joj1004

 

/*#include<stdio.h>
main()
{
    double n;
    double sum=0;

    while(scanf("%lf",&n)!=EOF)
    {
        sum+=n;
    }
    if(sum>(int)sum)
    printf("%.2lf\n",sum);
    else
    printf("%d\n",sum);
}*/
/*#include<iostream>
using namespace std;
int main()
{
 float a,b;
 float sum=0;
 while(cin>>a>>b)
 {
  sum+=a+=b;
 }
 cout.setf(ios::fixed);
 cout.precision(2);
 cout<<sum<<endl;
 return 0;
}*/
/*#include<iostream>
#include<string>
int main()
{
 char str[1000],q[1000];
 int len,k=1,j=0;
 
 while(gets(str))
 {
  double s=0,m;
  len=strlen(str);
  int t=8;
  for(int i=2;i<len;i++)
  {
   q[j++]=(str[i]-'0')*m;
   for(int j=1;j<k;j++)
   t*=8;
   m=1/t;
   k++;
   s+=q[j++];
  } 
 printf("%s",char(s));
 }
 return 0;
}*/
/*#include<stdio.h>
#include<string.h>
#define N 300
#define M N*3int main()
{
 char s[N]; int a[M];
 
 int b[10]={0,125,250,375,500,625,750,875,1000,1125};
 int i,j,len1,len2,k,x;
 while(scanf("%s",s)==1)
 {      
  for(i=0;i<M;i++)       
   a[i]=0;      
  len1=strlen(s);      
  len2=(len1-2)*3;      
  a[M-1]=b[s[len1-1]-'0'];
  x=M-4;      
  for(i=len1-2;i>1;i--)
  {       
   a[x]+=s[i]-'0';       
  x-=3;       
  for(j=M-1;j>M-1-len2;j--)       
  {
   a[j-1]+=a[j]/10;        
   a[j]%=10;      
  }       
  for(j=M-1;j>M-1-len2;j--)        
   a[j]=b[a[j]];      
  }      
  for(j=M-1;j>M-1-len2;j--)      
  {        a[j-1]+=a[j]/10;     
  a[j]%=10;      
  }      
  printf("%s [8] = 0.",s);   
  for(k=M-1;a[k]==0;k--);
  for(i=M-len2;i<=k;i++)      
   
   printf("%d",a[i]);     
  printf(" [10]\n");
}
 return 0;
}

*/

/*joj 1004: Octal Fractions
分类： joj 数学相关 2011-04-13 20:15 58人阅读 评论(0) 收藏 举报
Result TIME Limit MEMORY Limit Run Times AC Times JUDGE
 3s 8192K 3686 828 Standard
Fractions in octal (base 8) notation can be expressed exactly in decimal notation. For example, 0.75 in octal is 0.963125 (7/8 + 5/64) in decimal. All octal numbers of n digits to the right of the octal point can be expressed in no more than 3n decimal digits to the right of the decimal point. Write a program to convert octal numerals between 0 and 1, inclusive, into equivalent decimal numerals. The input to your program will consist of octal numbers, one per line, to be converted. Each input number has the form 0.d1d2d3 ... dk, where the di are octal digits (0..7). There is no limit on k. Your output will consist of a sequence of lines of the form

0.d1d2d3 ... dk [8] = 0.D1D2D3 ... Dm [10]

where the left side is the input (in octal), and the right hand side the decimal (base 10) equivalent. There must be no trailing zeros, i.e. Dm is not equal to 0.

SAMPLE INPUT
0.75 0.0001 0.01234567
SAMPLE OUTPUT
0.75 [8] = 0.953125 [10] 0.0001 [8] = 0.000244140625 [10] 0.01234567 [8] = 0.020408093929290771484375 [10]
/*

孔牛用的秦九韶算法，我这里贴的算法是利用千进制数作为中间媒介来达到八进制与十进制之间的转换，时间上一样，其实算法的本质也是一样（早期的代码，比较乱）；

*/

/*#include <iostream>
#include <stdio.h>
#include <memory>
#include <string.h>
using namespace std;
const int maxn=2000;
double oct[8]={0,125,250,375,500,625,750,875};
int main ()
{
 char str[maxn];
 int oc[maxn],dec[3*maxn];
 while (memset(str,'/0',sizeof(str)),cin>>str)
 {
  int i,len=strlen(str),j;
  for( i=0 ,j=len-1;str[j]!='.' ; i++,j--)
   oc[i]=str[j]-48;
   cout<<str<<" [8] = 0.";
  for( i=0 ;i<len-2 ; i++)
   for( j=0 ; j<len-2-i ; j++)
    {
     oc[j]=oc[j]*125;
     if( oc[j]>999)
     {
      oc[j+1]+=oc[j]/1000;
      oc[j]=oc[j]%1000;
     }
    }//千进制 （将原数×125存到新的数组里）
  for(i=len-3 ; i>0 ;i--)
  {
   if(oc[i]>99)
   cout<<oc[i];
   else if(oc[i]<100 && oc[i]>9)
   cout<<"0"<<oc[i];
   else if(oc[i]<10)
   cout<<"00"<<oc[i];
  }// 3位一输出，不够的加前导零；
 
  if (oc[0]%10)cout<<oc[0];
  else if(oc[0]%100)cout<<oc[0]/10;
  else if(!(oc[0]%100))cout<<oc[0]/100;
  //清后缀零
  cout<<" [10]/n" ;
 }
 return 0;
}*/

#include <iostream>
#include <stdio.h>
#include <memory>
#include <string.h>
using namespace std;
const int maxn=2000;
double oct[8]={0,125,250,375,500,625,750,875};
int main ()
{
 char str[maxn];
 int oc[maxn],dec[3*maxn];
 while (memset(str,'/0',sizeof(str)),cin>>str)
 {
  int i,len=strlen(str),j;
  for( i=0 ,j=len-1;str[j]!='.' ; i++,j--)
   oc[i]=str[j]-48;
   cout<<str<<" [8] = 0.";
  for( i=0 ;i<len-2 ; i++)
   for( j=0 ; j<len-2-i ; j++)
    {
     oc[j]=oc[j]*125;
     if( oc[j]>999)
     {
      oc[j+1]+=oc[j]/1000;
      oc[j]=oc[j]%1000;
     }
    }
  for(i=len-3 ; i>0 ;i--)
  {
   if(oc[i]>99)
   cout<<oc[i];
   else if(oc[i]<100 && oc[i]>9)
   cout<<"0"<<oc[i];
   else if(oc[i]<10)
   cout<<"00"<<oc[i];
  }
 
  if (oc[0]%10)cout<<oc[0];
  else if(oc[0]%100)cout<<oc[0]/10;
  else if(!(oc[0]%100))cout<<oc[0]/100;
  cout<<" [10]\n" ;
 }
 return 0;
}