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2008年12月30日



不管巴以谁对谁错,我们都不得不向这些手无寸铁的青年致敬。他们不曾拥有梦幻式的童年和青少年,迎接他们的是残酷的现实:每天生与死的考验。
但愿身处异乡的中华子民,居安思危,不论过去还是将来,面对敌人,我们都要拥有勇气,以各种方式向敌人投掷“石块”

posted @ 2008-12-30 23:08 赵均泽 阅读(106) | 评论 (0)编辑 收藏

2008年11月30日

Problem Statement

***Note:  Please keep programs under 7000 characters in length.  Thank you


Class Name: SquareDigits
Method Name: smallestResult
Parameters: int
Returns: int

Define the function S(x) as the sum of the squares of the digits of x.  
For example: S(3)=3*3=9 and S(230)=2*2+3*3+0*0=13.

Define the set T(x) to be the set of unique numbers that are produced by
repeatedly applying S to x.  That is: S(x), S(S(x)), S(S(S(x))), etc...
For example, repeatedly applying S to 37:
S(37)=3*3+7*7=58. 
S(58)=5*5+8*8=89.
S(89)=145.
S(145)=42.
S(42)=20.
S(20)=4.
S(4)=16.
S(16)=37.
Note this sequence will repeat so we can stop calculating now and:
T(37)={58,89,145,42,20,4,16,37}.
However, note T(x) may not necessarily contain x. 

Implement a class SquareDigits, which contains a method smallestResult.  The
method takes an int, n, as a parameter and returns the smallest int, x, such
that T(x) contains n.

The method signature is (be sure your method is public):
int smallestResult(int n);

TopCoder will ensure n is non-negative and is between 0 and 199 inclusive.

Examples:
If n=0: S(0) = 0, so T(0)={0}, so the method should return 0.

If n=2: T(0) through T(10) do not contain the value 2.  If x=11, however:
S(11)=1*1+1*1=2, so T(11) contains 2, and the method should return 11.

If n=10: T(0) through T(6) do not contain 10.  If x=7:
S(7)=49.
S(49)=97.
S(97)=130.
S(130)=10.
S(10)=1.
and it starts to repeat...
so T(7) is {49,97,130,10,1}, which contains 10, and the method should return 7.

n=1 -> x=1
n=19 -> x=133
n=85 -> x=5
n=112 -> x=2666
Definition
????
Class:
SquareDigits
Method:
smallestResult
Parameters:
int
Returns:
int
Method signature:
int smallestResult(int param0)
(be sure your method is public)

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.


posted @ 2008-11-30 14:26 赵均泽 阅读(290) | 评论 (1)编辑 收藏

TopCoder has decided to automate the process of assigning problem difficulty
levels to problems.  TopCoder developers have concluded that problem difficulty
is related only to the Average Word Length of Words in the problem statement:

If the Average Word Length is less than or equal to 3,  the problem is a 250
point problem.
If the Average Word Length is equal to 4 or 5, the problem is a 500 point
problem.
If the Average Word Length is greater than or equal to 6, the problem is a 1000
point problem.
 
Definitions:
Token - a set of characters bound on either side by spaces, the beginning of
the input String parameter or the end of the input String parameter.
Word - a Token that contains only letters (a-z or A-Z) and may end with a
single period. A Word must have at least one letter.
Word Length - the number of letters in a Word. (NOTE: a period is NOT a letter)

The following are Words :
"ab",  "ab."

The following are not Words :
"ab..", "a.b", ".ab", "a.b.", "a2b.", "."

Average Word Length - the sum of the Word Lengths of every Word in the problem
statement divided by the number of Words in the problem statement.  The
division is integer division. If the number of Words is 0, the Average Word
Length is 0.
 
Implement a class HowEasy, which contains a method pointVal.  The method takes
a String as a parameter that is the problem statement and returns an int that
is the point value of the problem (250, 500, or 1000). The problem statement
should be processed from left to right.
 
Here is the method signature (be sure your method is public):
int pointVal(String problemStatement);
 
problemStatement is a String containing between 1 and 50 letters, numbers,
spaces, or periods.  TopCoder will ensure the input is valid.
 
Examples:
 
If problemStatement="This is a problem statement", the Average Word Length is
23/5=4, so the method should return 500.
If problemStatement="523hi.", there are no Words, so the Average Word Length is
0, and the method should return 250.
If problemStatement="Implement a class H5 which contains some method." the
Average Word Length is 38/7=5 and the method should return 500.
If problemStatement=" no9 . wor7ds he8re. hj.." the Average Word Length is 0,
and the method should return 250.
Definition

Class:
HowEasy
Method:
pointVal
Parameters:
string
Returns:
int
Method signature:
int pointVal(string param0)
(be sure your method is public)

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

class HowEasy
{
    
private:
    
int HowEasy::nextToken(string pChar,bool &bInvalidToken,int &idx);
public:    
    HowEasy();
    
int pointVal(string param0);

}
;
HowEasy::HowEasy()
{
}
;

int HowEasy::nextToken(string pChar,bool &bInvalidToken,int &idx)
{
    
int result = 0;
    bInvalidToken 
= false;

    
while( pChar[idx] == ' ' )
        idx 
= idx+1;
        
    
while(pChar[idx] !=' ' && pChar[idx] !='\0')
    
{
        
if( ( 'a' <=pChar[idx] && pChar[idx] <='z' )
            
|| ('A' <= pChar[idx] && pChar[idx] <='Z'
            
|| ('.' == pChar[idx] && pChar[idx+1== '\0') )
        
{
        }

        
else
        
{
            result 
= 0;
            bInvalidToken 
= true;            
        }

        
if!bInvalidToken )
        
{
            result
++;
        }

        
if( pChar[idx] !='\0')
            idx
++;
    }

    
if( pChar[idx] ==' ' )
        idx
++;
    
return result;
}

int HowEasy::pointVal(string param0)
{
    
int nTotalChars = 0;
    
int nTotoalWords = 0;
    
int idx = 0;
    
bool bInvalidToken = false;
    
    
    
while( param0[idx] != '\0' )
    
{
        
        nTotalChars 
+= this->nextToken(param0,bInvalidToken,idx);
        
if!bInvalidToken )
            nTotoalWords
++;
        /*
           else
        
{
            nTotoalWords 
= 0;
            
break;
        }
  */
    }

    
    
if( nTotoalWords == 0 )
    
{
        
return 250;
    }

    
else if( nTotalChars/nTotoalWords <=3 )
    
{
        
return 250;
    }

    
else if( nTotalChars/nTotoalWords >=4 && nTotalChars/nTotoalWords <=5 )
    
{
        
return 500;
    }

    
else if( nTotalChars/nTotoalWords >= 6 )
    
{
        
return 1000;
    }

    
return 250;
}
;

posted @ 2008-11-30 12:42 赵均泽 阅读(254) | 评论 (4)编辑 收藏