最近做的几条题(2)Fraction

 Description:

Consider the following set:

  S = {  p/q  |   w <= p <= x,  y <= q <= z  }

That means S is the set of all rational numbers whose numerators are between w and x inclusive, and whose denominators are between y and z inclusive.

 

Given w, x, y, and z return the number of distinct elements in S.

 

Notes:

1. p, q, w, x, y and z are all integers.

2. x and z will be between 1 and 100, inclusive.

3. w will be between 1 and x, inclusive.

4. y will be between 1 and z, inclusive.

Input

The input consists of separate lines containing w,x,y,z described above

The last line of the input contains four 0s.(0 0 0 0).

Output

For each test case, output the number of distinct elements in S in separate lines.

Sample Input

1 1 1 1

1 10 1 1

1 2 1 2

2 4 2 4

1 100 1 100

0 0 0 0

Sample Output

1

10

3

7

6087

Some Tips for the Sample Output:

1) Returns:1

Only the value 1/1 is being considered.

2) Returns: 10

Here S contains the values 1,...,10.

3) Returns: 3

Here the values are 1/1, 1/2, 2/1, and 2/2. Since 2/2 = 1/1 the answer is 3.

 

 

程序：

#include<iostream>
#include<vector>
using namespace std;

bool Find(double x,vector<double>check)
{
 for(int i = 0;i<check.size();i++)
 {
  if(x == check[i])
  {
   return true;
  }
 }
 return false;
}

int calu(int w,int x,int y,int z)
{
 int Num = 0;
 double ans = 0;
 vector<double>check;
 for(int i = y;i <= z;i++)
 {
  for(int j = w;j <= x;j++)
  {
   ans = double(j)/double(i);
   if(!Find(ans,check))
   {
    Num++;
    check.push_back(ans);
   }
  }
 }
 return Num;
}

 

void main()
{
 vector<int>Answer;
 int w(0),x(0),y(0),z(0);
 while(1)
 {
  cin>>w>>x>>y>>z;
  if(w<1||x<1||y<1||z<1||w>x||y>z)
  {
   break;
  }
  else
  {
   Answer.push_back(calu(w,x,y,z));
  }
 }
 for(int i = 0;i<Answer.size();i++)
 {
  cout<<Answer[i]<<endl;
 }
}