Answer / patrick

//This works for strings of a fixed length(here 10)..Its
written in assumption that strings are given in same case
completely(ie,india,INdia will be false witout modifican..

#include<stdio.h>
#include<string.h>
main(){
char a1[10],a2[10],cn[10]={0,0,0,0,0,0,0,0,0,0};
int i,j,k,l,c=0;
printf("enter string B\n");
scanf("%s",a1);
printf("enter string A\n");
scanf("%s",a2);
for(i=0;i<strlen(a1);i++){
for(j=0;j<strlen(a2);j++){
if(a1[i]==a2[j]){
c++;
cn[i]=1;
}}}
for(k=0;k<strlen(a1);k++){
if(cn[k]==0){
printf("FALSE\n");exit(0);}}
printf("TRUE\n");}

Answer / latter

why u all waisting time... First understand problem then solve.
Problem is like that

suppose B = "iwantto"
A = "Hiwannachatoopsthatiswhy"

Now here all character of B is in A string...

Okay!!!

Answer / sravanthi

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#include<string.h>
int main()
{
char *a;
a=(char*)malloc(20*sizeof(char));
char *b;
b=(char*)malloc(20*sizeof(char));
int i,j,flag,count,n;
i=j=count=flag=0;
printf("enter the string 1");
gets(a);
printf("enter the string 2");
gets(b);
n=strlen(b);
while(b[j]!='\0')
{
i=0;
flag=0;
while(a[i]!='\0')
{
if(a[i]==b[j])
{
flag=1;
break;
}
else
{
i++;
}
}

if(flag==1)
{
count=count+1;
}
j++;
}
if(count==n)
{
printf("true");

}
else
{
printf("false");
}
getch();
}

Answer / raghuram

#include<iostream.h>
#include<stdio.h>
#include<string.h>
int main()
{
char str1[100],str2[100];
int flag1=0,flag=0;
cout<<"\n\nenter 1st string:";
cin>>str1;
cout<<"\n\nenter 2nd string:";
cin>>str2;
int i=0,j=0;
for(i=0;i<strlen(str1);i++)
{ flag=0;
for(j=0;j<strlen(str2);j++)
{
if(str1[i]==str2[j])
{flag=1;
str2[j]=' ';
break;
}
}
if(flag==0)
{
flag1=1;
cout<<"\n\nnot anagrams";
break;
}
}
if(flag1!=1)
cout<<"\nanagrams";


return 0;

}

Answer / sam

private bool IsPartOf(string B, string A)
{
if(A==null || B==null) return false;
if(B.Length>A.Length) return false;
A = A.ToLower();
B = B.ToUpper();

for(int i=0; i<A.Length;i++)
{
if((i<B.Length)&&(A[i]==B[i]))
{
bool found = true;
for(int j=i;j<B.Length;j++)
{
if(A[j]!=B[j])
{
found = false;
break;
}
}
if(found) return true;
}
}
return false;
}

Answer / satish nerlekar

//assuming both the strings in same case

int main()
{

string A;
string B;

int count = 0;
char repeatChar='$';

cout<<"Enter string A";
cin>>A;
cout<<"Enter the string B";
cin>>B;

if(A.length()<B.length())
{
cout<<"Length of string A is small than B, so all the charaters of the B can't be there in A";
exit(0);
}


int A_length = A.length();
int B_length = B.length();

for(int i=0; i<B_length; i++)
{
for(int j=0; j<A_length; j++)
{
if(A[j] == B[i])
{
if(repeatChar == A[j])
;
else
{
count++;
repeatChar = A[j];
break;
}

}
}
}

if(count==B_length)
cout<<"string A has all the chars of string B";
else
cout<<"string A doesn't have all chars of string B";

return 0;

}

Answer / raghuram.a

#include<iostream.h>
#include<string.h>
int h(char letter)
{

return(letter-97);
}

int main()

{

char str1[100],str2[100];

int i=0,count[26];
for(i=0;i<26;i++)

count[i]=0;

clrscr();
cout<<"\n\nenter 1st string:";

cin>>str1;

cout<<"\n\nenter 2nd string:";

cin>>str2;
for(i=0;i<strlen(str1);i++)

{

count[h(str1[i])]++;

}
for(i=0;i<strlen(str2);i++)

{

count[h(str2[i])]--;
} int flag=0;

for(i=0;i<26;i++)

{if(count[i]!=0)
{cout<<"\n\nnot anagrams";

flag=1;

break;

}
}

if(flag==0)

cout<<"\n\nanagrams";

return 0;

}

Answer / defcon

#include<iostream.h>
void main()
{
char a,b;
cout<<"Entr the values of a & b";
cin>>a>>b;
if(strcmp(a,b))
cout<<"Both a & b are the same";
else
cout<<"A & b are different";
}

Answer / tushar

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

int main()
{
char str1[10],str2[10],temp[10];
char *p,*q;
int flag=1;

cout<<"enter the two strings"<<endl;
cin>>str1;
cout<<"and"<<endl;
cin>>str2;

p=str1;

//strcpy(str2,temp);
//cout<<temp;
if(strlen(str1)!=strlen(str2))
flag=0;

while(*p)
{
char ch=*p;
q=str2;
while(*q)
{
if(ch!=*q)
q++;
else
{
*q='*';
break;
}
}
if((*q)=='\0')
{flag=0;
break;
}
p++;

}

if(flag)
cout<<"anagrams"<<endl;
else
cout<<"not anagrams"<<endl;

getch();
return 0;
}

Answer / somisetty

char B_str[] = "aaaaaa";
char A_str[] = "a"
// For the above case, this function should return true

char B_str[] = "aB";
char A_str[] = "aaaaaBaaaaaa"
// For the above case, this function should return true

int B_chars[256]; //for all unique chars in B_str

/* Init B_chars[] */
for( int i = 0; i < 256; i++ )
B_chars[i] = 0;

/* Get characters in B_str */
for( int i = 0; i < strlen(B_str); i++ )
B_chars[(int) B_str[i]]++;

/* check if A_str has which of them */
for( int i = 0; i < strlen(A_str); i++ )
B_chars[(int) A_str[i]] = 0;

/* If any of the B_chars[] has non-zero entry */
for( int i = 0; i < 256; i++ )
{
if( B_chars[i] != 0 )
{
return (false);
}
}

/* Otherwise return true */
return (true);

I need your help, i need a Turbo C code for this problem.. hope u'll help me guys.? Your program will have a 3x3 array. The user will input the sum of each row and each column. Then the user will input 3 values and store them anywhere, or any location or index, temporarily in the array. Your program will supply the remaining six (6) values and determine the exact location of each value in the array. Example: Input: Sum of row 1: 6 Sum of row 2: 15 Sum of row 3: 24 Sum of column 1: 12 Sum of column 2: 15 Sum of column 3: 18 Value 1: 3 Value 2: 5 Value 3: 6 Output: Sum of Row 1 2 3 6 4 5 6 15 7 8 9 24 Sum of Column 12 15 18 Note: Your program will not necessary sort the walues in the array Thanks..

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