SPOJ 5450. Tiles of Tetris, Not! November 30, 2009

SPOJ 5450. Tiles of Tetris, Not!
Problem code: ANARC09B

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

long long int gcd(long long int n1,long long int n2)
{
  if(n1==0) return n2;
  else if(n2==0) return n1;

  if(n1>n2)
    return gcd(n1%n2,n2);
  else
    return gcd(n1,n2%n1);

return 0;

}

int main()
{
  long long int i,j,k,lcm;
  long long int W,H;
  
  while(1)
  {
   scanf("%lld%lld",&W,&H);
   if(W==0&&H==0) break;
  
   if(W==H) 
       printf("1\n");
   else if(W>H)
   {
        if(W%H==0) 
           printf("%lld\n",W/H);
        else
          {
           lcm=(W*H)/gcd(W,H);
           printf("%lld\n",lcm);
          }
    }
    else if(H>W)
    {
        if(H%W==0)
          printf("%lld\n",H/W);
        else
          {
            lcm=(W*H)/gcd(W,H);
            printf("%lld\n",lcm);
          }
     }
   
   }

return 0;
}

SPOJ 5447. Probability One November 30, 2009

SPOJ 5447. Probability One
Problem code: ANARC09E

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

int main()
{
  int i,j;
  int k,B,Q;
  k=0;
  while(1)
  {
    
    scanf("%d",&B);
    if(B==0) break;
    if(B%2==0)
      printf("%d. even %d\n",++k,B/2);
    else
      printf("%d. odd %d\n",++k,(B-1)/2);
  }
return 0;
}

TJU 2199. A+B Problem November 29, 2009

TJU 2199. A+B Problem
<em>–TLE–

----its giving TLE----
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(x,y,z) for(x=y;x<=z;x++)
int main()
{
  int n,a[1002],b[1002];
  int i,j,k,l,max,tmp,flag;

  while(1)
  {
    scanf("%d",&n);
    if(n==0) break;
    for(i=1;i<=n;i++)
     scanf("%d",&a[i]);

    max=-1;
    FOR(i,1,n)
    {
      flag=0;
      FOR(j,1,n)
      {
        FOR(k,1,n)
        {
         if(i!=j&&i!=k&&j!=k&&a[i]==(a[j]+a[k]))
             {tmp=a[i];flag=1;goto down;}
         }
        }
       down:
       if(tmp>max&&flag==1)
        max=tmp;
      }
    
        printf("%d\n",max);
   }

return 0;
}

TJU 1389. Forests November 28, 2009

TJU 1389. Forests

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

int main()
{
  int i,j,k,count;
  int a[201][201],T,P,flag[201];

  scanf("%d%d",&P,&T);
  for(i=0;i<201;i++)
  memset(a,0,sizeof(a[i]));
  memset(flag,0,sizeof(flag));
  while(1)
  {
    if(scanf("%d%d",&i,&j)==-1) break;
     a[i][j]=1;
  }
   count=0;
  for(i=1;i<=P;i++)
  {
     if(flag[i]==1) 
         continue;
     ++count;
     for(j=i;j<=P;j++)
     {
       for(k=1;k<=T;k++)
        if(a[i][k]!=a[j][k])
          break;
       if(k==T+1)
         flag[j]=1;
     }
   
   }
      printf("%d\n",count);

return 0;
}

TJU 1643. Golf November 28, 2009

TJU 1643. Golf

#include<stdio.h>
#include<iostream>
using namespace std;

int main()
{
  int i=0,j,k;
  int p,s;
  
  while(1)
  {
    scanf("%d%d",&p,&s);
    if(p==0) break;
    
    printf("Hole #%d\n",++i);
    if(s==1)
       printf("Hole-in-one.\n");
    else if(s==p)
       printf("Par.\n");
    else if(s==p-1)
       printf("Birdie.\n");
    else if(s==p+1)
       printf("Bogey.\n");
    else if(s==p-2)
       printf("Eagle.\n");
    else if(s==p-3)
       printf("Double Eagle.\n");
    else if(s==p+2)
       printf("Double Bogey.\n");
    else 
       printf("Double Bogey.\n");
    printf("\n");
   }

