编程实现求a+b的和
Problem Description
Now let’s calculate the answer of a + b ~
Input
The input will consist of a set of pairs of integers for a and b(-10^1000 <= a, b <= 10^1000). The input is ended by EOF.
Output
For each test case you should output the answer of a + b.
Sample Input
1 1
1 -1
Sample Output
2
0

大家可以讨论下，要求执行时间限制为1s，内存尽可能小，

[解决办法]
MS就是一个大数相加吧？

C/C++ code

#include <string>#include <iostream>#include <algorithm>using namespace std;string PLUS(string number1,string number2) {    int i;    int length1 = number1.size();    int length2 = number2.size();    string result="";    reverse(number1.begin(), number1.end());    reverse(number2.begin(), number2.end());    for(i = 0; i < length1 && i < length2; i++) {        char c = (char)(number1[i] + number2[i] - 48);        result = result + c;    }    while(i < length1) {        result = result + number1[i];        ++i;    }    while(i < length2) {        result = result + number2[i];        ++i;    }    int carry = 0;    for(i = 0; i < (int)result.size(); ++i) {        int value = result[i] - 48 + carry;        result[i] = (char)(value % 10 + 48);        carry = value / 10;    }    if(carry !=0 ) {        result = result + (char)(carry + 48);    }    for(i = result.size() - 1; i >= 0; i--) {        if(result[i] != '0') break;    }    result = result.substr(0, i + 1);    reverse(result.begin(), result.end());    if(result.length() == 0) result = "0";    return result;}void main(){    string add1,add2;    while(cin>>add1>>add2!=0||add1.size()<1001||add2.size()<1001)    {        cout<<PLUS(add1,add2);    }    }