这道题十分有十二分之困难(easy)

算法一：直接算

#include<iostream>

using namespace std;

int main(){
    int a , b;
    cin >> a >> b;
    cout << a + b;
    return 0;
}

算法二：高精度（详见算法基础课）

#include <iostream>
#include <vector>

using namespace std;

const int N = 10e6 + 10;

vector<int> add(vector<int> &A, vector<int> &B)  // C = A + B, A >= 0, B >= 0
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(t);
    return C;
}

int main()
{
    string a , b;
    vector<int> A , B;
    cin >> a >> b ;
    for (int i = a.size() - 1 ; i >= 0 ; i --) A.push_back(a[i] - '0');
    for (int i = b.size() - 1 ; i >= 0 ; i --) B.push_back(b[i] - '0');
    auto C = add(A , B);
    for (int i = C.size() - 1 ; i >= 0 ; i --) cout << C[i];
    return 0;
}

如果你不会用vector还可以这么做：

#include <iostream>

using namespace std;

const int N = 10e6 +10;
int A[N] , B[N] , C[N];

int Add(int a[],  int b[] , int c[] , int cnt){
    int t = 0 ;
    for (int i = 1 ; i <= cnt ; i ++){
        t += a[i] + b[i];
        c[i] = t % 10;
        t = t / 10;
    }
        if (t) c[++cnt] = 1;
        return cnt;
}

int main()
{
    string a , b;
    cin >> a >> b;

    int cnt1 = 0 , cnt2;
    for (int i = a.size() - 1 ; i >= 0 ; i --) A[++cnt1] = a[i] - '0';
    for (int i = b.size() - 1 ; i >= 0 ; i --) B[++cnt2] = b[i] - '0';

    int tot = Add(A , B , C, max(cnt1 , cnt2));

    for (int i = tot ; i >= 1 ; i --) cout << C[i];
    return 0;
}