思路：题目已经告诉另一个数是原始数据(>=11)，那么意味着我们需要得到的是剩下n-1个数的和再对11取模。需要注意当n==2时，就不需要取模了，不然就只能通过90%的测试点。
代码实现：
#include<bits/stdc++.h>
// #define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {
{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
inline void read(int &x){
char t=getchar();
while(!isdigit(t)) t=getchar();
for(x=t^48,t=getchar();isdigit(t);t=getchar()) x=x*10+(t^48);
}
int n, m;
signed main()
{
IOS;
while(cin >> n >> m){
int sum = 0, flag = 0;
for(int i = 1; i <= n; i ++){
int x; cin >> x;
sum += x;
}
cout << (n==2? sum-m:(sum-m)%11) << endl;
}
return 0;
}