PAT甲级1001

【题目】

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

【题意】

输入两个整数a，b然后让其相加并将其的和每3个用“,”隔开

【注意】

1.这个逗号隔开的形式是比如 :   1111+1111 那么应该输出的结果是2,222而不是222,2（因为我起初这么想然后错了........）

2.还要注意形如 :

• 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         //处理负号
11         if (ans < 0) cout << "-";
12         ans = abs(ans);
13
14         //用栈来存储
15         stack<int> s;
16         int cnt=0;
17
18         //如果ans是0就直接输出了吧
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;   //标记前面第几个digit会有逗号用的,因为这边不一定是第三个
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             //为了防止最后会多出个逗号就这么来了,本来还想再用个cnt2来判断
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 }

原文地址：https://www.cnblogs.com/drake233/p/10191532.html