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- #include <iostream>
- #include <cstdlib>
- #include <string>
- /*int n;
- int twon = 2 * n;
- int * a;
- int * b;
- int * sums = malloc(2 * sizeof(b));
- int pos = 0;
- for (int i = 0; i < twon; i++)
- {
- for (int j = (i + 1); j < twon; j++, pos++)
- {
- sums[pos] = b[i] + b[j];
- }
- }
- sort c;
- for (int i = 0; i < twon; i++)
- {
- for (int j = (i + 1); j < twon; j++, pos++)
- {
- if sums.search(a[i] + a[j]) return true;
- }
- }
- return false;
- ================*/
- int n;
- std::cin >> n; // get the length of the array we are manipulating
- int * a = malloc(n * sizeof(int));
- int * b = malloc(n * sizeof(int));
- for (int i = 0; i < n; ++i)
- {
- // get each line which is 2 ints, put into A and B
- std::string input;
- std::getline(std::cin, input);
- std::stringstream nums(input);
-
- nums >> a[i];
- nums >> b[i];
- }
- /*
- int clen = 2 * n;
- int * C = malloc(clen * sizeof(int)); // to hold the list of possible sums
- int tempcount = 0;
- for (int i = 0; i < n; i++)
- {
- for (int j = (i + 1); j < n; ++j, ++tempcount)
- {
- C[tempcount] = (B[i] + B[j]);
- C[clen - tempcount - 1] = (A[i] + A[j]);
- }
- } // n squared
- sort(C, C + clen); // n^2 log n squared == n^2 log n */
- int clen = (n * (n - 1)) / 2; // total possible sums in an array of length n
- int * c = malloc(clen * sizeof(int));
- int tempcount = 0;
- // Store within c the possible sums inside b
- for (int i = 0; i < n; ++i)
- {
- for (int j = (i + 1); j < n; ++j, ++tempcount)
- {
- c[tempcount] = b[i] + b[j];
- }
- }
- // sort c, with time complexity n squared log (n squared), which reduces to n squared log n
- std::sort(c, c + clen);
- // for reach sum in a, see if it is in c. n squared checks of log (n squared), or n squared log n
- for (int i = 0; i < clen; ++i)
- {
- for (int j = (i + 1); j < n; ++j)
- {
- if (std::binary_search(c, c + clen), (a[i] + a[j]))
- {
- free(a); free (b); free(c);
- cout << 1 << endl;
- return true;
- }
- }
- }
- free(a); free(b); free(c);
- cout << 0 << endl;
- return false;
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