2. A+B——the first step

1000 1001

A + B Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 704421    Accepted Submission(s): 217867

Problem Description

Calculate A + B.

 

Input

Each line will contain two integers A and B. Process to end of file.

 

Output

For each case, output A + B in one line.

 

Sample Input

  
  

   
   1 1
  
  

 

Sample Output

  
  

   
   2
  
  

 

Author

HDOJ

#include<iostream>   
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iomanip>
#include<cstdio>
using namespace std;
int main()
{
    /*ios::sync_with_stdio(false);
    cin.tie(0);
理论上，scanf 和 printf 要比cin 和 cout 快，但是加上这两行之后就会使
cin cout比 scanf printf 都要快。
但是用了“
    ios::sync_with_stdio(false);
    cin.tie(0);
    “
    之后，scanf 和 printf 都不能使用了，只能使用cin 和 cout。
    详见http://blog.youkuaiyun.com/jinzhilong580231/article/details/7697608
    http://www.cnblogs.com/bofengyu/p/6720041.html
    等其他相关操作*/
    int a,b,sum=0;
    while(cin>>a>>b)
    {
         sum=a+b;
        cout<<sum<<endl;
    }
 return 0;  
}

Sum Problem

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 502828    Accepted Submission(s): 127638

Problem Description

Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).

In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.

 

Input

The input will consist of a series of integers n, one integer per line.

 

Output

For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.

 

Sample Input

   
   

    
    1
100
   
   

 

Sample Output

   
   

    
    1

5050
   
   

 

Author

DOOM III

#include <iostream>
using namespace std;

int main()
{
    long long n;
    while( cin>>n )
    {
        cout<<( n*(n+1) ) / 2<<endl<<endl;
    }
    return 0;
}