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PAT 1001

时间:2015-12-06 15:49:45      阅读:133      评论:0      收藏:0      [点我收藏+]

1001. A+B Format (20)

时间限制
400 ms
内存限制
65536 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.

Sample Input
-1000000 9
Sample Output
-999,991

解析:略

code:
#include <iostream>
using namespace std;

int main(){
  int a,b;
  cin>>a>>b;
  
  int c = a + b;
  if(c < 0){
      cout<<"-";
      c = -c;
  }
  
  
  if(c >= 1000000){
      cout<<c/1000000<<","<<(c/1000)%1000<<","<<c%1000;
  }
  else if(c >= 1000){
      cout<<c/1000<<","<<c%1000;
  }
  else{
      cout<<c;
  }
  
  return 0;
}

原文:http://alvenxuan.iteye.com/blog/1450533

我自己的方法:

 

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main(){
    int a,b,c;
    vector<char> cache;
    cin>>a>>b;
    c = a + b;
    if(c < 0)
    {
        cout<<-;
        c = -c;
    }
    else if(c == 0)
    {
        cout<<c;
        return 0;
    }
    
    int count = 0;
    while(c!=0){
        count++;
        cache.push_back(c%10 + 0);
        
        c /= 10;
        if(c != 0 && count%3 == 0)
        cache.push_back(,);
    }
    
    for(int i=cache.size()-1; i>=0; i--){
        cout<<cache[i];
    } 
    
    
    return 0;
}

PAT 1001

原文:http://www.cnblogs.com/RookieCoder/p/5023622.html

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