140730summer training .txt

1. addition and subtraction of large numbers
Train of Thought Analysis:
1. Use data as a string input (gets (s ))
2. Convert the memory type to an integer type and store it in reverse order.
Char? Int I = 0, j = len-1, int [I ++] = char [j --]
3. Starting from the first place,
If sum> 9, int [I] = sum % 10, int [I + 1] + = sum/10;
4. Input
1> judge whether int [Len] is 0. If yes, Skip. If not, output.
2> output int [(len-1) --];
Exercise questions hdoj1002
# Include <stdio. h>
# Include <string. h>
Int main ()
{
Int T;
Char A [1100], B [1100], Beitai [1100];
Int C [1100], d [1100], sum [1100];
Int I, j, BT, k = 1;
Int Lena, lenb;
Scanf ("% d", & T );
Getchar ();
While (t --)
{
Scanf ("% S % s", a, B );
Getchar ();
Lena = strlen ();
Lenb = strlen (B );
Memset (C, 0, sizeof (c ));
Memset (D, 0, sizeof (d ));
Memset (sum, 0, sizeof (SUM ));
For (I = 0, j = lena-1; I <Lena; I ++, j --)
C [J] = A [I]-'0 ';
For (I = 0, j = lenb-1; I <lenb; I ++, j --)
D [J] = B [I]-'0 ';
If (Lena <lenb)
{
Bt = Lena;
Lena = lenb;
Lenb = Bt;
}
For (I = 0; I <= Lena; I ++)
{
Sum [I] = C [I] + d [I] + sum [I];
If (sum [I]/10> 0)
{
Sum [I + 1] = sum [I]/10;
Sum [I] % = 10;
}
}
Printf ("case % d: \ n", k );
K ++;
Printf ("% S + % s =", a, B );
If (sum [Lena]! = 0)
Printf ("% d", sum [Lena]);
For (I = lena-1; I> = 0; I --)
Printf ("% d", sum [I]);
Printf ("\ n ");
If (T! = 0)
Printf ("\ n ");
}
Return 0;
}