P1957 口算练习题

题目链接：https://www.luogu.com.cn/problem/P1957
　　眼高手低系列
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int len(int number) { //返回number这个数长度
 4     int x = 0; //x是number这个数的长度
 5     if (number == 0) {
 6         return 1;
 7     }
 8     if (number > 0)
 9         while (number) {
10             number /= 10;
11             x++;
12         }
13     if (number < 0) {
14         x = 1; //负号的长度也要算上
15         number = -number; //转为正数
16         while (number) {
17             number /= 10;
18             x++;
19         }
20     }
21     return x;
22 }
23 int main() {
24     int n;
25     cin >> n;
26     int a, b; //操作数
27     string c; //操作符号
28     char l_c; //当本次没输入操作符号时，用上一次输入的操作符号
29     while (n--) {
30         cin >> c;
31         int x; //x是长度
32         switch (c[0]) {
33             case 'a': //如果是+的话 
34                 cin >> a >> b; //读入操作数
35                 cout << a << "+" << b << "=" << a + b << endl;
36                 x = len(a + b) + 2;
37                 l_c = c[0];
38                 break;
39             case 'b': 
40                 cin >> a >> b;
41                 cout << a << "-" << b << "=" << a - b << endl;
42                 x = len(a - b) + 2;
43                 l_c = c[0];
44                 break;
45             case 'c': 
46                 cin >> a >> b;
47                 cout << a << "*" << b << "=" << a * b << endl;
48                 x = len(a * b) + 2;
49                 l_c = c[0];
50                 break;
51             default: //此时字符串c存储的就是第一个操作数
52                 cin >> b; //输入第二个操作数
53                 a = 0; //然后把字符串c的值，转为int，存入a
54                 int l = c.size();
55                 for (int i = 0; i < l; i++) {
56                     a = a * 10 + c[i] - '0';
57                 }
58                 switch (l_c) {
59                     case 'a':
60                         cout << a << "+" << b << "=" << a + b << endl;
61                         x = len(a + b) + 2;
62                         break;
63                     case 'b':
64                         cout << a << "-" << b << "=" << a - b << endl;
65                         x = len(a - b) + 2;
66                         break;
67                     case 'c':
68                         cout << a << "*" << b << "=" << a * b << endl;
69                         x = len(a * b) + 2;
70                         break;
71                     default:
72                         break;
73                 }    
74                 break;  
75         }
76         x = len(a) + len(b) + x;
77         cout << x << endl;
78     }
79     return 0;
80 }