HDU1720（输入输出涨姿势）

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A+B Coming

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5953 Accepted Submission(s): 3901

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1720

Problem Description

Many classmates said to me that A+B is must needs.
If you can’t AC this problem, you would invite me for night meal. ^_^

Input

Input may contain multiple test cases. Each case contains A and B in one line.
A, B are hexadecimal number.
Input terminates by EOF.

Output

Output A+B in decimal number in one line.

Sample Input

  1 9
A B
a b

Sample Output

  10
21
21

解题思路：

输入输出的姿势，头一次看到cin、cout可以这样用，新技能get。
11在十进制中等于10 + 1 ，1F在十六进制中等于16 + F = 31。

完整代码：

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;
typedef double DB;
typedef unsigned uint;
typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const DB EPS = 1e-9;
const DB OO = 1e20;
const DB PI = acos(-1.0); //M_PI;int main()
{#ifdef DoubleQfreopen("in.txt","r",stdin);#endif__int64 a , b;while(cin >> hex >> a >> b){cout << dec << a + b << endl;}
}

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