【題目】

1001 A+B Format （20 point(s)）

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 ?10?6??≤a,b≤10?6??. 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

【題意】

輸入兩個整數a，b然後讓其相加並將其的和每3個用“,”隔開

【註意】

1.這個逗號隔開的形式是比如 : 1111+1111 那麽應該輸出的結果是2,222而不是222,2（因為我起初這麽想然後錯了........）

2.還要註意形如 :

• 111,
• ,111
• 111,

 1 #include<iostream>
 2 #include<cmath>
 3 #include<stack>
 4 using namespace std;
 5 int main() {
 6     int a,b;
 7     while (cin >> a >> b) {
 8         int ans = a + b;
 9         
10         //處理負號 
11         if (ans < 0
 
) cout << "-";              
12         ans = abs(ans);
13         
14         //用棧來存儲 
15         stack<int> s;
16         int cnt=0;
17         
18         //如果ans是0就直接輸出了吧 
19         if (ans == 0) {
20             cout << 0 << endl;
21             continue;
22         }
23         
24         while (ans) {
25             s.push(ans % 10);
26             ans /= 10;
27         }
28         int mk;   //標記前面第幾個digit會有逗號用的,因為這邊不一定是第三個 
29         int array[100000];
30         while (!s.empty()) {
31             array[++cnt] = s.top();
32             s.pop();
33         }
34 
35         mk = cnt % 3;
36         for (int i = 1; i<=cnt; ++i) {
37             //為了防止最後會多出個逗號就這麽來了,本來還想再用個cnt2來判斷 
38             if ((i-mk-1>0)&&(!((i-mk-1)%3))) cout<<",";    
39             else if (i-1==mk&&i-1>0)cout<<",";
40             cout << array[i];
41         }
42         cout << endl;
43     }
44 }

PAT甲級1001