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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long int
  5. const int MOD = 1000000007;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX / 2;
  8. const int MAXN = 100000;
  9. int primes[1000000];
  10.  
  11. /*void seive() {
  12.   fill(primes, primes + 1000000, 1);
  13.   primes[0] = primes[1] = 0;
  14.   for (int i = 2; i * i < 1000000; i++) {
  15.   if (primes[i]) {
  16.   for (int j = i * i; j < 1000000; j += i) {
  17.   primes[j] = 0;
  18.   }
  19.   }
  20.   }
  21. }
  22.  
  23. bool isPrime(int n) {
  24.   if (n <= 1) return false;
  25.   for (int i = 2; i * i <= n; i++) {
  26.   if (n % i == 0) return false;
  27.   }
  28.   return true;
  29. }
  30.  
  31. int gcd(int a, int b) {
  32.   if (a == 0) return b;
  33.   return gcd(b % a, a);
  34. }*/
  35.  
  36. int power(int a, int b, int mod) {
  37. int res = 1;
  38. a %= mod;
  39. while (b > 0) {
  40. if (b & 1) res = res * a % mod;
  41. a = a * a % mod;
  42. b >>= 1;
  43. }
  44. return res;
  45. }
  46.  
  47. // nCr % MOD for n < MOD
  48. int nCrModP(int n, int r) {
  49. if (r > n) return 0;
  50. if (r == 0 || r == n) return 1;
  51.  
  52. int numerator = 1, denominator = 1;
  53. for (int i = 0; i < r; i++) {
  54. numerator = (numerator * (n - i)) % MOD;
  55. denominator = (denominator * (i + 1)) % MOD;
  56. }
  57. return (numerator * power(denominator, MOD - 2, MOD)) % MOD;
  58. }
  59.  
  60. // Lucas's Theorem
  61. int lucas(int n, int r) {
  62. if (r == 0) return 1;
  63. return (lucas(n / MOD, r / MOD) * nCrModP(n % MOD, r % MOD)) % MOD;
  64. }
  65. void solve() {
  66. int n,m;
  67. cin>>n>>m;
  68. int A[n],B[m];
  69. unordered_map<int,int>m1,m2;
  70. for(int i = 0 ; i<n ; i++){
  71. cin>>A[i];
  72. m1[A[i]]++;
  73. }
  74. for(int i = 0 ; i<m ; i++){
  75. cin>>B[i];
  76. m2[B[i]]++;
  77. }
  78. int sum = 0;
  79. for(auto i:m2){
  80. int d = i.first;
  81. for(int j = 1 ; j<=sqrt(d) ; j++){
  82. if(d%j==0){
  83. sum += m1[j]*m2[d];
  84. if(j!=(d/j)){
  85. int r = d/j;
  86. sum += m1[r]*m2[d];
  87. }
  88. }
  89. }
  90. }
  91. cout<<sum<<endl;
  92. }
  93.  
  94. signed main() {
  95. ios::sync_with_stdio(false); cin.tie(NULL);
  96. //int t;
  97. //cin >> t;
  98. //while (t--) {
  99. solve();
  100. //}
  101. return 0;
  102. }
  103.  
Success #stdin #stdout 0.01s 5288KB
stdin
4 2
2 4 3 5
4 8
stdout
4