PAT甲级1001. A+B Format

最新推荐文章于 2022-03-31 10:41:10 发布

原创最新推荐文章于 2022-03-31 10:41:10 发布 · 235 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT

PAT 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一个简单的C++程序，该程序接收两个整数作为输入，并以标准格式输出它们的和，即当和大于999时，每三位数字间插入逗号以方便阅读。

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include <queue>
#include <stack>
using namespace std;
//ios::sync_with_stdio(false);
//typedef long long ll;
const int MAXN = 100005;
//const int inf = 0x3f3f3f3f;


int main()
{
    int a,b;
    cin>>a>>b;
    int sum=a+b;
    int s;
    if(sum<0)
        s=-sum;
    else
        s=sum;
    stack<int> x;
    int t=0,h=0;
    if(sum<=999&&sum>=-999)
    {
        cout<<sum<<endl;
        return 0;
    }
    while(s!=0)
    {
        x.push(s%10);
        s/=10;
        h++;//记录数字个数
    }
    if(sum<0)
        cout<<"-";
    int flag=0;//判断剩余数字个数是否为3的倍数
    while(!x.empty())
    {
        if(h%3==0)
            flag=1;
        if(h%3!=0&&flag==0)
        {
            h--;
        }
        if(t==3&&flag==1)
        {
            cout<<",";
            t=0;
        }
        cout<<x.top();
        x.pop();
        if(flag==0&&h%3==0)
                cout<<",";
        if(flag==1)
        t++;
    }
    cout<<endl;

    return 0;
}