给定三个整数X ， Y和N ，任务是找到长度为N的最大可能数，仅由X和Y作为其数字组成，这样，其中X的计数可以被Y整除，反之亦然.如果不能形成这样的数字，则打印-1 。
例子：
Input: N = 3, X = 5, Y = 3
Output: 555
Explanation:
Count of 5’s = 3, which is divisible by 3
Count of 3’s = 0
Input: N = 4, X = 7, Y = 5
Output: -1
方法：
请按照以下步骤解决问题：
- 将X和Y 中较大的视为X ，将较小的视为Y 。
- 由于数字需要长度为N ，因此执行以下两个步骤直到 N ≤ 0：
- 如果 N 可被 Y 整除，则将 X 附加到答案中 N 次并将 N 减少为零。
- 否则，将 N 减少 X 并将 Y、X 次附加到答案。
- 完成上述步骤后，如果N<0，则需要的多个类型是不可能的。打印-1 。
- 否则，打印答案。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to generate and return
// the largest number
void largestNumber(int n, int X, int Y)
{
int maxm = max(X, Y);
// Store the smaller in Y
Y = X + Y - maxm;
// Store the larger in X
X = maxm;
// Stores respective counts
int Xs = 0;
int Ys = 0;
while (n > 0) {
// If N is divisible by Y
if (n % Y == 0) {
// Append X, N times to
// the answer
Xs += n;
// Reduce N to zero
n = 0;
}
else {
// Reduce N by X
n -= X;
// Append Y, X times
// to the answer
Ys += X;
}
}
// If number can be formed
if (n == 0) {
while (Xs-- > 0)
cout << X;
while (Ys-- > 0)
cout << Y;
}
// Otherwise
else
cout << "-1";
}
// Driver Code
int main()
{
int n = 19, X = 7, Y = 5;
largestNumber(n, X, Y);
return 0;
} Java
// Java program to implement the
// above approach
import java.util.*;
class GFG{
// Function to generate and return
// the largest number
public static void largestNumber(int n, int X,
int Y)
{
int maxm = Math.max(X, Y);
// Store the smaller in Y
Y = X + Y - maxm;
// Store the larger in X
X = maxm;
// Stores respective counts
int Xs = 0;
int Ys = 0;
while (n > 0)
{
// If N is divisible by Y
if (n % Y == 0)
{
// Append X, N times to
// the answer
Xs += n;
// Reduce N to zero
n = 0;
}
else
{
// Reduce N by X
n -= X;
// Append Y, X times
// to the answer
Ys += X;
}
}
// If number can be formed
if (n == 0)
{
while (Xs-- > 0)
System.out.print(X);
while (Ys-- > 0)
System.out.print(Y);
}
// Otherwise
else
System.out.print("-1");
}
// Driver code
public static void main (String[] args)
{
int n = 19, X = 7, Y = 5;
largestNumber(n, X, Y);
}
}
// This code is contributed by divyeshrabadiya07Python3
# Python3 program to implement
# the above approach
# Function to generate and return
# the largest number
def largestNumber(n, X, Y):
maxm = max(X, Y)
# Store the smaller in Y
Y = X + Y - maxm
# Store the larger in X
X = maxm
# Stores respective counts
Xs = 0
Ys = 0
while (n > 0):
# If N is divisible by Y
if (n % Y == 0):
# Append X, N times to
# the answer
Xs += n
# Reduce N to zero
n = 0
else:
# Reduce N by x
n -= X
# Append Y, X times to
# the answer
Ys += X
# If number can be formed
if (n == 0):
while (Xs > 0):
Xs -= 1
print(X, end = '')
while (Ys > 0):
Ys -= 1
print(Y, end = '')
# Otherwise
else:
print("-1")
# Driver code
n = 19
X = 7
Y = 5
largestNumber(n, X, Y)
# This code is contributed by himanshu77C#
// C# program to implement the
// above approach
using System;
class GFG{
// Function to generate and return
// the largest number
public static void largestNumber(int n, int X,
int Y)
{
int maxm = Math.Max(X, Y);
// Store the smaller in Y
Y = X + Y - maxm;
// Store the larger in X
X = maxm;
// Stores respective counts
int Xs = 0;
int Ys = 0;
while (n > 0)
{
// If N is divisible by Y
if (n % Y == 0)
{
// Append X, N times to
// the answer
Xs += n;
// Reduce N to zero
n = 0;
}
else
{
// Reduce N by X
n -= X;
// Append Y, X times
// to the answer
Ys += X;
}
}
// If number can be formed
if (n == 0)
{
while (Xs-- > 0)
Console.Write(X);
while (Ys-- > 0)
Console.Write(Y);
}
// Otherwise
else
Console.Write("-1");
}
// Driver code
public static void Main (String[] args)
{
int n = 19, X = 7, Y = 5;
largestNumber(n, X, Y);
}
}
// This code is contributed by shivanisinghss2110Javascript
输出：
7777755555555555555时间复杂度： O(N)
辅助空间： O(1)
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。