c++实现大数相加，hdu1002解析

题目描述：

Problem DescriptionI 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 Input21 2112233445566778899 998877665544332211 Sample OutputCase 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

#include 
    
     using namespace std;string CalculateTwoBigNum(string a,string b){    int i,j,k;    int a_Length = a.length();    int b_Length = b.length();    string BigerNum = a; //默认两个数字中更大为a    string SmallerNum = b;    if(a_Length<=b_Length){        BigerNum = b;        SmallerNum = a;    }    string result = BigerNum;    for(i=BigerNum.length()-1,j=SmallerNum.length()-1;j>=0;i--,j--){        result[i] = BigerNum[i] + SmallerNum[j]-'0';    }    for(k=BigerNum.length()-1;k>=0;k--){        if(result[k]>'9'){            result[k] = result[k]-10;            if(k-1>=0){                result[k-1] =result[k-1]+1;            }            else                result = "1" + result;        }    }    return result;}int main(){    int Frequence; //计算次数    string a,b; //参与计算的两个数    string result; //计算结果    cin>>Frequence;    int Flag = 1; // case Flag    while (Frequence--) {        cin>>a>>b;        result = CalculateTwoBigNum(a,b);        printf("Case %d:\n",Flag++);        cout<
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