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1002 -- 487-3279

Richeir — Wed, 01 Jul 2009 05:16:00 GMT

487-3279

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 104347		Accepted: 17506

Description

企业喜欢用容易被��C��的电话号码。让电话��L��Ҏ��被记住的一个办法是��它写成一个容易记住的单词或者短语。例如，你需要给滑铁卢大学打电话�Ӟ��可以拨打TUT-GLOP。有�Ӟ��只将电话��L��中部分数字拼写成单词。当你晚上回到酒店，可以通过拨打310-GINO来向Gino's订一�?/span>pizza。让电话��L��Ҏ��被记住的另一个办法是以一�U�好记的方式对号码的数字�q�行分组。通过拨打必胜客的“三个�?/span>”��L��3-10-10-10�Q�你可以从他们那里订pizza�?/span>

电话��L��的标准格式是七位十进制数�Q��ƈ在第三、第四位数字之间有一个连接符。电话拨��L��提供了从字母到数字的映射�Q�映��关�p�d��下：
A, B, �?/span>C 映射�?/span> 2
D, E, �?/span>F 映射�?/span> 3
G, H, �?/span>I 映射�?/span> 4
J, K, �?/span>L 映射�?/span> 5
M, N, �?/span>O 映射�?/span> 6
P, R, �?/span>S 映射�?/span> 7
T, U, �?/span>V 映射�?/span> 8
W, X, �?/span>Y 映射�?/span> 9

Q�?/span>Z没有映射��C�Q何数字，�q�字�W�不需要拨��P��可以��L��d��和删除�?/span> TUT-GLOP的标准格式是888-4567�Q?/span>310-GINO的标准格式是310-4466�Q?/span>3-10-10-10的标准格式是310-1010�?/span>

如果两个��L��有相同的标准格式�Q�那么他们就是等同的�Q�相同的拨号�Q?/span>

你的公司正在为本地的公司�~�写一个电话号码薄。作��量控制的一部分�Q�你惌��查是否有两个和多个公司拥有相同的电话��L��?/span>

Input

输入的格式是�Q�第一行是一个正整数�Q�指定电话号码薄中号码的数量�Q�最�?/span>100000�Q�。余下的每行是一个电话号码。每个电话号码由数字�Q�大写字母（除了Q�?/span>Z�Q�以及连接符�l�成。每个电话号码中只会刚好�?/span>7个数字或者字母�?/span>

Output

对于每个出现重复的号码��生一行输出，输出是号码的标准格式紧跟一个空格然后是它的重复�ơ数。如果存在多个重复的��L��Q�则按照��L��的字典升序输出。如果输入数据中没有重复的号码，输出一行：
No duplicates.

Sample Input

4873279

ITS-EASY

888-4567

3-10-10-10

888-GLOP

TUT-GLOP

967-11-11

310-GINO

F101010

888-1200

-4-8-7-3-2-7-9-

487-3279

Sample Output

310-1010 2

487-3279 4

888-4567 3

Source

East Central North America 1999

Translator

北京大学�E�序设计实习2007

Richeir 2009-07-01 13:16 发表评论

1001 -- Exponentiation

Richeir — Wed, 01 Jul 2009 05:15:00 GMT

求高�_�ֺ��q?/span>

Time Limit: 500MS		Memory Limit: 10000K
Total Submissions: 52024		Accepted: 12083

Description

�Ҏ��值很大、精度很高的数进行高�_�ֺ�计算是一�c�d��分常见的问题。比如，对国��行计��就是属于这�c�问题�?/span>

现在要你解决的问题是�Q�对一个实�?/span>R( 0.0 < R < 99.999 )�Q�要求写�E�序�_��计算 R �?/span> n �ơ方(Rⁿ)�Q�其�?/span>n 是整数�ƈ�?/span> 0 < n <= 25�?/span>

Input

T输入包括多组 R �?/span> n�?/span> R 的值占�W?/span> 1 到第 6 列，n 的值占�W?/span> 8 和第 9 列�?/span>

Output

对于每组输入�Q�要求输��Z��行，该行包含�_��?/span> R �?/span> n �ơ方。输出需要去掉前导的 0 后不要的 0 。如果输出是整数�Q�不要输出小数点�?/span>

Sample Input

95.123 12

0.4321 20

5.1234 15

6.7592 9

98.999 10

1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721

.00000005148554641076956121994511276767154838481760200726351203835429763013462401

43992025569.928573701266488041146654993318703707511666295476720493953024

29448126.764121021618164430206909037173276672

90429072743629540498.107596019456651774561044010001

1.126825030131969720661201

Source

East Central North America 1988

Translator

北京大学�E�序设计实习�Q?/span>Xie Di

Richeir 2009-07-01 13:15 发表评论

1000 -- A+B Problem

Richeir — Wed, 01 Jul 2009 02:03:00 GMT

A+B Problem

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 129751		Accepted: 70436

Description

Calculate a+b

Input

Two integer a,b (0<=a,b<=10)

Output

Output a+b

Sample Input

1 2

Sample Output

Hint

Q: Where are the input and the output?

A: Your program shall always read input from stdin (Standard Input) and write output to stdout (Standard Output). For example, you can use 'scanf' in C or 'cin' in C++ to read from stdin, and use 'printf' in C or 'cout' in C++ to write to stdout.

