给定维度为N * M的矩阵mat[][] ，
任务是打印从左上角单元格(0, 0)到右下角单元格(N – 1, M – 1)的路径中的矩阵元素乘积中可以获得的最大尾随零数给定矩阵的。只有来自任何单元格(i, j) 的可能移动是(i + 1, j)或(i, j + 1)。
例子：
Input: N = 3, M = 4, mat[][] = {{6, 25, 4, 10}, {12, 25, 1, 15}, {7, 15, 15, 5}}
Output: 4
Explanation: Among all possible paths from top left to bottom right, the path (6, 25, 4, 10, 15, 5} has product (= 450000) with maximum number of trailing 0s. Therefore, the count of zeros is 4.
Input: N = 3, M = 3, mat[][] = {{2, 5, 2}, {10, 2, 40}, {5, 4, 8}}
Output: 2
朴素的方法：这个想法是递归地生成从给定矩阵的左上角单元格(0, 0)到右下角单元格(N – 1, M – 1)的所有可能路径，并计算每个元素的乘积小路。打印产品中尾随零的最大数量。请按照以下步骤解决问题：
- 初始化一个变量，比如product以存储从左上角单元格(0, 0)到右下角单元格(N – 1, M – 1)的路径上所有可能元素的乘积。
- 下面的递推关系计算从左上角单元格(0, 0)到右下角单元格(N – 1, M – 1)的所有可能路径的maxZeros值。
maxZeros(i, j, productValue) = max(maxZeros(i – 1, j, product*mat[i][j]), maxZeros(i, j – 1, product*mat[i][j]))
where,
maxZeros() is the function that returns the maximum number of trailing zeros.
- 从上面的递推关系递归地生成所有可能的路径，当任何路径到达单元格(N – 1, M – 1) 时，然后找到直到该路径的产品的尾随零的计数。
- 当每个递归调用结束时，最大化该递归调用返回的零计数。
- 在上述步骤之后打印尾随零的最大数量。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define N 3
#define M 4
// Stores the maximum count of zeros
int zeros = 0;
// Function that counts the trailing
// zeros in the given number num
int countZeros(int num)
{
// Stores the count of zeros
int count = 0;
// Iterate digits of num
while (num > 0 && num % 10 == 0)
{
num /= 10;
count++;
}
// Return the count
return count;
}
// Function to count maximum trailing
// zeros in product of elements in a
// path from top-left to bottom-right
void maxZeros(int mat[][M], int i,
int j, int product)
{
// If reached end of matrix
if (i == N - 1 && j == M - 1)
{
// Count the no of zeros product
product *= mat[i][j];
zeros = max(zeros, countZeros(product));
return;
}
// If out of bounds, return
if (i >= N)
return;
if (j >= M)
return;
// Recurse with move (i+1, j)
maxZeros(mat, i + 1, j,
product * mat[i][j]);
// Recurse with move(i, j+1)
maxZeros(mat, i, j + 1,
product * mat[i][j]);
}
// Function to print the maximum
// count of trailing zeros obtained
void maxZerosUtil(int mat[][M], int i,
int j, int product)
{
// Function Call
maxZeros(mat, 0, 0, 1);
// Print the maximum count
cout << zeros << endl;
}
// Driver Code
int main()
{
// Given matrix
int mat[N][M] = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZerosUtil(mat, 0, 0, 1);
}
// This code is contributed by bolliranadheer Java
// Java program for the above approach
import java.util.*;
class GFG {
// Stores the maximum count of zeros
static int zeros = 0;
// Function to count maximum trailing
// zeros in product of elements in a
// path from top-left to bottom-right
public static void maxZeros(int[][] mat,
int i, int j,
int product)
{
// If reached end of matrix
if (i == mat.length - 1
&& j == mat[0].length - 1) {
// Count the no of zeros product
product *= mat[i][j];
zeros = Math.max(zeros,
countZeros(product));
return;
}
// If out of bounds, return
if (i >= mat.length)
return;
if (j >= mat[0].length)
return;
// Recurse with move (i+1, j)
maxZeros(mat, i + 1, j,
product * mat[i][j]);
// Recurse with move(i, j+1)
maxZeros(mat, i, j + 1,
product * mat[i][j]);
}
// Function that counts the trailing
// zeros in the given number num
public static int countZeros(int num)
{
// Stores the count of zeros
int count = 0;
// Iterate digits of num
while (num > 0 && num % 10 == 0) {
num /= 10;
count++;
}
// Return the count
