在给定两个整数N和M的情况下,将N划分为M个整数,以使该划分所获得的最大整数和最小整数之差尽可能小。
打印M个数字A1,A2 …. Am ,这样:
- sum(A)=N。
- max(A)-min(A)被最小化。
例子:
Input : N = 11, M = 3
Output : A[] = {4, 4, 3}
Input : N = 8, M = 4
Output : A[] = {2, 2, 2, 2}
为了使术语之间的差异最小化,我们应该使它们彼此尽可能接近。假设我们可以打印任何浮点值而不是整数,那么答案是0(打印N / MM次)。但是由于我们需要打印整数,因此我们可以将其分为2个部分,即floor(N / M)和floor(N / M)+1,这将最多为我们提供答案1。
每种类型需要打印多少个术语?
假设我们打印了Floor(N / M) M次,总和等于N –(N%M) 。因此,我们需要选择N%M个项并将其增加1。
下面是上述方法的实现:
C++
// C++ program to partition N into M parts
// such that difference Max and Min
// part is smallest
#include
using namespace std;
// Function to partition N into M parts such
// that difference Max and Min part
// is smallest
void printPartition(int n, int m)
{
int k = n / m; // Minimum value
int ct = n % m; // Number of (K+1) terms
int i;
for (i = 1; i <= ct; i++)
cout << k + 1 << " ";
for (; i <= m; i++)
cout << k << " ";
}
// Driver Code
int main()
{
int n = 5, m = 2;
printPartition(n, m);
return 0;
}
Java
// Java program to partition N into M parts
// such that difference Max and Min
// part is smallest
import java.io.*;
class GFG {
// Function to partition N into M parts such
// that difference Max and Min part
// is smallest
static void printPartition(int n, int m)
{
int k = n / m; // Minimum value
int ct = n % m; // Number of (K+1) terms
int i;
for (i = 1; i <= ct; i++)
System.out.print( k + 1 + " ");
for (; i <= m; i++)
System.out.print( k + " ");
}
// Driver Code
public static void main (String[] args) {
int n = 5, m = 2;
printPartition(n, m);
}
}
// This code is contributed by anuj_67..
Python3
# Python 3 program to partition N into M parts
# such that difference Max and Min
# part is the smallest
# Function to partition N into M parts such
# that difference Max and Min part
# is smallest
def printPartition(n, m):
k = int(n / m)
# Minimum value
ct = n % m
# Number of (K+1) terms
for i in range(1,ct+1,1):
print(k + 1,end= " ")
count = i
for i in range(count,m,1):
print(k,end=" ")
# Driver Code
if __name__ == '__main__':
n = 5
m = 2
printPartition(n, m)
# This code is contributed by
# Surendra_Gangwar
C#
// C# program to partition N into M parts
// such that difference Max and Min
// part is smallest
using System;
class GFG
{
static void printPartition(int n, int m)
{
int k = n / m; // Minimum value
int ct = n % m; // Number of (K+1) terms
int i;
for (i = 1; i <= ct; i++)
Console.Write( k + 1 + " ");
for (; i <= m; i++)
Console.Write( k + " ");
}
// Driver Code
static public void Main ()
{
int n = 5, m = 2;
printPartition(n, m);
}
}
// This code is contributed by Sachin
PHP
Javascript
输出:
3 2
时间复杂度: O(M)
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