Problem D.Exponential Form
Input: form.in
Output: Standard Output
Time limit: 5 second
Memory limit: 64 megabytes
Every positive number can be presented by the exponential form.For example,
137 = 2^7 + 2^3 + 2^0

Let's present a^b by the form a(b).Then 137 is presented by 2(7)+2(3)+2(0).
Since 7 = 2^2 + 2 + 2^0 and 3 = 2 + 2^0 , 137 is finally presented by 2(2(2)+2
+2(0))+2(2+2(0))+2(0).


Given a positive number n,your task is to present n with the exponential form
which only contains the digits 0 and 2.
Input
The input file contains a positive integer n (n<=20000).
Output
You should output the exponential form of n an a single line.Note that,there s
hould not be any additional white spaces in the line.
Example
form.in
137
Stardard Output
2(2(2)+2+2(0))+2(2+2(0))+2(0)
form.in
1315
Stardard Output
2(2(2+2(0))+2)+2(2(2+2(0)))+2(2(2)+2(0))+2+2(0)
2006 年上机试题参考答案:
//A.cpp
//18 lines
#include "iostream"
#include "fstream"
using namespace std;
int fib(int n)
{

6

if (n==0) return 0;
else if (n==1) return 1;
else return fib(n-1)+fib(n-2);


}

void main()

{
fstream f("fib.in");
int n;
f>>n;
cout<<fib(n)<<endl;

}


//B.cpp
//28 lines
#include "iostream"
#include "fstream"
#include "string"
#include "stdio.h"
using namespace std;


char table[100]="`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";


void main()


{
fstream f("wertyu.in");
char c[1000];
string s;
int tablelen = strlen(table);


while (!f.getline(c,1000).eof())

{
s = c;
int n = s.length();
for (int i=0;i<n;i++)
{


for (int j=0;j<tablelen;j++)
{
if (s[i]==table[j]) s[i]=table[j-1];

}
}
cout<<s<<endl;

}
}

//C.cpp
//24 lines
#include "iostream"
#include "fstream"
#include "string"
using namespace std;


int match(string s,string t)


{
int count=0;
string::size_type index = -1;
while ((index = s.find(t,index+1))!=string::npos) count++;
return count;

}


void main()

{
fstream f("matching.in");
string s,t;
while (!f.eof())
{

f>>s;
f>>t;
cout<<match(s,t)<<endl;


}
}

//D.cpp
//60 lines
#include "iostream"
#include "fstream"
#include "string"
using namespace std;


int getminex(int n)


{
int k=0;
int l=1;
while (1)
{

if (n<l) break;
else
{


l=l*2;
k++;

}
}
return k-1;

}

int getremain(int n)

{
int t=getminex(n);
int s=1;
for (int i=0;i<t;i++)
{

s=s*2;
}
return n-s;

}

string getform(int n)

{
if (n==0) return "";
else if (n==1) return "2(0)";
else if (n==2) return "2";
else if (n==4) return "2(2)";
else
{

string t,s;

int e = getminex(n);
if (e == 1) t= "";


else t = "("+getform(e)+")";
s = "2"+t;
int r = getremain(n);
if (r != 0) s = s+"+"+getform(r);
return s;
}
}
void main()
{
fstream f("form.in");
int n;
f>>n;
cout<<getform(n)<<endl;
}
参考试题1:
Fire Net
Suppose that we have a square city with straight streets. A map of a city is
a square board with n rows and n columns, each representing a street or a
piece of wall.
A blockhouse is a small castle that has four openings through which to
shoot. The four openings are facing North, East, South, and West,