Puzzle:There are 10 numbers, each has values between 0 to 99. Write computer program to find the average of those 10 numbers, such that the total must never exceed 99 (at any point during or after iteration).Answer:/* * First Approach * Programming language: Java */ int[]number=new int[10]; // initialize number int absolute_average=0,sum_of_remainders=0; for(int i = 0 ; i < 10 ; ++i) { absolute_average+=number[i]/10; sum_of_remainders+=number[i]%10; if(sum_of_remainders>10) { absolute_average+=sum_of_remainders/10; /* or better: absolute_average+=1; (as sum_of_remainders cannot be more than 18 with this if-block in place.) */ sum_of_remainders%=10; } } float average=absolute_average+(sum_of_remainders/10.0F); // or : //float average=((float)absolute_average)+(((float)sum_of_remainders)/10); /* * Alternate Approach * Programming language: Java */ byte[]number=new byte[10]; // initialize number double average=0.0; for(int i = number.length ; --i > -1 ; ) average+=number[i]/10.0;Reason:(x[1]+x[2]+.....+x[n])/n = (x[1]/n)+(x[2]/n)+.....+(x[n]/n)Need for this puzzle:Programming in similar way can be useful in preventing bit-overflow in a computer program. More Explanation: For example, int in Java can store a value between Integer.MIN_VALUE and Integer.MAX_VALUE, and any attempt to assign a number out of those bounds (which can happen during addition, subtraction, or multiplication) will result in bit-overflow and loss of precision. So is with byte(between Byte.MIN_VALUE and Byte.MAX_VALUE), short(between Short.MIN_VALUE and Short.MAX_VALUE), and long(between Long.MIN_VALUE and Long.MAX_VALUE).

## Wednesday, 10 October 2012

### Average Without Overflow

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