杭电hdu上的链接http://acm.hdu.edu.cn/showproblem.php?pid=1002
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
解题思路：刚开始做这道题的时候，以为是A+B的题也难不到哪去，谁知道改了好几个小时。开始试了int，不行，又改long，不行，然后改long long，还是不行。只能另外想一种方法了：用大位数加法，用数组装数字，然后从个位开始逐位相加，过十进一。开始的时候想肯定是用数组了，肯定不会是字符数组。但是做着做着发现不行，数组的长度无法确定，想起来字符数组有个strlen的用法，就改用字符数组了。
以下是具体代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define N 1005 5 using namespace std; 6 7 int count = 0; 8 int a[N]; 9 int b[N]; 10 int he[N]; 11 char sza[N]; 12 char szb[N]; 13 14 void calcuation() { 15 int flag = 1; 16 int len1,len2; 17 len1 = strlen(sza); 18 len2 = strlen(szb); 19 int i,j,k,l; 20 i = 0; 21 while(sza[i] != ' ') { 22 a[i] = sza[i] - '0'; 23 i++; 24 } 25 i = 0; 26 while(szb[i] != ' ') { 27 b[i] = szb[i] - '0'; 28 i++; 29 } 30 if(len1 < len2) {l = len1-1;flag = 0;} 31 if(len1 > len2) {l = len2-1;flag = -1;} 32 if(len1 == len2) l = len1-1; 33 k = 0; 34 int num = 0; 35 int m,n; 36 m=len1-1; 37 n=len2-1; 38 while(l >= 0) { 39 he[k] += a[m--] + b[n--]; 40 41 if(he[k] >= 10) { 42 num = he[k] / 10; 43 he[k] = he[k] % 10; 44 he[k+1] = num; 45 num = 0; 46 } 47 k++; 48 l--; 49 } 50 i = k; 51 j = k; 52 if(flag == 0) {//len1 < len2 53 for(k = len2-1-i;k >= 0; k--) { 54 he[j] += b[k]; 55 j++; 56 } 57 } 58 if(flag == -1) {//len1 > len2 59 for(k = len1-1-i;k >= 0; --k) { 60 he[j] += a[k]; 61 j++; 62 } 63 } 64 } 65 66 void find_print() { 67 int loc; 68 calcuation(); 69 printf("Case %d: ",count); 70 // cout << "Case";//此处考虑到用cout输出需要的行数多，故使用printf输出 71 // cout << count; 72 // cout << ": "; 73 int i; 74 i = 0; 75 while(sza[i]!=' ') { 76 printf("%c",sza[i]); 77 i++; 78 } 79 cout << " + "; 80 i = 0; 81 while(szb[i]!=' ') { 82 printf("%c",szb[i]); 83 i++; 84 } 85 cout << " = "; 86 for(i = N; i >= 0; i--) { 87 if(he[i] != 0) { 88 loc = i; 89 break; 90 } 91 } 92 for(i = loc;i >= 0; i--) { 93 printf("%d",he[i]); 94 } 95 cout << endl; 96 } 97 98 int main() { 99 int n; 100 cin >> n; 101 while(n--) { 102 count++; 103 memset(he,0,sizeof(he)); 104 memset(a,0,sizeof(a)); 105 memset(b,0,sizeof(b)); 106 cin >> sza >> szb; 107 find_print(); 108 if(n != 0)///根据题意最后用换行 109 cout << endl; 110 } 111 return 0; 112 }
方法仅供参考，我知道肯定不是最简单的方法，有时间会再改善的，谢谢码友支持，欢迎评论。