#include #include #include /*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;