if(hou[i] == '-' ) cal.push(t2-t1);                  //注意数被取出来的先后顺序(-和/运算有影响).                if(hou[i] == '*' ) cal.push(t1*t2);                if(hou[i] == '/' ) cal.push(t2/t1);            }        }        printf("%.2f\n",cal.top());        cal.pop();    }}