这一段时间一直都在刷OJ，这里建一个博客合集，用以记录和分享算法学习的进程。
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
AC CODE
//这一版AC代码，由于题目对于给定的输入输出太多限定，所以代码有很多冗余的部分，
//核心思想就是使用堆栈的方法，使用三个堆栈实现了大数的加法
#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;
int count = ;//声明一个count全局变量，用作进位计算
int main()
{
int i, n;
vector<string> vec1;
vector<string> vec2;
string str1, str2;
stack<char> stk1;
stack<char> stk2;
stack<int> rstk;
cin >> n;
//完成大数字符串的输入
for(i = ; i < n; ++i)
{
cin >> str1 >> str2;
vec1.push_back(str1);
vec2.push_back(str2);
}
for(i = ; i < n; ++i)
{
//将数字以char字符形式读入并push进堆栈中
for(const auto x : vec1[i])
{
stk1.push(x);
}
for(const auto y : vec2[i])
{
stk2.push(y);
}
//取堆栈元素进行加法运算
while(!stk1.empty() && !stk2.empty())
{
int temp = stk1.top() - '' + stk2.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk1.pop();
stk2.pop();
}
//进行多余高位的push
while(!stk1.empty())
{
int temp = stk1.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk1.pop();
}
while(!stk2.empty())
{
int temp = stk2.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk2.pop();
}
while(count == )
{
rstk.push();
count = ;
}
cout << "Case " << i+ << ":" << endl;
cout << vec1[i] << " " << "+" << " " << vec2[i] << " = ";
//最终结果打印出来即可
while(!rstk.empty())
{
cout << rstk.top();
rstk.pop();
}
cout << endl;
while(i < n-)
{
cout << endl;
break;
}
}
return ;
}
精简代码
/*这一版是简化版的大数加法代码
*去除了一部分输入输出规则，以核心代码为主
*/
#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;
int count = ;//声明一个count全局变量，用作进位计算
int main()
{
int i = ;
string str1, str2;
stack<char> stk1;
stack<char> stk2;
stack<int> rstk;
while(cin >> str1 >> str2)
{
//将数字以char字符形式读入并push进堆栈中
for(const auto x : str1)
{
stk1.push(x);
}
for(const auto y : str2)
{
stk2.push(y);
} //取堆栈元素进行加法运算
while(!stk1.empty() && !stk2.empty())
{
int temp = stk1.top() - '' + stk2.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk1.pop();
stk2.pop();
}
//进行多余高位的push
while(!stk1.empty())
{
int temp = stk1.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk1.pop();
}
while(!stk2.empty())
{
int temp = stk2.top() - '' + count;
if(temp > )
{
temp %= ;
count = ;
}
else
{
count = ;
}
rstk.push(temp);
stk2.pop();
}
while(count == )
{
rstk.push();
count = ;
}
cout << "Case " << i+ << ":" << endl;
cout << str1 << " " << "+" << " " << str2 << " = ";
//最终结果打印出来即可
while(!rstk.empty())
{
cout << rstk.top();
rstk.pop();
}
cout << endl << endl;
++i;
}
return ;
}
PS：网上OJ的输入输出的形式规范实在是太蛋疼了！！！