TJU 3181. Stock Exchange June 10, 2010

TJU 3181. Stock Exchange
ACCEPTED

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;

int main()
{
  char s[100],s1[100],s2[100];
  int i,j,k;
  
  vector<double>price;
 
  vector<string>agent,buysell;
  while(1){
   int n;
   scanf("%d%s",&n,s);
   string str=s,str1;
   double d;
   if(n==0&&str=="END") break;
     agent.clear();buysell.clear();price.clear();
      for(int i=0;i<n;i++){
        scanf("%s%s%lf",s1,s2,&d);
        str1=s1;agent.push_back(str1);
        str1=s2;buysell.push_back(str1);
        price.push_back(d);
        }
       vector< vector<string> > v;
       vector<string> tmp;
       vector< vector<string> >::iterator x;
       //for(x=v.begin();x<v.begin();x++)
         // (*x).clear();
       v.clear(); 
       tmp.clear();

       vector<string>::iterator it1,it2,it3,it4;
       it3=buysell.begin();it4=buysell.begin();
       vector<double>::iterator it5,it6;
       it5=price.begin();it6=price.begin();
      for(it1=agent.begin();it1<agent.end();it1++,it3++,it5++){
              //cout<<"yes";
              tmp.clear();
              int flag=0;
              for(it2=agent.begin(),it4=buysell.begin(),it6=price.begin();it2<agent.end();it2++,it4++,it6++){
                   if(it1!=it2){
                          if(*it3!=*it4){
                              if(*it3=="buy"){
                                      if(*it6<=*it5){
                                        tmp.push_back(*it2);
                                        flag=1;
                                        }
                                   }
                               else if(*it3=="sell"){
                                           if(*it5<=*it6){
                                              tmp.push_back(*it2);
                                              flag=1;
                                              }
                                   }
                               }
                          }
                    }
                 if(flag==0) {tmp.push_back("NO-ONE");
                              v.push_back(tmp);
                             }
                 else
                           v.push_back(tmp);
               }  
                    
         printf("%s\n",s); 
        vector<string>::iterator it=agent.begin();
         vector< vector<string> >::iterator ii;
         vector<string>::iterator ii2;
         ii=v.begin();                                   
         for(int i=0;i<n;i++,it++,ii++){
           cout<<*it<<":";
           for(ii2=(*ii).begin();ii2<(*ii).end();ii2++){
              cout<<" "<<(*ii2);
              }
            cout<<endl;
           }
       
       }

return 0;
}

TJU 3249. Happy Coins June 9, 2010

TJU 3249. Happy Coins
ACCEPTED

Earlier i was trying to code what was written in the problem statement. Then i found that its hard to decide programatically. So i took paper & pen started finding the repeating pattern.

I found this pattern
if we take
lxh->0
hhb->1
Then
=================
when n=2
input1 input2 Ans
0 0 1(hhb)
0 1 0(lxh)
1 0 0(lxh)
1 1 1(hhb)
=====================
when n=3
input1 input2 input3 Ans
0 0 0 0(lxh)
0 0 1 1(hhb)
0 1 0 1(hhb)
0 1 1 0(lxh)
1 0 0 1(hbb)
1 0 1 0(lxh)
1 1 0 0(lxh)
1 1 1 1(hhb)
==================
when n=4
inp1 inp2 inp3 inp4 ans
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 0
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 1
1 0 1 1 0
1 1 0 0 1
1 1 0 1 0
1 1 1 0 0
1 1 1 1 1
====================

my AC code

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
using namespace std;

int main()
{
   int cases;
   cin>>cases;
   int n,turn,zero,one;
   string s;
   
   while(cases--){
        s="";    
       cin>>n;
       zero=0;one=0;
       int tmp=n;   
       while(tmp--){
         cin>>s;
         if(s=="lxh") ++zero;
         else if(s=="hhb") ++one;

