菜鸟上路，杭电OJ1002之大数相加

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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

21 2112233445566778899 998877665544332211

Sample Output

Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110

刚开始想用数组做，后来想到了栈，瞬间简化了不少，数据结构的正确选择还是很重要的！！

我的代码：

#include <iostream>#include <stack>#include <string>using namespace std;int main(){    string s1,s2;    int n;    int i=0;int j=1;    cin>>n;    stack<char> A;    stack<char> B;    stack<char> C;    int c=0;    int num=0;    while(n--)    {c=0;i=0;        cin>>s1;        cin>>s2;while(!A.empty())A.pop();while(!B.empty())B.pop();if(j!=1)cout<<endl;        while(s1[i]!='\0')            A.push(s1[i++]);        i=0;        while(s2[i]!='\0')            B.push(s2[i++]);        while(!A.empty()||!B.empty())        {            if(!A.empty()&&B.empty())            {               num=A.top()-'0'+c;               A.pop();            }            else if(A.empty()&&!B.empty())            {                num=B.top()-'0'+c;                B.pop();            }            else if(!A.empty()&&!B.empty())            {                num=A.top()-'0'+B.top()-'0'+c;                A.pop();                B.pop();            }            if(num >=10)            {                c=1;                C.push(num%10+'0');            }            else            {                c=0;                C.push(num+'0');            }                   }/*if(C.top()=='0')C.push('1');*/cout<<"Case "<<j++<<":"<<endl;cout<<s1<<" + "<<s2<<" = "; while(!C.empty())         {             cout<<C.top();             C.pop();          } cout<<endl;    }}

注意：题目要求输出的两个结果之间应当有一个空行，但是这并不代表我们就可以在代码的最后写cout<<endl<<endl;这样的表述，这会导致PE错误！

正确的做法是：

if(j!=1)cout<<endl;

再在代码最后来一个endl就可以了

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