• 问题描述
  • The bored Three-God get another boring question.

    This problem is ask you plus two big nubmer, please help him, when you solve this problem you can

    speak to Three-God,"Oh, Please give me a diffculte one, this one is too easy".

  • 输入

  • There are muti-case
    For each case, there are two integers, n, m (0 < n, m < 10^10000).

  • 输出

  • Calculate two integers' sum

  • 样例输入
  • 1 1
2 3
10000 100000
  • 样例输出
  • 2
5
110000
  • 提示
  • 无
  • 来源
Three-God

参考代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<map>
#include<set>
using namespace std;
#define N 10005
string a,b;
int A[N],B[N],C[N];
void init()
{
    memset(A,0,sizeof(A));
    memset(B,0,sizeof(B));
    memset(C,0,sizeof(C));
    int j=0;
    for(int i=a.size()-1; i>=0; i--)
        A[j++]=a[i]-'0';
    j=0;
    for(int i=b.size()-1; i>=0; i--)
        B[j++]=b[i]-'0';
}
int main()
{
    while(cin>>a>>b)
    {
        init();
        int len=max(a.size(),b.size());
        int dell=0;
        for(int i=0; i<len; i++)
        {
            int temp=A[i]+B[i]+dell;
            dell=temp/10;
            C[i]=temp%10;
        }
        if(dell)
            C[len++]++;
        for(int i=len-1; i>=0; i--)
            cout<<C[i];
        cout<<endl;
    }
    return 0;
}

另外附上其他大数操作：

大数减法：

大数乘法：

大数除法：

大数取余：