         }
       if(n%2==0){
              if(one%2==0) printf("hhb\n");
              else printf("lxh\n");
           }
       else{
              if(one%2!=0) printf("hhb\n");
              else printf("lxh\n");
           }
      }

return 0;
}

TJU 3513. Triangle June 9, 2010

TJU 3513. Triangle
This problem require effort to tackle Precision errors

ACCEPTED

My AC code

//precision error upto 5 digits after decimal taken 
//ACCEPTED
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;

int main()
{
  long double PI=3.14159265/2.000000000,PI2=3.14159265;
  PI=(int)(PI*100000);PI2=(int)(PI2*100000);
  PI/=100000;PI2/=100000;
  //cout<<PI<<endl;
  int x1,y1,x2,y2,x3,y3;
  int cases,k=0;
  cin>>cases;
  while(cases--){
     cin>>x1>>y1>>x2>>y2>>x3>>y3;
     long double a,b,c;
     a=sqrt(pow((x1-x2),2.00000)+pow((y1-y2),2.00000));
     b=sqrt(pow((x2-x3),2.00000)+pow((y2-y3),2.00000));
     c=sqrt(pow((x3-x1),2.00000)+pow((y3-y1),2.00000));
     long double A,B,C;
     A=acos((b*b+c*c-a*a)/(2.00000*b*c));
     B=acos((a*a+c*c-b*b)/(2.00000*a*c));
     C=acos((b*b+a*a-c*c)/(2.00000*b*a));
     A=(int)(A*100000); B=(int)(B*100000); C=(int)(C*100000);
     A/=100000;B/=100000;C/=100000;
    //cout<<A<<" "<<B<<" "<<C<<endl;
   printf("Case #%d: ",++k);
if(x1==x2&&y1==y2||x2==x3&&y2==y3||x3==x1&&y3==y1||A==0.00000||B==0.00000||C==0.00000||A==PI2||B==PI2||C==PI2) printf("not a triangle\n");
else if(A<PI&&B<PI&&C<PI){
         
         if(a==b&&b!=c||b==c&&b!=a||c==a&&a!=b) printf("isosceles acute triangle\n");
         else if(a!=b&&b!=c&&c!=a) printf("scalene acute triangle\n");                          
         }
else if(A==PI||B==PI||C==PI){
         if(a==b&&b!=c||b==c&&b!=a||c==a&&a!=b) printf("isosceles right triangle\n");
        else if(a!=b&&b!=c&&c!=a) printf("scalene right triangle\n");                          
         }
else if(A>PI||B>PI||C>PI){
        if(a==b&&b!=c||b==c&&b!=a||c==a&&a!=b) printf("isosceles obtuse triangle\n");
        else if(a!=b&&b!=c&&c!=a) printf("scalene obtuse triangle\n");                          
         }
       
     
}
return 0;
}

TJU 1701. Stamps June 4, 2010

TJU 1701. Stamps
AC

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

int main()
{
  int cases;
  cin>>cases;
  int n,a,arr[1009],i,j,k=0;
  while(cases--){
      
      cin>>n>>a;
      for(i=0;i<a;i++)
          cin>>arr[i];

      sort(arr,arr+a);
      int total=0;
         i=0;
       while(total<n){
         if(i==a) break;
         total+=arr[a-1-i];
         i++;
         }

      down:
        printf("Scenario #%d:\n",++k);
        if(total<n)
             printf("impossible\n\n");
        else
             printf("%d\n\n",i);
     }


return 0;
}

TJU 2199. A+B Problem May 30, 2010

TJU 2199. A+B Problem
AC
This problem requires Sorting and Searching(Binary) approch

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;


int main()
{
 int i,j,k;
 int a[1009],n;
 int tmp,diff;
 int low,mid,high,flag,num;
 while(1){
    scanf("%d",&n);
    if(n==0) break;
    for(i=0;i<n;i++)
       scanf("%d",&a[i]);
    
    sort(a,a+n);//sort
    
    for(i=n-1;i>=0;i--){
       tmp=a[i];
       for(j=i-1;j>=0;j--){
         diff=tmp-a[j];
         high=j-1;low=0;
         flag=0;
         mid=(low+high)/2;
         while(low<=high){//binary search
            mid=(low+high)/2;
            
            if(a[mid]==diff){ flag=1;num=tmp;goto down;}
            else if(diff<a[mid]){high=mid-1;}
            else if(diff>a[mid]){low=mid+1;}
         }
       }
      
