原题链接：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
用字符串实现：
#include<iostream> #include<string> #include<cstring> using namespace std; string add(string s1,string s2); int main() { int n,t=1; string s1,s2,sum; cin>>n; while(n--) { cin>>s1>>s2; sum = add(s1,s2); cout << "Case " << t << ":" << endl ; cout << s1 << " + " << s2 << " = " << sum << endl; t++; if(n>0)cout<<endl; } return 0; } string add(string s1,string s2) { if (s1 == "0" && s2 == "0") return "0"; //0的处理 if (s1 == "0") return s2;//0的处理 if (s2 == "0") return s1;//0的处理 string max = s1,min = s2; if (s1.length()<s2.length()) { max = s2; min = s1; } int a = max.length()-1,b = min.length()-1; for(int i=b;i>=0;i--) {max[a--] += min[i] -'0';} for(int i = max.length()-1;i>0;i--) { if(max[i]>'9') { max[i] -= 10; max[i-1]++; } } if(max[0]>'9')//位数相同，最高位大于9 {max[0] -=10; max = '1'+max;} return max; }