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Someone's coursework?! This function is for an array of integers, the length of the array is Count.
int MaxInt(NumArray as *int, Count as int)
{
int Inst;
int Result;
Result = NumArray[0];
for (Inst=0;Inst
{
if(NumArray[Inst] > Result)
{
Result = NumArray[Inst];
}
}
return Result;
}
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Write an Algorithm for multiplication table of an integer?
int firstNumber,secondNumber for(firstNumber = min; firstNumber <= max; firstNumber++); { for(secondNumber = min; secondNumber <=max; secondNumber++); int result firstNumber * secondNumber; }
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** pseudo code ** if array length == 1, return first element else if array length > 1 max = first element for second item to last item if item > max max = item return max else // array length is 0 error
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To find the largest of three numbers you must first find the largest of two numbers: int max (int a, int b) { return a>b?a:b; // or, equivalently: if (a>b) return a; else return b; } Now we can use this function to find the maximum of three: int max3 (int a, int b, int c) { return max (max (a,b), c); }
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Max = 0For K = 1 to NIf Number(K) > Max then Max = Number(K)Next KPrint Max
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To find the largest of three numbers, first find the largest of two numbers: int max (int x, int y) { return x<y?y:x; } Now you can use this one function to find the largest of three numbers: int max (int x, int y, int z) { return max (max (x, y), z); }
Write a c program to find the sum of numbers between given limits?
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int max (int a, int b) { return a>b?a:b; }
What is the algorithm for average of three numbers?
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You can find out what your vo2 max is in many different places. There are various websites dedicated to finding these numbers for a more effective workout.
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Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }Assuming the numbers are integers, and they are stored in an array called myArray. The following has not been tested, but gives the general idea: int max; max = myArray[0]; for (int i = 1; i < myArray.length(); i++) } if (myArray[i] > max) max = myArray[i]; System.out.println(max); }
