题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 409136 Accepted Submission(s): 79277
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
思路：
采用字符数组储存两个加数，模拟小学的加法竖式计算
注意点：
1.俩个加数长度不等的时候，长度短的加数前面加0，0的个数为二者长度相减的绝对值
2.输出格式问题，只有输出最后一组数据的结果的时候，一个回车，其余都是两个回车
代码如下：
#include<bits/stdc++.h> int main() { int n; scanf("%d",&n); int y=1; while(y<=n) { char a[1005]= {‘0‘},b[1005]= {‘0‘},C[1005],A[1005],B[1005]; getchar(); scanf("%s %s",a,b); int l1=strlen(a); int l2=strlen(b); if(l1>l2) { int k=l1-l2; char d[k]; for(int i=0; i<k; i++) d[i]=‘0‘; d[k]=‘\0‘; B[0]=‘\0‘; strcat(B,d); strcat(B,b); A[0]=‘\0‘; strcat(A,a); } else if(l1<l2) { int k=l2-l1; char d[k]; for(int i=0; i<k; i++) d[i]=‘0‘; d[k]=‘\0‘; A[0]=‘\0‘; strcat(A,d); strcat(A,a); B[0]=‘\0‘; strcat(B,b); } else if(l2==l1) { A[0]=‘\0‘; B[0]=‘\0‘; strcat(A,a); strcat(B,b); } l1=strlen(A); l2=strlen(B); int k=0,cc=0; for(int i=l1-1,j=l2-1; i>=0&&j>=0; i--,j--) { int t=A[i]-‘0‘+B[j]-‘0‘+cc; if(t>=10) { cc=1; t=t-10; } else { cc=0; } C[k]=t+‘0‘; k++; } if(cc==1) { C[k]=‘1‘; C[k+1]=‘\0‘; } else { C[k]=‘\0‘; } int l3=strlen(C); printf("Case %d:\n",y); printf("%s + %s = ",a,b); for(int i=l3-1; i>=0; i--) { printf("%c",C[i]); } printf("\n"); if(y!=n) printf("\n"); y++; } return 0; }
