词法分析 —— 计算表达式值

2011-02-26 wcdj

 

题目：
在1 2 3 4 5 6 7 8 9九个数字中插入“+”或“-”符号使得结果为100，编程实现所有的组合。
注意：数字的顺序不能改变。
如：
123 - 45 - 67 + 89 = 100
12 - 3 -4 + 5 - 6 + 7 + 89 = 100
……

一种算法思路：
(1) 使用递归实现所有组合情况。
(2) 删除字符串中空格的方法。
(3) 对表达式进行词法分析，设计一个类来实现“+”和“-”两种算术运算。

 

#include <iostream>using namespace std;class CAL{enum{NUMBER, OPERATOR, END, UNKNOWN};class Token{public:int value;int type;};const char* buff;int curr;Token tk;public:int Evalue(const char* exp){int value = 0;int sign = 1;buff = exp;curr = 0;do{MoveToNextToken();switch (tk.type){case NUMBER: value += sign * tk.value; break;case OPERATOR: sign = tk.value == '+' ? 1 : -1;}}while (tk.type==NUMBER||tk.type==OPERATOR);return value;}void MoveToNextToken(){static char temp[256];tk.type = UNKNOWN, tk.value = 0;if (!buff[curr]) {tk.type = END;return;}if (buff[curr]>='0'&&buff[curr]<='9'){int value = 0;while (buff[curr]&&buff[curr]>='0'&&buff[curr]<='9')value = value * 10 + buff[curr++] - '0';// 若两位以上的数字要逐位相加tk.value = value;tk.type = NUMBER;return;}switch (buff[curr++]){case '+':tk.type = OPERATOR, tk.value = '+';return;case '-':tk.type = OPERATOR, tk.value = '-';return;}tk.type = UNKNOWN;return;}};void DelSpace(char* str){char* curr = str;while (*curr=*str++)if (*curr!=' ') curr++;}void GetExpression(int a[], int n){static char token[] = {'+','-',' '};static char buff[256];static CAL cal;sprintf(buff, "1%c2%c3%c4%c5%c6%c7%c8%c9",token[a[0]], token[a[1]], token[a[2]], token[a[3]],token[a[4]], token[a[5]],token[a[6]], token[a[7]]);DelSpace(buff);//cout << buff << endl;int result = cal.Evalue(buff);if (result == 100)cout << buff << '=' << result << endl;}void DoEnum(int n, int curr, int to[]){if (curr == n){GetExpression(to, n);}else{for (int i = 0; i < 3; ++i){to[curr] = i;DoEnum(n, curr+1, to);}}}int main(){int to[8];DoEnum(8, 0, to);return 0;}/*1+2+3-4+5+6+78+9=1001+2+34-5+67-8+9=1001+23-4+5+6+78-9=1001+23-4+56+7+8+9=10012+3+4+5-6-7+89=10012+3-4+5+67+8+9=10012-3-4+5-6+7+89=100123+4-5+67-89=100123+45-67+8-9=100123-4-5-6-7+8-9=100123-45-67+89=100Press any key to continue*/

 

问题来源：
http://topic.csdn.net/u/20080504/15/57909a01-1989-43d2-bff0-d104e34e5a97.html