return count;
}
// Function to print the maximum
// count of trailing zeros obtained
public static void maxZerosUtil(
int[][] mat, int i, int j, int product)
{
// Function Call
maxZeros(mat, 0, 0, 1);
// Print the maximum count
System.out.println(zeros);
}
// Driver Code
public static void main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int mat[][] = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZerosUtil(mat, 0, 0, 1);
}
}Python3
# Python3 program for the
# above approach
N = 3
M = 4
# Stores the maximum count
# of zeros
zeros = 0
# Function that counts the
# trailing zeros in the
# given number num
def countZeros(num):
# Stores the count of
#zeros
count = 0
# Iterate digits of
# num
while (num > 0 and
num % 10 == 0):
num //= 10
count += 1
# Return the count
return count
# Function to count maximum
# trailing zeros in product
# of elements in a path from
# top-left to bottom-right
def maxZeros(mat, i,
j, product):
global M
global N
# If reached end of
# matrix
if (i == N - 1 and
j == M - 1):
# Count the no of
# zeros product
product *= mat[i][j]
global zeros
zeros = max(zeros,
countZeros(product))
return
# If out of bounds,
# return
if (i >= N):
return
if (j >= M):
return
# Recurse with move
# (i+1, j)
maxZeros(mat, i + 1, j,
product * mat[i][j])
# Recurse with move
# (i, j+1)
maxZeros(mat, i, j + 1,
product * mat[i][j])
# Function to print the
# maximum count of trailing
# zeros obtained
def maxZerosUtil(mat, i,
j, product):
# Function Call
maxZeros(mat, 0, 0, 1)
# Print the maximum
# count
print(zeros)
# Driver Code
if __name__ == "__main__":
# Given matrix
mat = [[6, 25, 4, 10],
[12, 25, 1, 15],
[7, 15, 15, 5]]
# Function Call
maxZerosUtil(mat, 0, 0, 1)
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
class GFG{
// Stores the maximum count of zeros
static int zeros = 0;
// Function to count maximum trailing
// zeros in product of elements in a
// path from top-left to bottom-right
public static void maxZeros(int[,] mat,
int i, int j,
int product,
int N, int M)
{
// If reached end of matrix
if (i == N - 1 && j == M - 1)
{
// Count the no of zeros product
product *= mat[i, j];
zeros = Math.Max(zeros,
countZeros(product));
return;
}
// If out of bounds, return
if (i >= mat.GetLength(0))
return;
if (j >= mat.GetLength(1))
return;
// Recurse with move (i+1, j)
maxZeros(mat, i + 1, j,
product * mat[i, j], N, M);
// Recurse with move(i, j+1)
maxZeros(mat, i, j + 1,
product * mat[i, j], N, M);
}
// Function that counts the trailing
// zeros in the given number num
public static int countZeros(int num)
{
// Stores the count of zeros
int count = 0;
// Iterate digits of num
while (num > 0 && num % 10 == 0)
{
num /= 10;
count++;
}
// Return the count
return count;
}
// Function to print the maximum
// count of trailing zeros obtained
public static void maxZerosUtil(int[,] mat, int i,
int j, int product,
int N, int M)
{
// Function Call
maxZeros(mat, 0, 0, 1, N, M);
// Print the maximum count
Console.WriteLine(zeros);
}
// Driver Code
public static void Main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int [,]mat = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZerosUtil(mat, 0, 0, 1, N, M);
}
}
// This code is contributed by Amit KatiyarJavascript
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
// Create a class pair to store
// count of 2's and 5's
class pair {
int x, y;
// Parameterized Constructor
pair(int x, int y)
