// 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.

 // Use hash to achieve O(n) time complexity. Duplicates pairs are skipped.

 #include <cstdio>

 #include <unordered_map>

 #include <vector>

 using namespace std;

 int main()

 {

     vector<int> v;

     unordered_map<int, int> um;

     int n, i;

     int x, y;

     int target;

     while (scanf("%d", &n) ==  && n > ) {

         scanf("%d", &target);

         v.resize(n);

         for (i = ; i < n; ++i) {

             scanf("%d", &v[i]);

         }

         // duplicate pairs are skipped

         for (i = ; i < n; ++i) {

             um[v[i]] = um[v[i]] + ;

         }

         unordered_map<int, int>::iterator it, it2;

         for (it = um.begin(); it != um.end(); ++it) {

             x = it->first;

             y = target - x;

             if (x > y) {

                 continue;

             }

             --it->second;

             if ((it2 = um.find(y)) != um.end() && it2->second > ) {

                 printf("(%d, %d)\n", x, y);

             }

             ++it->second;

         }

         v.clear();

         um.clear();

     }

     return ;

 }

 // 17.12 Given an array of integers and target value, find all pairs in the array that sum up to the target.

 // O(n * log(n) + n) solution.

 #include <algorithm>

 #include <cstdio>

 #include <vector>

 using namespace std;

 int main()

 {

     vector<int> v;

     int n, i;

     int ll, rr;

     int target;

     while (scanf("%d", &n) ==  && n > ) {

         scanf("%d", &target);

         v.resize(n);

         for (i = ; i < n; ++i) {

             scanf("%d", &v[i]);

         }

         sort(v.begin(), v.end());

         ll = ;

         rr = n - ;

         int sum;

         while (ll < rr) {

             sum = v[ll] + v[rr];

             if (sum < target) {

                 while (ll +  < rr && v[ll] == v[ll + ]) {

                     ++ll;

                 }

                 ++ll;

             } else if (sum > target) {

                 while (rr -  > ll && v[rr] == v[rr - ]) {

                     --rr;

                 }

                 --rr;

             } else {

                 printf("(%d, %d)\n", v[ll], v[rr]);

                 while (ll +  < rr && v[ll] == v[ll + ]) {

                     ++ll;

                 }

                 ++ll;

             }

         }

         v.clear();

     }

     return ;

 }