PAT 甲级 1001 A+B Format（20）

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −106 ≤ a，b ≤ 106. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

Experiential Summing-up 

把 a+b 转化为字符串，除第一位为负号时（特判），当 (数组下标+1)%3 == 总和位数%3 ，且遍历的此数不是最后一位时，输出“,”。

Accepted Code

 1 /*单纯的笨办法*/
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int a, b, sum;
 8     cin >> a >> b;
 9     sum = a + b;
10     if(sum < 0)
11     {
12         sum = -sum;
13         cout << "-" ;
14     }
15     if(sum == 0)
16     {
17         cout << 0;
18         return 0;
19     }
20 
21     int count = 0, temp = sum, temp_t = sum;//count为sum的位数
22     while(temp_t != 0)
23     {
24         temp_t /= 10;
25         count ++;
26     }
27 
28     int q[10], ct = count;
29     while(temp != 0)
30     {
31         int g = temp % 10;
32         q[count] = g;
33         temp /= 10;
34         count --;
35     }
36 
37     if(sum < 100 && sum > -100)
38     {
39         for(int i = 1; i <= ct; i ++)
40             cout << q[i];
41         return 0;
42     }
43     if(ct % 3 == 0)
44     {
45         int k = 0;
46         for(int i = 1; i <= ct; i ++)
47         {
48             if(k == 3)
49             {
50                 cout << ",";
51                 k = 0;
52             }
53             cout << q[i];
54             k ++;
55         }
56     }
57     if(ct % 3 == 1)
58     {
59         cout << q[1] << ",";
60         int k = 0;
61         for(int i = 2; i <= ct; i ++)
62         {
63             if(k == 3)
64             {
65                 cout << ",";
66                 k = 0;
67             }
68             cout << q[i];
69             k ++;
70         }
71     }
72     if(ct % 3 == 2)
73     {
74         cout << q[1] << q[2] << ",";
75         int k = 0;
76         for(int i = 3; i <= ct; i ++)
77         {
78             if(k == 3)
79             {
80                 cout << ",";
81                 k = 0;
82             }
83             cout << q[i];
84             k ++;
85         }
86     }
87 
88     return 0;
89 }
 1 /*简单聪明的办法*/
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 
 5 int main()
 6 {
 7     int a, b;
 8     cin >> a >> b;
 9     string s = to_string(a + b);
10     int len = s.length();
11     for(int i = 0; i < len; i ++)
12     {
13         cout << s[i];
14         if(s[i] == '-') continue;
15         if((i + 1) % 3 == len % 3 && i != len - 1)
16             cout << ",";
17     }
18     return 0;
19 }

 

posted @ 2023-02-18 18:49  爱吃虾滑  阅读(12)  评论(0)  编辑  收藏  举报