return 0;
}

TJU 3232. kangaroo November 28, 2009

TJU 3232. kangaroo

#include<stdio.h>
#include<iostream>
using namespace std;

int main()
{
  int diff1,diff2;
  int a,b,c,res;
  scanf("%d%d%d",&a,&b,&c);
  diff1=b-a;
  diff2=c-b;
  if(diff1>diff2)
    printf("%d\n",(b-a-1));
  else
    printf("%d\n",(c-b-1));

return 0;
}

How to use binary_search() November 28, 2009

How to use binary_search()

#include <algorithm>
#include <iostream>
#include <utility>
using namespace std;
#define N 20
int a[N];
int main(){
    sort(a,a+N);
    if(binary_search(a,a+N,1)){
        cout<<"OK";
    }
}

How to use sort() November 28, 2009

How to use sort()

#include <algorithm>
using namespace std;
#define N 20
int a[N];
bool cmp(int x,int y){
    return x<y;
}
int main(){
    sort(a,a+N,cmp);
    return 0;
}

TJU 2391. Ancient Cipher November 28, 2009

TJU 2391. Ancient Cipher

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

int main()
{
  int i,j,k;
  char s1[101],s2[101];
  int a[26],b[26];
  int n,l,temp,flag;
  while(1)
  {
  if(scanf("%s",s1)==-1) break;
  scanf("%s",s2);
  memset(a,0,sizeof(a));
  memset(b,0,sizeof(b));

  for(i=0;s1[i]!='\0';i++)
  {
     a[s1[i]-'A']++;
     b[s2[i]-'A']++;
  }

  for(i=0;i<26;i++)
  {
    for(j=i;j<26;j++)
    {
      if(a[i]>a[j])
      {
        temp=a[i];a[i]=a[j];a[j]=temp;
      }
      if(b[i]>b[j])
      {
        temp=b[i];b[i]=b[j];b[j]=temp;
      }
    }
  }

    flag=1;
   for(i=0;i<26;i++)
   {
    if(a[i]!=b[i])
    {
     flag=0;
     break;
    }
   }
    if(flag==1) printf("YES\n");
    else printf("NO\n");
  }

return 0;
}
     

TJU 2674. Java vs C++ November 28, 2009

2674. Java vs C++
1163. Java vs C ++ Problem code: JAVAC

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

int main()
{

  int i,j,k,flag;
  char s[201],c,res[201];
  while(1)
  {
    if(scanf("%s",s)==-1) break;;
    for(i=0;i<201;i++)
     res[i]='\0'; 
    if(strchr(s,'_'))
    {
      flag=0;
      for(i=0;s[i]!='\0';i++)
       if(s[i]>='A'&&s[i]<='Z')
       {
         flag=1;
         break;
       }
      if(flag==1||s[0]=='_'||s[strlen(s)-1]=='_'||s[0]>'A'&&s[0]<='Z'||strstr(s,"__")) 
         printf("Error!\n");
      else
        {
          j=0;
          for(i=0;s[i]!='\0';)
          {
            if(s[i]=='_')
            {
             res[j++]=s[i+1]-32;
             i=i+2;
            }
            else
            {
             res[j++]=s[i];
             i=i+1;
            }
          }
         res[j]='\0';

       printf("%s\n",res); 
      }
  
   }
    else 
    {
     if(s[0]>='A'&&s[0]<='Z')
       printf("Error!\n");
     else
     {
     j=0;
     for(i=0;s[i]!='\0';)
     {
      if(s[i]>='A'&&s[i]<='Z')
      {
        res[j++]='_';
        res[j++]=s[i]+32;
        i++;
      }
     else
      {
        res[j++]=s[i];
        i++;
      }
    }
   
