给定两个整数N和X ,任务是找到总和N为单位数字X的最小整数计数。如果不存在这样的表示形式,则打印-1 。
例子:
Input: N = 38, X = 9
Output: 2
Explanation:
Minimum two integers are required with unit digit as X to represent as a sum equal to 38.
38 = 19 + 19 or 38 = 29 + 9
Input: N = 6, X = 4
Output: -1
Explanation:
No such representation of
方法:
请按照以下步骤解决问题:
- 获取N的单位数字,并检查单位数字为X的数字总和是否达到该数字。
- 如果可能,检查是否N ? X *(需要加上单位数字X的数字的最小次数以获得总和N ) 。
- 如果满足上述条件,则需要打印最少的次数,该数字需要加上单位数字X以获得总和N。否则,打印-1。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (int times = 1; times <= 10;
times++) {
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else {
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
int main()
{
int N = 58, X = 7;
cout << getNum(N, X) << endl;
return 0;
}
Java
// Java Program to implement
// the above approach
class GFG{
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (times = 1; times <= 10;
times++)
{
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else
{
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
public static void main(String []args)
{
int N = 58, X = 7;
System.out.println( getNum(N, X));
}
}
// This code is contributed by Ritik Bansal
Python3
# Python3 program to implement
# the above approach
# Function to calculate and return
# the minimum number of times a number
# with unit digit X needs to be added
# to get a sum N
def check(unit_digit, X):
# Calculate the number of additions
# required to get unit digit of N
for times in range(1, 11):
digit = (X * times) % 10
if (digit == unit_digit):
return times
# If unit digit of N
# cannot be obtained
return -1
# Function to return the minimum
# number required to represent N
def getNum(N, X):
# Stores unit digit of N
unit_digit = N % 10
# Stores minimum addition
# of X required to
# obtain unit digit of N
times = check(unit_digit, X)
# If unit digit of N
# cannot be obtained
if (times == -1):
return times
# Otherwise
else:
# If N is greater than
# or equal to (X*times)
if (N >= (times * X)):
# Minimum count of numbers
# that needed to represent N
return times
# Representation not
# possible
else:
return -1
# Driver Code
N = 58
X = 7
print(getNum(N, X))
# This code is contributed by Sanjit_Prasad
C#
// C# Program to implement
// the above approach
using System;
class GFG{
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
int times, digit;
// Calculate the number of
// additions required to get unit
// digit of N
for (times = 1; times <= 10;
times++)
{
digit = (X * times) % 10;
if (digit == unit_digit)
return times;
}
// If unit digit of N
// cannot be obtained
return -1;
}
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
int unit_digit;
// Stores unit digit of N
unit_digit = N % 10;
// Stores minimum addition
// of X required to
// obtain unit digit of N
int times = check(unit_digit, X);
// If unit digit of N
// cannot be obtained
if (times == -1)
return times;
// Otherwise
else
{
// If N is greater than
// or equal to (X*times)
if (N >= (times * X))
// Minimum count of numbers
// that needed to represent N
return times;
// Representation not
// possible
else
return -1;
}
}
// Driver Code
public static void Main()
{
int N = 58, X = 7;
Console.Write(getNum(N, X));
}
}
// This code is contributed by Code_Mech
输出:
4
时间复杂度: O(1)
辅助空间: O(1)