{
this.x = x;
this.y = y;
}
// Function to covert into Strings
public String toString()
{
return "(" + this.x + ", "
+ this.y + ")";
}
}
class GFG {
// Function to get maximum no of
// zeros in product of path from
// topleft to bottom right
public static void maxZeros(
int[][] mat, int n, int m)
{
// Base Case
if (n == 0 || m == 0)
return;
// Store the maximum count of
// zeros till ith and jth index
pair dp[][] = new pair[n + 1][m + 1];
// Initialize the (0, 0)
dp[0][0] = new pair(countTwos(mat[0][0]),
countFives(mat[0][0]));
// Initialize the first row
// and column explicitly
for (int i = 1; i < n; i++)
dp[i][0] = add(
dp[i - 1][0],
new pair(
countTwos(mat[i][0]),
countFives(mat[i][0])));
for (int i = 1; i < m; i++)
dp[0][i] = add(
dp[0][i - 1],
new pair(
countTwos(mat[0][i]),
countFives(mat[0][i])));
// Iterate through all the cells
for (int i = 1; i < n; i++) {
for (int j = 1; j < m; j++) {
// Get the pair from the
// top and from left
pair top = dp[i - 1][j];
pair left = dp[i][j - 1];
pair curr = new pair(
countTwos(mat[i][j]),
countFives(mat[i][j]));
top = add(top, curr);
left = add(left, curr);
// If there are more number
// of 0s from top or left
if (check(top, left))
dp[i][j] = top;
else
dp[i][j] = left;
}
}
// Print the no of zeros
// min(no of 2's, no of 5's)
System.out.println(
Math.min(dp[n - 1][m - 1].x,
dp[n - 1][m - 1].y));
}
// Function to calculate no of zeros
public static boolean check(
pair one, pair two)
{
int top = Math.min(one.x, one.y);
int left = Math.min(two.x, two.y);
if (top > left)
return true;
else
return false;
}
// Function to calculate no of 2's
public static int countTwos(int num)
{
int count = 0;
while (num != 0 && num % 2 == 0) {
num /= 2;
count++;
}
// Return the final count
return count;
}
// Function to calculate no of 5's
public static int countFives(int num)
{
int count = 0;
while (num != 0 && num % 5 == 0) {
num /= 5;
count++;
}
// Return the final count
return count;
}
// Function to add pairs
public static pair add(pair one,
pair two)
{
pair np = new pair(one.x + two.x,
one.y + two.y);
// Return the resultant pair
return np;
}
// Driver Code
public static void main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int mat[][] = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZeros(mat, N, M);
}
}C#
// C# program for the above approach
using System;
// Create a class pair to store
// count of 2's and 5's
public class pair
{
public int x, y;
// Parameterized Constructor
public pair(int x, int y)
{
this.x = x;
this.y = y;
}
// Function to covert into Strings
public String toString()
{
return "(" + this.x + ", " +
this.y + ")";
}
}
class GFG{
// Function to get maximum no of
// zeros in product of path from
// topleft to bottom right
public static void maxZeros(int[,] mat, int n,
int m)
{
// Base Case
if (n == 0 || m == 0)
return;
// Store the maximum count of
// zeros till ith and jth index
pair [,]dp = new pair[n + 1, m + 1];
// Initialize the (0, 0)
dp[0, 0] = new pair(countTwos(mat[0, 0]),
countFives(mat[0, 0]));
// Initialize the first row
// and column explicitly
for(int i = 1; i < n; i++)
dp[i, 0] = add(dp[i - 1, 0],
new pair(
countTwos(mat[i, 0]),
countFives(mat[i, 0])));
for(int i = 1; i < m; i++)
dp[0, i] = add(dp[0, i - 1],
new pair(
countTwos(mat[0, i]),
countFives(mat[0, i])));
// Iterate through all the cells
for(int i = 1; i < n; i++)
{
for(int j = 1; j < m; j++)
{
// Get the pair from the
