0 + 0 = 0 1 + 0 = 1 0 + 1 = 1 1 + 1 = 10 1 + 1 + 1 = 11
Just as with decimal addition, when the sum in one column is a two-bit (two-digit) number, the least significant figure is written as part of the total sum and the most significant figure is “carried” to the next left column. Consider the following examples:
11 1 --- Carry bits ----- 11 1001101 1001001 1000111 + 0010010 + 0011001 + 0010110 --------- --------- --------- 1011111 1100010 1011101
The addition problem on the left did not require any bits to be carried, since the sum of bits in each column was either 1 or 0, not 10 or 11. In the other two problems, there definitely were bits to be carried, but the process of addition is still quite simple.
INPUT
The first line of input contains an integer N , (1<=N<=1000) , which is the number of binary addition problems that follow. Each problem appears on a single line containing two binary values separated by a single space character. The maximum length of each binary value is 80 bits (binary digits). Note: The maximum length result could be 81 bits (binary digits).
OUTPUT
For each binary addition problem, print the problem number, a space, and the binary result of the addition. Extra leading zeroes must be omitted.
Sample Input
3
1001101 10010
1001001 11001
1000111 1010110
Sample Output
1 1011111
2 1100010
3 10011101
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 | import java.util.*; import java.lang.*; import java.math.*; public class Q2{ public static void main(String[] args){ Scanner scan = new Scanner(System.in); String line = System.getProperty("line.separator"); scan.useDelimiter(line); int cases = scan.nextInt(); String [] ans = new String[cases]; for(int x=0;x<cases;x++){ String[] get = scan.next().split(" "); long val1 = Long.parseLong(get[0],2); long val2 = Long.parseLong(get[1],2); long sum = val1+val2; ans[x] = ""+Long.toBinaryString(sum); } for(int x=0;x<cases;x++){ System.out.println((x+1)+" "+ans[x]); } } } |
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