Two numbers for S

Enter an incremental sorting array and a number S, and search for two numbers in the array. Yes, their sum is exactly S. If there are multiple pairs of numbers and the sum is equal to S, the product of the two numbers is the smallest. Input: each test case contains two rows: the first row contains an integer n and k, and n represents the number of elements in the array, and k represents the sum of the two. Where 1 <= n <= 10 ^ 6, k is the second line of int contains n integers, each array is of the int type. Output: two numbers are output for each test case. A small number is output first. If not, output the "-1-1" sample input: 6 151 2 4 7 11 15 sample output: 4 11 idea: because the order has been sorted, if there are two numbers, it may be a relatively large number, a relatively small number, or a relatively equal number! So let's set a n1 = min (that is, a [0]), n2 = max (that is, a [n-1]), which is actually n1 = a [low], low = 0 start, n2 = a [high], high = n-1 start, then n1 + n2 compare with S (We Need To sum), if>, n2 = a [-- high]; If <, n1 = a [++ low], and if =, use mul to save the product, and save two numbers, because there may be more than one pair of numbers, as long as the product is small! Scan until low> = high ends! Code AC: [cpp] # include <stdio. h> # include <stdlib. h> int main () {long int n, I, low, high; long int mul; long int s, * data, m1, m2, flag; while (scanf ("% ld", & n, & s )! = EOF) {data = (long int *) malloc (sizeof (long int) * n); for (I = 0; I <n; I ++) // {scanf ("% ld", & data [I]);} m1 =-1; m2 =-1; low = 0; high = n-1; mul =-1; flag = 1; while (low