// top and from left
pair top = dp[i - 1, j];
pair left = dp[i, j - 1];
pair curr = new pair(
countTwos(mat[i, j]),
countFives(mat[i, j]));
top = add(top, curr);
left = add(left, curr);
// If there are more number
// of 0s from top or left
if (check(top, left))
dp[i, j] = top;
else
dp[i, j] = left;
}
}
// Print the no of zeros
// min(no of 2's, no of 5's)
Console.WriteLine(
Math.Min(dp[n - 1, m - 1].x,
dp[n - 1, m - 1].y));
}
// Function to calculate no of zeros
public static bool check(pair one, pair two)
{
int top = Math.Min(one.x, one.y);
int left = Math.Min(two.x, two.y);
if (top > left)
return true;
else
return false;
}
// Function to calculate no of 2's
public static int countTwos(int num)
{
int count = 0;
while (num != 0 && num % 2 == 0)
{
num /= 2;
count++;
}
// Return the readonly count
return count;
}
// Function to calculate no of 5's
public static int countFives(int num)
{
int count = 0;
while (num != 0 && num % 5 == 0)
{
num /= 5;
count++;
}
// Return the readonly count
return count;
}
// Function to add pairs
public static pair add(pair one,
pair two)
{
pair np = new pair(one.x + two.x,
one.y + two.y);
// Return the resultant pair
return np;
}
// Driver Code
public static void Main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int [,]mat = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZeros(mat, N, M);
}
}
// This code is contributed by Amit Katiyar输出
4时间复杂度： O(2 N*M )
辅助空间： O(1)
使用自底向上方法的动态规划：可以使用辅助数组dp[][]减少上述方法中的递归调用，并计算自底向上方法中每个状态的值。请按照以下步骤操作：
- 创建一个大小为N*M的辅助数组dp[][] 。
- DP [i] [j]表示没有击掌和二进制补码的直到第i行和第j列。
- 遍历矩阵并将dp[][]数组的每个状态更新为：
dp[i][j] = max(dp[i – 1][j], dp[i][j – 1])
- 打印上述步骤后(2s，5s)各自计数的最小值作为结果。
下面是上述方法的实现：
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
// Create a class pair to store
// count of 2's and 5's
class pair {
int x, y;
// Parameterized Constructor
pair(int x, int y)
{
this.x = x;
this.y = y;
}
// Function to covert into Strings
public String toString()
{
return "(" + this.x + ", "
+ this.y + ")";
}
}
class GFG {
// Function to get maximum no of
// zeros in product of path from
// topleft to bottom right
public static void maxZeros(
int[][] mat, int n, int m)
{
// Base Case
if (n == 0 || m == 0)
return;
// Store the maximum count of
// zeros till ith and jth index
pair dp[][] = new pair[n + 1][m + 1];
// Initialize the (0, 0)
dp[0][0] = new pair(countTwos(mat[0][0]),
countFives(mat[0][0]));
// Initialize the first row
// and column explicitly
for (int i = 1; i < n; i++)
dp[i][0] = add(
dp[i - 1][0],
new pair(
countTwos(mat[i][0]),
countFives(mat[i][0])));
for (int i = 1; i < m; i++)
dp[0][i] = add(
dp[0][i - 1],
new pair(
countTwos(mat[0][i]),
countFives(mat[0][i])));
// Iterate through all the cells
for (int i = 1; i < n; i++) {
for (int j = 1; j < m; j++) {
// Get the pair from the
// top and from left
pair top = dp[i - 1][j];
pair left = dp[i][j - 1];
pair curr = new pair(
countTwos(mat[i][j]),
countFives(mat[i][j]));
top = add(top, curr);
left = add(left, curr);
// If there are more number
// of 0s from top or left
if (check(top, left))
dp[i][j] = top;
else
dp[i][j] = left;
}
}
// Print the no of zeros
// min(no of 2's, no of 5's)
System.out.println(
Math.min(dp[n - 1][m - 1].x,
dp[n - 1][m - 1].y));
}
// Function to calculate no of zeros
public static boolean check(
pair one, pair two)
{
int top = Math.min(one.x, one.y);
int left = Math.min(two.x, two.y);
if (top > left)
return true;
else
