zoukankan      html  css  js  c++  java
  • 1001. A+B Format (20)

     

    时间限制
    400 ms
    内存限制
    32000 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CHEN, Yue

    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.




    C++:

     1 #include <iostream>
     2 #include <string>
     3 #include <sstream>
     4 using namespace std;
     5 int main()
     6 {
     7 int n1,n2,sum;
     8  
     9 while(cin>>n1)
    10 {
    11   cin>>n2;
    12        sum=n1+n2;
    13   stringstream ss; string s;
    14   ss<<sum;
    15   ss>>s;
    16   int i;
    17   char ch[20];
    18   int len=0;int num=0;
    19  
    20        for(i=s.length()-1;i>=0;i--)
    21  {
    22         ch[len++]=s[i];
    23              num++;
    24 if(sum>=0&&num==3&&i!=0)   
    25 {
    26   ch[len++]=',';
    27   num=0;
    28 }
    29  if(sum<0&&num==3&&i!=0&&i!=1)   
    30 {
    31   ch[len++]=',';
    32   num=0;
    33 }
    34  }
    35  
    36  
    37         for(i=len-1;i>=0;i--)
    38 cout<<ch[i];
    39  
    40 cout<<endl;
    41  
    42 }
    43    return 0;
    44 }

     Python:

     1 a,b = raw_input().split( )
     2 a = int(a)
     3 b = int(b)
     4 c = a+b
     5 c = str(c)
     6 n = len(c) - 1 
     7 a = ""
     8 while n >= 0:
     9     for x in range(0,3):
    10         a = c[n] + a
    11         n-=1
    12         if n < 0:
    13             break
    14     if n > 0:
    15         a = ',' + a
    16     if n == 0 and c[0] != '-':
    17         a = ',' + a
    18 print a

    Java

     1 import java.util.*;
     2 
     3 public class Main {
     4     public static void main(String args[])
     5     {
     6         Scanner in = new Scanner(System.in);
     7         int a = in.nextInt();
     8         int b = in.nextInt();
     9         in.close();
    10         int c = a + b;
    11         String str = Integer.toString(c);
    12         for(int i = str.length() -3 ; i > 0 ; i = i -3)
    13         {
    14             if(!(i==1 && c < 0))
    15                 str = str.substring(0,i) + ',' + str.substring(i,str.length());
    16         }
    17         System.out.println(str);
    18     }
    19 }
  • 相关阅读:
    2 安装驱动出现异常
    1 Yoga3 系统装机总结.
    6 关于 Oracle NULL栏位和PL./SQL执行实验
    4 C#和Java 的比较
    3 委托、匿名函数、lambda表达式
    2 跨线程访问控件InvokeHelper类
    2 跨线程访问控件InvokeHelper类
    1 Winform 异步更新控件
    1 Winform 异步更新控件
    C# DataTable的詳細用法
  • 原文地址:https://www.cnblogs.com/xiaoyesoso/p/4263589.html
Copyright © 2011-2022 走看看