A+B Format (20)_format >20-优快云博客

本文介绍了一种方法，用于将两个整数相加后的结果按照英文习惯进行格式化输出，即每三位数字用逗号分隔。提供了两种实现方案，一种使用字符串和链表操作，另一种采用算术运算直接格式化。

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

题意理解：

本题实质就是按英文习惯输出两个数的和

我是先把两个数的和转化为字符串，然后倒着把每个元素依次放到链表里，每读到三位加一个‘，’，最后输出链表各元素

#include<iostream> 
#include<string>
#include<list> 
using namespace std;
list<char>l;
int main()  
{  
  int sa,sb,k=1,sum;
  cin>>sa>>sb;
  sum=sa+sb;
  string s=to_string(sum);
    for(int i=s.size()-1;i>=0;i--)
    {
      l.push_front(s[i]);
      if((k++)%3==0&&i!=0&&s[i-1]!='-')
        l.push_front(',');
    }
    for(list<char>::iterator i=l.begin();i!=l.end();i++)  
      cout<<*i;  
    return 0;  
}

当然还有一种更简单的方法，按照不同数字大小，每次直接读三位

#include<stdio.h>  
int main()  
{  
  int a,b;  
  int sum;  
  while(scanf("%d%d\n",&a,&b) != EOF){  
        sum = a+b;  
    if(sum < 0){  
    printf("-");  
    sum = -sum;  
    }  //注意负转正
    if(sum>=1000000){  
        printf("%d,%03d,%03d\n",sum/1000000, (sum/1000)%1000, sum%1000);  
    }  
    else if(sum >= 1000){  
        printf("%d,%03d\n",sum/1000,sum%1000);  
    } else{  
        printf("%d\n", sum);  
    }  
  }  
  return 0;  
}