return false;
}
// Function to calculate no of 2's
public static int countTwos(int num)
{
int count = 0;
while (num != 0 && num % 2 == 0) {
num /= 2;
count++;
}
// Return the final count
return count;
}
// Function to calculate no of 5's
public static int countFives(int num)
{
int count = 0;
while (num != 0 && num % 5 == 0) {
num /= 5;
count++;
}
// Return the final count
return count;
}
// Function to add pairs
public static pair add(pair one,
pair two)
{
pair np = new pair(one.x + two.x,
one.y + two.y);
// Return the resultant pair
return np;
}
// Driver Code
public static void main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int mat[][] = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZeros(mat, N, M);
}
}
C＃
// C# program for the above approach
using System;
// Create a class pair to store
// count of 2's and 5's
public class pair
{
public int x, y;
// Parameterized Constructor
public pair(int x, int y)
{
this.x = x;
this.y = y;
}
// Function to covert into Strings
public String toString()
{
return "(" + this.x + ", " +
this.y + ")";
}
}
class GFG{
// Function to get maximum no of
// zeros in product of path from
// topleft to bottom right
public static void maxZeros(int[,] mat, int n,
int m)
{
// Base Case
if (n == 0 || m == 0)
return;
// Store the maximum count of
// zeros till ith and jth index
pair [,]dp = new pair[n + 1, m + 1];
// Initialize the (0, 0)
dp[0, 0] = new pair(countTwos(mat[0, 0]),
countFives(mat[0, 0]));
// Initialize the first row
// and column explicitly
for(int i = 1; i < n; i++)
dp[i, 0] = add(dp[i - 1, 0],
new pair(
countTwos(mat[i, 0]),
countFives(mat[i, 0])));
for(int i = 1; i < m; i++)
dp[0, i] = add(dp[0, i - 1],
new pair(
countTwos(mat[0, i]),
countFives(mat[0, i])));
// Iterate through all the cells
for(int i = 1; i < n; i++)
{
for(int j = 1; j < m; j++)
{
// Get the pair from the
// top and from left
pair top = dp[i - 1, j];
pair left = dp[i, j - 1];
pair curr = new pair(
countTwos(mat[i, j]),
countFives(mat[i, j]));
top = add(top, curr);
left = add(left, curr);
// If there are more number
// of 0s from top or left
if (check(top, left))
dp[i, j] = top;
else
dp[i, j] = left;
}
}
// Print the no of zeros
// min(no of 2's, no of 5's)
Console.WriteLine(
Math.Min(dp[n - 1, m - 1].x,
dp[n - 1, m - 1].y));
}
// Function to calculate no of zeros
public static bool check(pair one, pair two)
{
int top = Math.Min(one.x, one.y);
int left = Math.Min(two.x, two.y);
if (top > left)
return true;
else
return false;
}
// Function to calculate no of 2's
public static int countTwos(int num)
{
int count = 0;
while (num != 0 && num % 2 == 0)
{
num /= 2;
count++;
}
// Return the readonly count
return count;
}
// Function to calculate no of 5's
public static int countFives(int num)
{
int count = 0;
while (num != 0 && num % 5 == 0)
{
num /= 5;
count++;
}
// Return the readonly count
return count;
}
// Function to add pairs
public static pair add(pair one,
pair two)
{
pair np = new pair(one.x + two.x,
one.y + two.y);
// Return the resultant pair
return np;
}
// Driver Code
public static void Main(String[] args)
{
// Given N & M
int N = 3, M = 4;
// Given matrix
int [,]mat = { { 6, 25, 4, 10 },
{ 12, 25, 1, 15 },
{ 7, 15, 15, 5 } };
// Function Call
maxZeros(mat, N, M);
}
}
// This code is contributed by Amit Katiyar
输出
4时间复杂度： O(N*M*log 10 (maxE))，其中 maxE 是给定矩阵中的最大值。
辅助空间： O(N*M)
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