A1001 A+B Format (20)

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本篇博客介绍了一个简单的程序设计问题：计算两个整数的和，并将结果以标准格式输出，即每三位数字用逗号分隔。程序通过读取两个整数，计算它们的和，并使用逆向输出的方法确保了输出格式正确。

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A+B Format (20)
时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
题目描述
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).

输入描述:
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.

输出描述:
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.

输入例子:
-1000000 9

输出例子:
-999,991

#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
int main(){

   int a,b,ans[10];
   scanf("%d %d",&a,&b);
   int sum=0,num=0;
   sum=a+b;
   if(sum<0)
   {
       sum=-sum;
       cout<<'-';
   }
   while(sum!=0)
   {
       ans[num++]=sum%10;
       sum=sum/10;
   }
   per(i,0,num)
   {
       cout<<ans[i];
       if(i%3==0&&i>0)cout<<",";
   }

}