数组中范围乘积的Java程序

给定一个大小为 N 的数组 A[]。解决 Q 个查询。在模 P 下找到范围 [L, R] 中的产品（P 是素数）。

例子：

Input : A[] = {1, 2, 3, 4, 5, 6} 
          L = 2, R = 5, P = 229
Output : 120

Input : A[] = {1, 2, 3, 4, 5, 6},
         L = 2, R = 5, P = 113
Output : 7 

蛮力
对于每个查询，遍历 [L, R] 范围内的每个元素并计算模 P 下的乘积。这将回答 O(N) 中的每个查询。

Java

// Product in range Queries in O(N)
import java.io.*;
 
class GFG
{
     
    // Function to calculate
    // Product in the given range.
    static int calculateProduct(int []A, int L,
                                int R, int P)
    {
         
        // As our array is 0 based as
        // and L and R are given as 1
        // based index.
        L = L - 1;
        R = R - 1;
     
        int ans = 1;
        for (int i = L; i <= R; i++)
        {
            ans = ans * A[i];
            ans = ans % P;
        }
     
        return ans;
    }
     
    // Driver code
    static public void main (String[] args)
    {
        int []A = { 1, 2, 3, 4, 5, 6 };
        int P = 229;
        int L = 2, R = 5;
        System.out.println(
            calculateProduct(A, L, R, P));
     
        L = 1;
        R = 3;
        System.out.println(
            calculateProduct(A, L, R, P));
    }
}
 
// This code is contributed by vt_m.

输出 ：

120
6

高效使用模乘逆：
由于 P 是素数，我们可以使用模乘逆。使用动态规划，我们可以计算模 P 下的前积数组，使得索引 i 处的值包含 [0, i] 范围内的乘积。类似地，我们可以计算模 P 下的预逆积。现在每个查询都可以在 O(1) 中得到回答。
逆积数组包含索引 i 处 [0, i] 范围内的逆积。因此，对于查询 [L, R]，答案将是 Product[R]*InverseProduct[L-1]
注意：我们不能将答案计算为 Product[R]/Product[L-1]，因为乘积是在模 P 下计算的。如果我们不计算模 P 下的乘积，总是有溢出的可能性。

Java

// Java program to find Product
// in range Queries in O(1)
class GFG
{
 
static int MAX = 100;
int pre_product[] = new int[MAX];
int inverse_product[] = new int[MAX];
 
// Returns modulo inverse of a
// with respect to m using extended
// Euclid Algorithm Assumption: a
// and m are coprimes,
// i.e., gcd(a, m) = 1
int modInverse(int a, int m)
{
    int m0 = m, t, q;
    int x0 = 0, x1 = 1;
 
    if (m == 1)
        return 0;
 
    while (a > 1)
    {
 
        // q is quotient
        q = a / m;
 
        t = m;
 
        // m is remainder now,
        // process same as
        // Euclid's algo
        m = a % m;
        a = t;
 
        t = x0;
 
        x0 = x1 - q * x0;
 
        x1 = t;
    }
 
    // Make x1 positive
    if (x1 < 0)
        x1 += m0;
 
    return x1;
}
 
// calculating pre_product array
void calculate_Pre_Product(int A[],
                           int N, int P)
{
    pre_product[0] = A[0];
 
    for (int i = 1; i < N; i++)
    {
        pre_product[i] = pre_product[i - 1] *
                                        A[i];
        pre_product[i] = pre_product[i] % P;
    }
}
 
// Calculating inverse_product array.
void calculate_inverse_product(int A[],
                               int N, int P)
{
    inverse_product[0] = modInverse(pre_product[0],
                                                P);
 
    for (int i = 1; i < N; i++)
        inverse_product[i] = modInverse(pre_product[i],
                                                     P);
}
 
// Function to calculate Product
// in the given range.
int calculateProduct(int A[], int L,
                     int R, int P)
{
    // As our array is 0 based as and
    // L and R are given as 1 based index.
    L = L - 1;
    R = R - 1;
    int ans;
 
    if (L == 0)
        ans = pre_product[R];
    else
        ans = pre_product[R] *
              inverse_product[L - 1];
 
    return ans;
}
 
// Driver code
public static void main(String[] s)
{
    GFG d = new GFG();
     
    // Array
    int A[] = { 1, 2, 3, 4, 5, 6 };
     
    // Prime P
    int P = 113;
 
    // Calculating PreProduct and
    // InverseProduct
    d.calculate_Pre_Product(A, A.length, P);
    d.calculate_inverse_product(A, A.length,
                                         P);
 
    // Range [L, R] in 1 base index
    int L = 2, R = 5;
    System.out.println(d.calculateProduct(A, L,
                                          R, P));
    L = 1;
    R = 3;
    System.out.println(d.calculateProduct(A, L,
                                          R, P));
         
}
}
 
// This code is contributed by Prerna Saini

输出 ：

7
6

有关更多详细信息，请参阅有关数组范围产品的完整文章！