Master Clever - C

Consider the following I/O samples. SAMPLE INPUT :  abcd cdba SAMPLE OUTPUT:   ANAGRAM C PROGRAM: #include<string.h> #include <stdio.h> int main() {     int i,arr[26]={0};     char s1[100],s2[100];     scanf("%s %s",s1,s2);     for(i=0;s1[i]!='\0';i++)         arr[s1[i]-'a']++;     for(i=0;s2[i]!='\0';i++)         arr[s2[i]-'a']--;     for(i=0;i<26;i++)     {         if(arr[i]!=0)         {             printf("Not a Anagram");             return 0;         }     }     printf("Anagram"); }   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article ...

Consider the following I/O samples. SAMPLE INPUT :  200 300 SAMPLE OUTPUT:   204 208 212 216 220 224 228 232 236 240 244 248                                             252 256  260 264 268 272 276 280 284 288 292 296 C PROGRAM: #include <stdio.h> int checkLeapYear(int year) {     if( (year % 400==0)||(year%4==0 && year%100!=0) )         return 1;     else         return 0; }   int main() {     int i,n,s;     scanf("%d %d",&s,&n);     printf("Leap years from %d to %d:\n\n",s,n);     for(i=s;i<=n;i++)     {         if(checkLeapYear(i))             printf("%d ",i);     }     return 0; }   OUTPUT:      The ...

Consider the following I/O samples. SAMPLE INPUT :  8000  200 SAMPLE OUTPUT:   200 8000 C PROGRAM: #include <stdio.h> int main() {      int a,b,t;     scanf("%d %d",&a,&b);     printf("\nBefore swapping : A = %d B = %d",a,b);     a^=b;                 b^=a;                 a^=b;       printf("\nAfter swapping : A = %d B = %d",a,b);     return 0; }   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article and want to improve this, please comment in the comment section.  

Consider the following I/O samples. SAMPLE INPUT :  600  1200 SAMPLE OUTPUT:   1200 600 C PROGRAM: #include <stdio.h> int main() {      int a,b,t;     scanf("%d %d",&a,&b);     printf("\nBefore swapping : A = %d B = %d",a,b);     a=a+b-(b=a);     printf("\nAfter swapping : A = %d B = %d",a,b);     return 0; } ANOTHER METHOD:                   a=a+b;                 b=a-b;                 a=a-b;   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article and want to improve this, please comment in the comment section.  

Consider the following I/O samples. SAMPLE INPUT :  9863 == 9+8+6+3=26 == 2+6=8 SAMPLE OUTPUT:   8 C PROGRAM: #include <stdio.h> int main() {     int n;     scanf("%d",&n);     if(n%9 == 0)         printf("9");     else         printf("%d",n%9);     return 0; }   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article and want to improve this, please comment in the comment section.  

Consider the following I/O samples. SAMPLE INPUT :  23 SAMPLE OUTPUT:   10011 C PROGRAM: #include <stdio.h> #include<math.h> int main() {    long int n, dec=0, bin=0, i=0;     scanf("%ld",&n);     while(n)     {         dec+=(n%10)*pow(8,i);         ++i;         n/=10;     }     i=1;     while(dec)     {         bin+=(dec%2)*i;         dec/=2;         i*=10;     }     printf("%ld",bin);     return 0; }   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article and want to improve this, please comment in the comment section.  

Consider the following I/O samples. SAMPLE INPUT :  32 SAMPLE OUTPUT:   26 C PROGRAM: #include <stdio.h> int main() {     long int n, base=1, rem, ans=0;     scanf("%ld",&n);     while(n)     {         rem=n%10;         ans=ans+rem*base;         base=base*8;         n=n/10;     }     printf("%ld",ans);     return 0; }   OUTPUT:      The given source code in C program is short and simple to understand. The source code is well tested in DEV-C++ and is completely error free.  If you have any feedback about this article and want to improve this, please comment in the comment section.