TJU 3004. Mispelling December 12, 2009

TJU 3004. Mispelling

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

int main()
{
  int i,j,k;
  int cases,cs=0;
  char s[1000];
  
  scanf("%d",&cases);
  while(cases--)
  {
    ++cs;
    scanf("%d %s",&i,s);
   printf("%d ",cs);  
  for(j=0;s[j]!='\0';j++)
     if(i-1!=j)
      printf("%c",s[j]);

   printf("\n");
  }

return 0;
}

TJU 2123. Head or Tail December 11, 2009

TJU 2123. Head or Tail

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

int main()
{
  int i,j,k;
  int n,v,M,J;

  while(1)
  {
    scanf("%d",&n);
    if(n==0) break;
    M=0;J=0;
    for(i=0;i<n;i++)
    {
      scanf("%d",&v);
      if(v==0)
      ++M;
      else
      ++J;
    }

    printf("Mary won %d times and John won %d times\n",M,J);
  }

return 0;
}

SPOJ 1028. Hubulullu December 11, 2009

SPOJ 1028. Hubulullu
Problem code: HUBULLU

Hard and Easy, depends which way u think
–AC–(easy way)

SPOJ 2528. Monkey Vines December 9, 2009

SPOJ 2528. Monkey Vines
Problem code: GNY07F

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

int main()
{
  int i,j,k;
  int N,max,test=0;
  char s[200],c;  

  scanf("%d",&N);
  getchar();
  while(N--)
  {
    i=0;
    while(1)
    {
      c=getchar();
      if(c=='\n')
      {
      s[i]='\0';
      break;
      }
     s[i++]=c;
    }
    ++test;
    j=0;
    if(s[0]=='\0'||s[0]=='\n')
    max=1;
    else
     max=0;
    for(i=0;s[i]!='\0';i++)
    {
      if(s[i]=='[')
         j++;
      else if(s[i]==']')
        j--;
      if(j>max)
       max=j;
   }
   
   if(max==1&&s[0]=='\0')
   printf("%d %d\n",test,max);
   else
   printf("%d %d\n",test,(int)pow(2,max));
 }

return 0;
}

SPOJ 4273. Train TimeTable December 9, 2009

SPOJ 4273. Train TimeTable
Problem code: TTTABLE
TODOLIST:

TJU 1516. Climbing Worm December 9, 2009

TJU 1516. Climbing Worm

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

int main()
{
  int i,j,k;
  int n,u,d,h,tmp,result,rem;

  while(1)
  {
    scanf("%d%d%d",&n,&u,&d);
   if(n==0) break;

   i=u-d;
   tmp=(n-u)/i;
   rem=(n-u)%i;
   if(rem!=0)
   tmp=tmp+1;
   result=2*tmp+1;  
   printf("%d\n",result);

  }

return 0;
}

SPOJ 1025. Fashion Shows December 8, 2009

SPOJ 1025. Fashion Shows
Problem code: FASHION
–AC–
sorting based problem(qsort implemnted)

TJU 3288. Stockholm Numbers December 8, 2009

TJU 3288. Stockholm Numbers
original thinking implemented
1->1*2+1=3 //1 has odd parity
2->2*2+1=5//2has odd parity
3->3*2=6//3has even parity
4->4*2+1=9//4has odd parity
5->5*2=10//5has even parity
and so on
……………….

--AC--
#include<stdio.h>
int main()
{
  long long int i,j,k,K,tmp,rem,cases;
  scanf("%lld",&cases);
  while(cases--)
  {
    scanf("%lld",&K);
    j=0;
    tmp=K;
    while(K>0)
    {
      rem=K%2;
      if(rem==1)
       ++j;
      K=K/2;
    }
    if(j%2==0)
     printf("%lld\n",2*tmp);
    else
     printf("%lld\n",2*tmp+1);
  }

return 0;
}

SPOJ 5240. Area of a Garden December 8, 2009

SPOJ 5240. Area of a Garden
Problem code: GARDENAR

–AC–
FORMULAE BASED
http://www.gogeometry.com/problem/p103_equilateral_triangle_heron_area_elearning.htm

SPOJ 2716. Maximal Quadrilateral Area December 8, 2009

SPOJ 2716. Maximal Quadrilateral Area
Problem code: QUADAREA
–AC–
0.05s
FORMULAE BASED

TJU 2775. Bee Tree December 8, 2009

TJU 2775. Bee Tree

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

int main()
{
  int T,n,i,j,k;
  int grandpa[50],grandma[50];
  
    grandpa[0]=1;grandma[0]=1;
    for(i=1;i<50;i++)
    {
      grandpa[i]=grandma[i-1];
      grandma[i]=grandpa[i-1]+grandma[i-1];
    }

   scanf("%d",&T);
   while(T--)
   {
     scanf("%d",&n);
     printf("%d %d\n",grandpa[n],grandma[n]);
   }
return 0;
}

      
     
    
  

SPOJ 8. Complete the Sequence! December 8, 2009

SPOJ 8. Complete the Sequence!
Problem code: CMPLS
TODOLIST:

SPOJ 127. Johnny Goes Shopping December 8, 2009

SPOJ 127. Johnny Goes Shopping (Challenge)
Problem code: JOHNNY
–AC– ( but less points)

proceedure:
find the sum half it and then sort the array and accordingly store the positions and then find the sum of the sorted array which is nearest to half the sum of weights.

TJU 2210. Adding Reversed Numbers December 7, 2009

SPOJ 42. Adding Reversed Numbers

TJU 2210. Adding Reversed Numbers

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

int main()	
{
  int i,j,k,l1,l2,l,result,n1,n2;
  char s1[10],s2[10],s[10],c,tmp;
  int cases;

  scanf("%d",&cases);
  while(cases--)
  {
   scanf("%s %s",s1,s2);
   l1=strlen(s1);
   l2=strlen(s2);

   for(i=0;i<l1/2;i++)
   {
     tmp=s1[i];
     s1[i]=s1[l1-i-1];
     s1[l1-i-1]=tmp;
   }

   for(i=0;i<l2/2;i++)
   {
     tmp=s2[i];
     s2[i]=s2[l2-i-1];
     s2[l2-i-1]=tmp;
   }

   n1=atoi(s1);
   n2=atoi(s2);
   result=n1+n2;
   sprintf(s1,"%d",result);
   l1=strlen(s1);
   for(i=0;i<l1/2;i++)
   {
     tmp=s1[i];
     s1[i]=s1[l1-i-1];
     s1[l1-i-1]=tmp;
   }
   result=atoi(s1);
   printf("%d\n",result);
  }

return 0;
} 

TJU 2218. Super Square December 7, 2009

TJU 2218. Super Square
EXPMOD implemented

--AC--
//expmod based program (m^m)%n
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
long long int expmod(long long int n, long long int p,long long  int m) {
   if (p == 0) return 1;
   int nm = n % m;
   long long int r = expmod(nm, p / 2, m);
   r = (r * r) % m;
   if (p % 2 == 0) return r;
   return (r * nm) % m;
}

int main()
{
  long long int i,j,k,N,n,r,sum,a,b,x,y,acc,m=2006;

  while(1)
  {
    scanf("%lld",&n);
    if(n==0) break;
   
    r = 0;
    r = (r + expmod(n, n, m)) % m;
   
    printf("%lld\n",r);
  }

 

return 0;
}