2 1 2 112233445566778899 998877665544332211
Case 1: 1 + 2 = 3 Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110
#include<iostream>
#include <string>
#include<algorithm>
using std::endl;
using std::cin;
using std::cout;
using std::string;
int main()
{
#ifdef LOCAL
freopen("input.txt" , "r" , stdin);
freopen("output.txt" , "w" , stdout);
#endif
int T;
cin >> T;
string a , b , sum;
for(int cases = 1; cases<=T; ++cases)
{
sum.clear();
cin >> a >> b;
cout << "Case " << cases << ":" << endl;
cout << a <<" + "<<b << " = ";
int c = 0 ;
reverse(a.begin() , a.end());
reverse(b.begin() , b.end());
int alength = a.length();
int blength = b.length();
int length = alength;
//使两个字符串的长度相等,补齐零
if(alength < blength)
{
for(int i=alength; i<blength; ++i)
{
a += '0';
}
length = blength;
}
if(blength < alength)
{
for(int i=blength; i<alength; ++i)
{
b +='0';
}
length = alength;
}
//进行加法计算,c为进位
for(int i=0; i<length; ++i)
{
int temp = (a[i] - '0') + (b[i] - '0') +c;
c = temp/10;
sum += (temp%10 + '0');
}
//最高位的进位如果不为0的话则进位
if(c!=0)
{
sum += (c +'0');
}
//结果转置
reverse(sum.begin() , sum.end());
cout << sum << endl;
//控制最后的那组数据不再输出空行
if(cases != T)
{
cout << endl;
}
}
}HDU1002--A + B Problem II,布布扣,bubuko.com
原文:http://blog.csdn.net/computer_liuyun/article/details/29221715