    }
     down:
          if(flag==1) printf("%d\n",tmp);
          else printf("-1\n");
   }    

return 0;
}

TJU 1547. To and Fro May 30, 2010

After a long time , playing TJU contest
TJU 1547. To and Fro
AC

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int main()
{
 int i,j,k;
 int n;
 char s[205];
 char str[100][21],tmp[100][21];
 while(1){

 scanf("%d",&n);
 if(n==0) break;
 scanf("%s",s);
 k=0;j=0;
 for(i=0;s[i]!='\0';){
  tmp[j][k]=s[i];
  i++;
  if(i%n==0)
     j=j+1;
    
  if(j%2==0&&i%n!=0) k=k+1;
  else if(j%2!=0&&i%n!=0) k=k-1;
  
  }
     
  for(i=0;i<n;i++){
      for(k=0;k<j;k++){
          printf("%c",tmp[k][i]);
      }
   }
    printf("\n");
      
 }


return 0;
}

TJU 1365. Hard to Believe, but True! February 25, 2010

TJU 1365. Hard to Believe, but True!

–AC–

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
  char sa[10],sb[10],sc[10];
  int i,j,k,la,lb,lc,a,b,sum;
  char c;
  while(1){
  i=0;
  while(1){ c=getchar(); if(c=='+') break;sa[i++]=c;}
  sa[i]='\0';
  i=0;
  while(1){ c=getchar();if(c=='=') break;sb[i++]=c;}
  sb[i]='\0';
  i=0;
  while(1){ c=getchar();if(c=='\n') break;sc[i++]=c;}
  sc[i]='\0';
  la=strlen(sa);lb=strlen(sb);lc=strlen(sc);
  for(i=0;i<la/2;i++){
    c=sa[i];sa[i]=sa[la-1-i];sa[la-1-i]=c;
   }
   for(i=0;i<lb/2;i++){
    c=sb[i];sb[i]=sb[lb-1-i];sb[lb-1-i]=c;
   }
   for(i=0;i<lc/2;i++){
    c=sc[i];sc[i]=sc[lc-1-i];sc[lc-1-i]=c;
   }
   a=atoi(sa);b=atoi(sb);sum=atoi(sc);
   if(a==0&&b==0&&sum==0) {printf("True\n");break;}
   if((a+b)==sum) printf("True\n");
   else printf("False\n");
   }

return 0;
}

TJU 1574. Financial Management February 11, 2010

TJU 1574. Financial Management

–AC–

#include<stdio.h>

int main()
{
  float in,avg=0.0;
  int i;
  for(i=1;i<=12;i++)
  {
    scanf("%f",&in);
    avg+=in;
  }
    avg=avg/12.0;
    printf("$%.2f\n",avg);

return 0;
}

TJU 1590. IBM Minus One February 11, 2010

TJU 1590. IBM Minus One

–AC–

#include<stdio.h>
#include<string.h>

int main()
{
  int cases,i,j=0;
  char s[100];

  scanf("%d",&cases);
  while(cases--){
  scanf("%s",s);
  for(i=0;s[i]!='\0';i++){
   s[i]='A'+(s[i]-'A'+1)%26;
  }
  printf("String #%d\n",++j);
  printf("%s\n\n",s);