   printf("%s\n",res);
   }
   }   
                 
  }

return 0;
}         

TJU 1357. Interpreter November 27, 2009

TJU 1357. Interpreter

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

int main()
{
  int i,j,k;
  static int reg[11],addr[1001];
  int a,b,c,ip;
  i=0;
  while(scanf("%d",&addr[i])!=-1) i++;
  j=1;ip=0;
  while(1)
  {
    a=addr[ip]/100;
    b=(addr[ip]/10)%10;
    c=addr[ip]%10;
    ip++;
    if(a==1)
        break;
    else if(a==2)
          reg[b]=c;
    else if(a==3)
          reg[b]=(reg[b]+c)%1000;
    else if(a==4)
          reg[b]=(reg[b]*c)%1000;
    else if(a==5)
          reg[b]=reg[c];
    else if(a==6)
          reg[b]=(reg[b]+reg[c])%1000;
    else if(a==7)
          reg[b]=(reg[b]*reg[c])%1000;
    else if(a==8)
          reg[b]=addr[reg[c]];
    else if(a==9)
          addr[reg[c]]=reg[b];
    else if(a==0&&reg[c]!=0)
          ip=reg[b];
    j++;
   }
    printf("%d\n",j);


return 0;
}

TJU 2698. Goldbach’s conjecture November 27, 2009

TJU 2698. Goldbach’s conjecture
[it is almost same as TJU 1171]

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



int compute(int n)
{
  int i,j,k,flag=0;
  for(j=2;j<=sqrt(n);j++)
    {
      if(n%j==0)
       {
        flag=1;
        break;
       }
      else
        flag=0;
     }

    if(flag==0)
      return 1;
    else
      return 0;    

return 0;
}

int main()
{
  int i,j,k,flg,count;
  int n;

  while(1)
  {
    scanf("%d",&n);
    count=0;
    if(n==0) break;
    flg=0;
     for(i=3;i<=n/2;i=i+2)
     {

       if(compute(i)&&compute(n-i)){
       ++count;
       }

     }
     printf("%d\n",count);

   }

return 0;
}

TJU 2843. Diamonds November 27, 2009

TJU 2843. Diamonds

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

int main()
{
  int t,n,i,j,k,l,big,max;
  int a[10002][4];

  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&n);
    for(i=1;i<=n;i++)
     for(j=1;j<=3;j++)
      scanf("%d",&a[i][j]);
    max=0;
    for(i=2;i<=n;i++)
    {
      for(j=1;j<=3;j++)
      {
        if(j==1) {k=2;l=3;}
        else if(j==2){k=1;l=3;}
        else if(j==3){k=1;l=2;}
        
        if(a[i-1][k]>a[i-1][l])
         big=a[i-1][k];
        else 
         big=a[i-1][l];

        a[i][j]=a[i][j]+big;
        if(a[i][j]>max)
          max=a[i][j];
        }
      }
    printf("%d\n",max);
   }
return 0;
}

TJU 1233. Number Steps November 27, 2009

TJU 1233. Number Steps

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

int main()
{
  int N,x,y;
  

  scanf("%d",&N);
  while(N--)
  {
    scanf("%d%d",&x,&y);
    
    if(x%2==0&&y%2==0&&abs(x-y)==2||x%2==0&&y%2==0&&x==y)
      printf("%d\n",x+y);
    else if(x%2!=0&&y%2!=0&&abs(x-y)==2||x%2!=0&&y%2!=0&&x==y)
      printf("%d\n",x+y-1);
    else 
      printf("No Number\n");
 }

return 0;
}

TJU 2841. Bitwise Reverse November 26, 2009

TJU 2841. Bitwise Reverse

[took 30 Mins to solve]

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

int main()
{
  int i,j,k;
  char s[1000002];
  int n,rem,res;
  int new_num,n_digit;

  while(1)
  {
    scanf("%d",&n);
    if(n==0) break;
    
    for(i=1;i<=20;i++)
    {
       if((pow(2,i-1))>n) 
       break;
    }
    n_digit=i-1;
   
   new_num=0;
   while(n>0)
   {
     rem=n%2;
     n=n/2;
     new_num=new_num+rem*pow(2,n_digit-1);
     n_digit--;
   }
   printf("%d\n",new_num);
  
  }

return 0;
}