链接：http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Problem Description
I have a very simple problem for you. Given two
integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer
T(1<=T<=20) which means the number of test cases. Then T lines follow,
each line consists of two positive integers, A and B. Notice that the integers
are very large, that means you should not process them by using 32-bit integer.
You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The
first line is "Case #:", # means the number of the test case. The second line is
the an equation "A + B = Sum", Sum means the result of A + B. Note there are
some spaces int the equation. Output a blank line between two test
cases.
Sample Input
2 1 2 112233445566778899 998877665544332211
Sample Output
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
这也是一个大数加法的模板
1 #include<stdio.h>
2 #include<string.h>
3 #define Max 10000
4 char sun[Max],str1[Max],str2[Max];
5 void jia( )
6 {
7 memset( sun,0,sizeof( sun ) );
8 int len1 = strlen( str1 ),len2 = strlen( str2 );
9 for( int i = 0; i < ( len1 > len2 ? len1 : len2 ); ++i )//转换成ASCII码
10 str1[i] -= '0',str2[i] -= '0';
11 for( int p = 0, q = len1 - 1; p < q; ++p , --q ) //逆序
12 {
13 char c = str1[q];
14 str1[q] = str1[p];
15 str1[p] = c;
16 }
17 for( int p = 0, q = len2 - 1; p < q; ++p,--q)
18 {
19 char c = str2[q];
20 str2[q] = str2[p];
21 str2[p] = c;
22 }
23 for( int i = 0,c = 0; i < ( len1 > len2 ? len1 : len2 ) || c; ++i )//相加 进位
24 {
25 if( i < len1 )
26 c += str1[i];
27 if( i < len2 )
28 c += str2[i];
29 sun[i] = c % 10;//不同的地方1
30 c /= 10;//不同的地方2
31 }
32 }
33 int main(){
34 int T;
35 scanf( "%d",&T );
36 int m=T;
37 while( T-- ){
38 scanf( "%s%s",str1,str2 );
39 char s1[Max],s2[Max];
40 strcpy( s1,str1 );
41 strcpy( s2,str2 );
42 jia( );
43 printf( "Case %d:\n",m-T );
44 printf( "%s + %s = ",s1,s2 );
45 int n=Max;
46 while( !sun[--n] );
47 for( ; n >= 0; --n )
48 printf( "%d",sun[n]);//不同的地方3
49 puts( "" );
50 if(T>0)
51 puts( "" );
52 }
53 }