思想： 运用字符串和整数的转换关系，然后模拟手写运算从后往前依次计算；详细参考【 大数加减乘除】
AC代码；
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#include <sstream>
#include <cctype>
#include <set>
#include <string>
#include <stack>
#include <vector>
using namespace std;
const int N = 1050;
typedef long long LL;
const int inf = 0x3f3f3f3f;
LL numA[N];
LL numB[N];
void resetnumA(string numAStr)
{
memset(numA,0,N*sizeof(LL));
for(int i=0;i<numAStr.length();i++)
{
numA[i]=numAStr[numAStr.length()-i-1]-'0';
}
}
void resetnumB(string numBStr)
{
memset(numB,0,N*sizeof(LL));
for(int i=0;i <numBStr.length();i++)
{
numB[i]=numBStr[numBStr.length()-i-1]-'0';
}
}
string getnumString(LL *num)
{
string numString;
int f=0;
for(int i=N-1;i>=0;i--)
{
if(num[i]!=0)
f=1;
if(f)
numString+=num[i]+'0';
}
return numString;
}
string plusn(string numAStr,string numBStr)
{
resetnumA(numAStr);
resetnumB(numBStr);
for(int i=0;i<N;i++)
{
numA[i]=numA[i]+numB[i];
if(numA[i]>9)
{
numA[i+1]++;
numA[i]-=10;
}
}
return getnumString(numA);
}
int main()
{
string numAStr,numBStr;
int t,y=1;
cin>>t;
while(t--)
{
cin>>numAStr>>numBStr;
cout<<"Case "<<y++<<":"<<endl;
cout<<numAStr<<" + "<<numBStr<<" = "<<plusn(numAStr,numBStr)<<endl;
if(t) cout<<endl;
}
return 0;
}
