Acm (UVa) : 713

Adding Reversed Number

Data type:
Character / String

// Critical Case:
After reversing a number, if there are preceding 0
Let, a number: 85210000, Reversed: 1258
After adding two number, if there r preceding 0 before result
Let, result = 0001, so the output will be: 1

// Technique:
This problem can be solved by without reversing anything
Start addition form Left most digit and move to right and pass the carry to the right.
Still u have to be very careful abt the critical cases.

Critical Input:
4
0005 36
9999 1111
1000 0001
963 369

Critical Output:
3605
1111
1001
2331

// Uva online Judge: ACM problem no: 713
// Author: Tausiq >>> CTE, UIU >>> Bangladesh.

#include <stdio.h>
#include <string.h>

int main ()
{
	long N, Len, I, LenA, LenB, Carry, Sum[210], X;
	char A [210], B [210], Temp [210];

	scanf("%ld", &N);

	while ( N ){

		// Initialize with 0
		for ( I = 0; I < 210; I++)
			A [I] = B [I] = 48;
		
		// Trim the former 0's of 1st number
		scanf("%s", Temp);
		Len = strlen ( Temp );

		while ( Temp [ Len-1 ] == '0' )
			Len--;

		LenA = 0;
		for ( I = 0; I < Len; I++)
			A [LenA++] = Temp [I];

		// Similarly from 2nd number
		scanf("%s", Temp);
		Len = strlen ( Temp );

		while ( Temp [ Len-1  ]== '0')
			Len--;

		LenB = 0;
		for ( I = 0; I < Len; I++)
			B [LenB++] = Temp [I];

		Len = LenA > LenB ? LenA : LenB; // Highest length

		// Addition
		Carry = X = 0;
		for ( I = 0; I < Len; I++){
			Carry = ( A [I] - 48 ) + ( B [I] - 48 ) + Carry;
			Sum [X++] = Carry % 10;
			//printf("%d", Carry%10);
			Carry /= 10;
		}
		while ( Carry != 0 ){ // if Carry is carrying sth
			Sum [X++] = Carry % 10;
			//printf("%d", Carry%10);
			Carry /= 10;
		}

		// cross out precedence 0
		I = 0;
		while ( Sum [I] == 0)
			I++;

		// output
		while ( I < X )
			printf("%d", Sum[I++]);
		printf("\n");

		N--;
	}
	return 0;
}