You shall not output any extra data to standard output other than that required by the problem, otherwise you will get a "Wrong Answer".

User programs are not allowed to open and read from/write to files. You will get a "Runtime Error" or a "Wrong Answer"if you try to do so.

Here is a sample solution to problem 1000 using C++/G++:

#include 

using namespace std;

int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

It's important that the return type of main() must be int when you use G++/GCC,or you may get compile error.

Here is a sample solution to problem 1000 using C/GCC:

#include 

int main()
{
    int a,b;
    scanf("%d %d",&a, &b);
    printf("%d\n",a+b);
    return 0;
}

Here is a sample solution to problem 1000 using Pascal:

program p1000(Input,Output);

var
  a,b:Integer;
begin
   Readln(a,b);
   Writeln(a+b);
end.

Here is a sample solution to problem 1000 using Java:

Now java compiler is jdk 1.5, next is program for 1000

import java.io.*;
import java.util.*;
public class Main
{
            public static void main(String args[]) throws Exception
            {
                    Scanner cin=new Scanner(System.in);
                    int a=cin.nextInt(),b=cin.nextInt();
                    System.out.println(a+b);
            }
}

Old program for jdk 1.4

import java.io.*;
import java.util.*;

public class Main
{
    public static void main (String args[]) throws Exception
    {
        BufferedReader stdin = 
            new BufferedReader(
                new InputStreamReader(System.in));

        String line = stdin.readLine();
        StringTokenizer st = new StringTokenizer(line);
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        System.out.println(a+b);
    }
}

Here is a sample solution to problem 1000 using Fortran:

	PROGRAM P1000
		IMPLICIT NONE
		INTEGER :: A, B
		READ(*,*) A, B
		WRITE(*, "(I0)") A + B
	END PROGRAM P1000

Source

POJ

a加b

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 129751		Accepted: 70436

Description

计算a加b�?/div>

Input

两个整数a和b�Q�其�?≤a, b≤10�?/div>

Output

输出a与b的和�?/div>

Sample Input

1 2

Sample Output

Hint

问：怎样输入输出�Q?
�{�：你的�E�序应该从标准输�?stdin)中读取数据，而将�l�果写到标准输出(stdout)。比如，在C语言中你可以使用“scanf”�Q�在C++语言中则可以使用“cin”来输入；输出可以使用C语言�?#8220;printf”或者C++语言�?#8220;cout”。其他的语言可以参照下面的程序�?

注意�Q�不要向标准输出写入题目要求输出�l�果之外的其他数据，否则你会被判�?#8220;Wrong Answer”�Q�即错误的运行结果�?

你的�E�序也不能试图读或写��M��文�g�Q�否则你可能被判�?#8220;Runtime Error”(�q�行旉��?�?#8220;Wrong Answer”(错误�l�果)�?

下面是这道题的一个C++或者G++的程序�?

#include 
using namespace std;
int main()
{
    int a,b;
    cin >> a >> b;
    cout << a+b << endl;
    return 0;
}

注意�Q�对于GCC或者G++�Q�main()函数的返回值必��L��int型，否则可能��D��“Compile Error”即编译错误�?

下面是这道题的一个C或者GCC的程序�?

#include 

int main()
{
    int a,b;
    scanf("%d %d",&a, &b);
    printf("%d\n",a+b);
    return 0;
}

下面是这道题的一个Pascal�E�序�?

program p1000(Input,Output);
var
  a,b:Integer;
begin
   Readln(a,b);
   Writeln(a+b);
end.

下面是这道题的一个Java�E�序�Q�这里Java�~�译器采用的是JDK 1.5�?

import java.io.*;
import java.util.*;
public class Main
{
            public static void main(String args[]) throws Exception
            {
                    Scanner cin=new Scanner(System.in);
                    int a=cin.nextInt(),b=cin.nextInt();
                    System.out.println(a+b);
            }
}

而老版本的JDK 1.4�E�序如下�Q?

import java.io.*;
import java.util.*;

public class Main
{
    public static void main (String args[]) throws Exception
    {
        BufferedReader stdin = 
            new BufferedReader(
                new InputStreamReader(System.in));

        String line = stdin.readLine();
        StringTokenizer st = new StringTokenizer(line);
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        System.out.println(a+b);
    }
}

下面是这道题的一个Fortran�E�序�?

	PROGRAM P1000
		IMPLICIT NONE
		INTEGER :: A, B
		READ(*,*) A, B
		WRITE(*, "(I0)") A + B
	END PROGRAM P1000

Source

POJ

Translator

Xie Di, xiedi@POJ

1/*
2Site: POJ
3Program Number: 1000
4Program Name: A+B Program
5Language: C++
6*/
7#include <iostream>
8using namespace std;
9
10int main()
11{
12    int a,b;
13    cin>>a>>b;
14    cout<<a+b<<endl;
15
16    return 0;
17}
18//Pogram By Richeir 04-07-09 00:50
19/*
205357421 Richeir 1000 Accepted 292K 16MS C++ 249B 2009-07-04 00:50:58
215357419 Richeir 1000 Accepted 768K 0MS G++ 249B 2009-07-04 00:50:32
22*/
23

Richeir 2009-07-01 10:03 发表评论