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,   

　　　　　

 

 

#include<iostream>
#include<cmath>
#include<stack>
using namespace std;
int main() {
    int a,b;
    while (cin >> a >> b) {
        int ans = a + b;
        
        //处理负号 
        if (ans < 0) cout << "-";              
        ans = abs(ans);
        
        //用栈来存储 
        stack<int> s;
        int cnt=0;
        
        //如果ans是0就直接输出了吧 
        if (ans == 0) {
            cout << 0 << endl;
            continue;
        }
        
        while (ans) {
            s.push(ans % 10);
            ans /= 10;
        }
        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 
        int array[100000];
        while (!s.empty()) {
            array[++cnt] = s.top();
            s.pop();
        }

        mk = cnt % 3;
        for (int i = 1; i<=cnt; ++i) {
            //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 
            if ((i-mk-1>0)&&(!((i-mk-1)%3))) cout<<",";    
            else if (i-1==mk&&i-1>0)cout<<",";
            cout << array[i];
        }
        cout << endl;
    }
}