satisfactory pairs hackerrank-pairs-again--pretreatment + indefinite array application __ Elementary Number Theory

Do you:
1: Input n, ask (A, b) to meet a < b and exist x, y so that x*a + y*b = N of the two-tuple number, require N, a, b, X, y are all positive integers
2 method: First preprocessing less than N of the number of divisors, and then the violent temptation + pruning, the test of the number of two groups, the time complexity 1/x from 1 to n integral, for NLN (n)

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The following is the accepted code

#include <bits/stdc++.h> using namespace std;

const int N = 3e5 + 4;
Vector <int> Y_b[n];

Vector <int>:: iterator B;

int book[n];
    int main () {int n, I, J, a, X, T, ans;
    scanf ("%d", &n); for (i = 1; I <= n; i++) {for (j = 1; j*j <= i; j +) {if (i%j = 0) {Y_b[i].push_ba
                CK (j);
                    if (I*i!= j) {t = i/j;
                Y_b[i].push_back (t);
    Sort (Y_b[i].begin (), Y_b[i].end (), greater<int> ());
    Ans = 0;
            for (a = 1; a < n; a++) {for (x = 1; x*a < n x + +) {t = n-x*a;
                for (b = Y_b[t].begin (), B!= y_b[t].end (); b++) {if ((*b) <= a) break;
                    if (Book[*b]!= a) {ans++;
                BOOK[*B] = A;
    }} printf ("%d\n", ans);
return 0;
 }

The following is the recommended reference code

#include <bits/stdc++.h> using namespace std;

const int N = 3e5 + 4;

Vector <int> y_b[n];/* Records the divisors of the current element/vector <int>:: iterator B;

int book[n];/* real-type global variable initialization defaults to 0*/BOOL CMP (int a, int b) {return a > B;}
    int main () {int n, I, J, a, X, T, ans;
    scanf ("%d", &n); for (i = 1; I <= n; i++) {for (j = 1; j*j <= i; j +) {if (i%j = 0) {Y_b[i].push_ba
                CK (j);
                    if (I*i!= j) {t = i/j;
                Y_b[i].push_back (t);
    Sort (Y_b[i].begin (), Y_b[i].end (), greater<int> ());
    Ans = 0;
            for (a = 1; a < n; a++) {for (x = 1; x*a < n x + +) {t = n-x*a; for (b = Y_b[t].begin (); b!= Y_b[t].end (); b++) {if (*b) <= a) *: A < b*/break
                ;
                    if (Book[*b]!= a) {/* Mark heavy */ans++;
    BOOK[*B] = A;            }} printf ("%d\n", ans);
return 0;
 }