What is the most efficient way to count the number of bits which are set in an integer?
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main() { char as[] = "\\0\0"; int i = 0; do{ switch( as[i++]) {case '\\' : printf("A"); break; case 0 : printf("B"); break; default : printf("C"); break; }} while(i<3); }
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a number is perfect if it is equal to the sum of its proper divisor.. 6 is perfect number coz its proper divisors are 1,2 and three.. and 1+2+3=6... a number is deficient if the sum of its proper divisor is less than the number.. sample: 8 is deficient, coz its proper divisors are 1,2 and 4, and 1+2+4=7. abundant number, if the sum of its proper divisor is greater than the number.. sample..12 is abundant coz 1+2+3+4+6=16 which is geater than 12. now write a program that prompts the user for a number, then determines whether the number is perfect,deficient and abundant..
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