  }

  return 0;
}

TJU 1644. Reverse Text February 11, 2010

TJU 1644. Reverse Text

–AC–

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main()
{
  int cases,i,j,k;
  char s[500],c;
 
  scanf("%d\n",&cases);
  while(cases--){
   gets(s);
   j=strlen(s);
   for(i=0;i<j/2;i++){
    c=s[i];
    s[i]=s[j-1-i];
    s[j-1-i]=c;
   }
   puts(s);
  }

return 0;
}
  

TJU 2945. HERMAN February 8, 2010

TJU 2945. HERMAN

-AC-

#include<stdio.h>
int main()
{
  double R;
  double a=3.14159265358979323846;
  scanf("%lf",&R);
  printf("%.6lf\n%.6lf\n",a*R*R,(2*R*R));

return 0;
}

TJU 2782. I am Lord Voldemort February 8, 2010

TJU 2782. I am Lord Voldemort

–AC–

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int main()
{

  int T,i,j,k;
  char s1[200],s2[200];
  int a[26],b[26],flag;
  
  scanf("%d",&T);
  while(T--){
  scanf("%s%s",s1,s2);

  for(i=0;i<26;i++){
     a[i]=0;
     b[i]=0;
  }
  for(i=0;s1[i]!='\0';i++){
     if(s1[i]>='A'&&s1[i]<='Z')
      a[s1[i]-'A']++;
     else
      a[s1[i]-'a']++;
   }
   
    for(i=0;s2[i]!='\0';i++){
     if(s2[i]>='A'&&s2[i]<='Z')
      b[s2[i]-'A']++;
     else
      b[s2[i]-'a']++;
   }
    
   flag=1;
   for(i=0;i<26;i++){
      if(a[i]==b[i])
        continue;
      else{ flag=0; break;}
   }
   
   if(flag==1)
     printf("Yes\n");
   else
     printf("No\n");
   }

return 0;
}       

TJU 2800. Cube February 8, 2010

TJU 2800. Cube

–AC–

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>

int main()
{
 
 long long int N,C,a[1002];
 int cases;
 scanf("%d",&cases);
 while(cases--){
 scanf("%lld",&N);
 C=N*N*N;

 printf("%lld %lld\n",(N*N)-N+1,(N*N)+N-1);
 }

return 0;
}

TJU 3098. Event Planning February 6, 2010

TJU 3098. Event Planning

–AC–

#include<stdio.h>

int main()
{
  int N,B,H,W,i,j,k;
  int p,a[1000];
  int min,flag,total;
  scanf("%d%d%d%d",&N,&B,&H,&W);
  min=100000000;
  for(i=1;i<=H;i++)
  {
    
    total=0;
    scanf("%d",&p);
    for(j=1;j<=W;j++)
    {
      scanf("%d",&a[j]);
       total+=a[j];
     }
     if(total>=N)
       if(min>(N*p))
         min=N*p;
     
     
   }

   if(min>B)
     printf("stay home\n");
   else
     printf("%d\n",min);

return 0;
}

TJU 1748. Power Digit January 27, 2010

TJU 1748. Power Digit
TODOLIST
Output Limit Exceed

#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;

class A{
   int m,n,i,j,k,rem,res;

  public:
    void input(){
       cin>>m>>n; 
       if(m==0&&n==0) exit(0);
    }

   void calc(){
     rem=m%10;
     switch(rem){
         case 1:
                 res=1;
                 break;
         case 2:
                 i=n%4;if(i==0) i=4;
                 res=pow(2,i);
                 break;
         case 3:
                 i=n%4;if(i==0) i=4;
                 res=pow(3,i);
                 break;
         case 4:
                 i=n%2;if(i==0) i=2;
                 res=pow(4,i);
                 break;
         case 5:
                 res=5;
                 break;
         case 6:
                 res=6;
                 break;
         case 7:
                 i=n%4;if(i==0) i=4;
                 res=pow(7,i);
                 break;
         case 8:
                 i=n%4;if(i==0) i=4;
                 res=pow(8,i);
                 break;
         case 9:
                 i=n%2;if(i==0) i=2;
                 res=pow(9,i);
                 break;
         case 0:
                 res=0;
                 break;
         }
      }

     void output(){
          cout<<(res%10)<<endl;
     }
};

int main()
{
  A obj;
    while(1){
      obj.input();
      obj.calc();
      obj.output();
    }

return 0;
}