Week 一 # A A + B Problem II
原题描述：
A - A + B Problem II
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
InputThe 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.
OutputFor 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
第二次看到这个题目了，第一次是在杭电acm上，那时侯想了用数组，不过没做出来。其实这个题目不算难，我直接定义了两个个字符型数组用来存储算数，定义一个整形数组存储值。先判断那个数组长度大.
两个字符型数组从后面相加，存储到整形数组上，如果值大于10再向前进1（不可能大于20）一直到短的那个数组为0的时候停止相加。直接把长的赋值到上面就可以了，记得进位！
用长的字符型数组长度判断整形数组的长度。然后就可以输出了。代码复杂，但是浅显易懂。
AC代码：
1 #include <iostream>
2 #include <string.h>
3 const int N=1000;
4 using namespace std;
5 int main()
6 {
7 int t,l,m,n,sum[N+1]={0};
8 int g=1;
9 char a[N],b[N];
10 cin>>t;
11 while(t--)
12 {
13 int i;
14 cin>>a>>b;
15 m=strlen(a);
16 n=strlen(b);
17 l=m>=n?m:n;
18 if(m<=N&&n<=N)
19 {
20 if(m>=n)
21 {
22 for(i=0;n>0;i++)
23 {
24 sum[i]=sum[i]+(a[m-1]-'0')+(b[n-1]-'0');
25 m--;
26 n--;
27 if(sum[i]>9)
28 {
29 sum[i]-=10;
30 sum[i+1]++;
31 }
32 }
33 for( ;m>0;i++,m--)
34 sum[i]=sum[i]+(a[m-1]-'0');
35 }
36 else
37 {
38 for(i=0;m>0;i++)
39 {
40 sum[i]=sum[i]+(a[m-1]-'0')+(b[n-1]-'0');
41 m--;
42 n--;
43 if(sum[i]>9)
44 {
45 sum[i]-=10;
46 sum[i+1]++;
47 }
48 }
49 for( ;n>0;i++,n--)
50 sum[i]=sum[i]+(b[n-1]-'0');
51 }
52 }
53 if(sum[i]==0)
54 i--;
55 cout<<"Case "<<g<<":"<<endl;
56 g++;
57 cout<<a<<" + "<<b<<" = ";
58 for( ;i>=0;i--)
59 cout<<sum[i];
60 for(i=0;i<=l;i++)
61 sum[i]=0;
62 if(t!=0)
63 cout<<endl<<endl;
64 else cout<<endl;
65 }
66 return 0;
67 }