这题没啥好说的。就是烂模拟，我的是这样想：
每次读三个数和两个运算符，如果第一个运算符是“+”或者“-”，后面的是“*”或者“/”那么就先将后面处理，之后将处理之后的值和还没用的数字再放回去，第一个运算符也同样放回。
其他情况便都是先算前面之后把处理之后的值和还没用的数字放回，将第二个运算符同样放回。（可以想想为啥）
一开始想用栈的但是用着用着发现太麻烦就换成双端队列了hhh有兴趣的可以去学一下，挺简单的。
注意：1.因为是字符数字读入，要注意数字不止一位，这个样例跑一下就知道了
2.题目要求精度较高需要用double，float会爆
1 #include <bits/stdc++.h> 2 #define maxn 200 + 10 3 using namespace std; 4 int main() 5 { 6 char s[maxn]; 7 while (gets(s) != NULL) { 8 int len = strlen(s); 9 if (len == 1 && s[0] == '0') break; 10 deque<char> q1; deque<double> q2; 11 for (int i = 0; i < len; i++) 12 { 13 int temp = 0; 14 if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/') 15 q1.push_back(s[i]); 16 else if (s[i] >= '0' && s[i] <= '9') 17 { 18 while (s[i] != ' ' && i < len) 19 temp = temp * 10 + s[i++]-'0'; 20 //printf("%d\n", temp); 21 q2.push_back(temp); 22 } 23 } 24 double temp; 25 while (q2.size() > 2) 26 { 27 char oper1 = q1.front(); q1.pop_front(); 28 char oper2 = q1.front(); q1.pop_front(); 29 double num1 = q2.front(); q2.pop_front(); 30 double num2 = q2.front(); q2.pop_front(); 31 double num3 = q2.front(); q2.pop_front(); 32 if ((oper1 == '+' || oper1 == '-') && (oper2 == '*' || oper2 == '/')) 33 { 34 if (oper2 == '*') 35 { 36 temp = num2 * num3; 37 //printf("%.2lf %.2lf %.2lf\n", num2, num3, temp); 38 q2.push_front(temp); 39 q2.push_front(num1); 40 q1.push_front(oper1); 41 42 } 43 else if (oper2 == '/') 44 { 45 temp = num2 / num3; 46 //printf("%.2lf %.2lf %.2lf\n", num2, num3, temp); 47 q2.push_front(temp); 48 q2.push_front(num1); 49 q1.push_front(oper1); 50 51 } 52 } 53 else 54 { 55 if (oper1 == '+') temp = num1 + num2, q2.push_front(num3), q2.push_front(temp), q1.push_front(oper2); 56 else if (oper1 == '-') temp = num1 - num2,q2.push_front(num3), q2.push_front(temp), q1.push_front(oper2); 57 else if (oper1 == '*') temp = num1 * num2, q2.push_front(num3), q2.push_front(temp), q1.push_front(oper2); 58 else temp = num1 / num2, q2.push_front(num3), q2.push_front(temp), q1.push_front(oper2); 59 } 60 61 } 62 double ans; 63 char oper = q1.front(); q1.pop_front(); 64 double num1 = q2.front(); q2.pop_front(); 65 double num2 = q2.front(); q2.pop_front(); 66 //printf("%.2lf\n", num1);printf("%.2lf\n", num2); 67 if (oper == '+') ans = num1 + num2; 68 else if (oper == '-') ans = num1 - num2; 69 else if (oper == '*') ans = num1 * num2; 70 else ans = num1 / num2; 71 printf("%.2lf\n", ans); 72 73 } 74 return 0; 75 }