#include <bits/stdc++.h>
using namespace std;
const int N = 1010;

int a[N], b[N];
int al, bl;

void add(int a[], int &al, int b[], int &bl) {
int t = 0;
al = max(al, bl);
for (int i = 0; i < al; i++) {
t += a[i] + b[i];
a[i] = t % 10;
t /= 10;
}
if (t) a[al++] = 1;
if(al > 1 && a[al - 1] == 0) al--;
}//到底是改用while还是if有待考究

int main() {
int T;
cin >> T;
for (int k = 1; k <= T; k++) {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
al = 0;
bl = 0;

string x, y;
cin >> x >> y;

for (int i = x.size() - 1; i >= 0; i--) a[al++] = x[i] - '0';
for (int i = y.size() - 1; i >= 0; i--) b[bl++] = y[i] - '0';

// 执行加法
add(a, al, b, bl);
cout << "Case " << k << ":" << endl;
cout << x << " + " << y << " = ";
for (int i = al - 1; i >= 0; i--) cout << a[i];
cout << endl;
if (k < T) cout << endl;

}
return 0;
}

有大佬用一个数组就可以存储，用空间换时间，但是我比较笨，不会简便。