Poj 1001 exponentiation High Precision

// Use high-precision multiplication for this question

# Include <iostream>

# Include <cstring>

Using namespace STD;

# Define x 140

Void multiply (INT num [], int A [], int m) // high-precision Multiplication

{

Int s [x];

Memset (S, 0, sizeof (s ));

Int I, J;

For (j = 0; j <m; j ++)

For (I = 0; I <126; I ++)

S [I + J] + = num [I] * A [J]; // CoreAlgorithmSave the multiplication result to the array s [].

 

Int carry = 0; // table carry

For (I = 0; I <X; I ++)

{

Num [I] = (carry + s [I]) % 10;

Carry = (carry + s [I])/10;

}

}

Int main ()

{

Freopen ("sum. In", "r", stdin );

Freopen ("sum. Out", "W", stdout );

Int N, I;

Char R [7];

While (CIN> r> N)

{

Int num [x];

For (I = 0; I <X; I ++)

Num [I] = 0;

 

Int A [7];

Int dot = 0; // records the decimal places

Int m = 0;

For (I = 5; I> = 0; I --) // converts a character array to an int array.

{

If (R [I]! = '.')

{

A [m] = R [I]-'0 ';

Num [M ++] = R [I]-'0 ';

}

Else

Dot = I; // record the decimal place

}

 

Int q = 0; // The number of zeros in the integer part of a record such as 20.000

For (I = dot-1; I> = 0; I --) // calculate from the first digit of the decimal point.

{

If (R [I] = '0 ')

Q ++;

Else

Break;

}

For (I = 5; I> = 0; I --)

If (R [I]! = '0 ')

Break;

Dot = I-dot;

Int Len = dot * n; // The number of digits in the fractional part of the entire number after calculation

 

For (I = 1; I <n; I ++) // because a number has been saved to num [], you only need to calculate n-1 times.

Multiply (Num, a, m); // use high-precision Multiplication

 

For (I = X-1; I> = 0; I --) // remove the rear zero

If (Num [I]! = 0)

Break;

Int rear = I; // zero position after record

 

For (I = 0; I <= rear; I ++) // remove the front zero

If (Num [I]! = 0)

Break;

Int front = I; // records the front zero position

 

If (rear-front + 1 <Len) // when the 0.0123 ^ 3 =. 000001860867 type, the zero before the decimal point has been removed, you need to add

{

Cout <"."; // 0 is not required for the 0.12 class

 

For (I = len-Rear + front-1; I> 0; I --) // Add the zero difference after the decimal point

Cout <0;

 

For (I = rear; I> = front; I --)

Cout <num [I];

 

Cout <Endl;

}

 

/* Else if (rear-front + 1 = Len) Thought About It. Can this be avoided,

This is because only when the index is set to a moment, but it will not enter this cycle. after submitting the index again, we will find that it is correct.

{

Cout <".";

For (I = rear; I> = front; I --)

Cout <num [I];

Cout <Endl;

}

*/

Else // The decimal part does not need to be filled with zero

{

For (I = rear; I> = front; I --)

{

Cout <num [I];

If (I-front = Len & Len) // If the decimal point is reached, enter.

{

Cout <".";

}

Else if (! Len & I-front = Len)

{// If the input is a 20.00 class, because the zero part of the integer is the same, you must add

For (Int J = 0; j <Q * n; j ++)

Cout <0;

}

}

Cout <Endl;

}

}

 

Return 0;

}