给定正整数N,找到可以由N形成的长度为a,b和c的最大片段数。
例子 :
Input : N = 7, a = 5, b, = 2, c = 5
Output : 2
N can be divided into 2 segments of lengths
2 and 5. For the second example,
Input : N = 17, a = 2, b = 1, c = 3
Output : 17
N can be divided into 17 segments of 1 or 8
segments of 2 and 1 segment of 1. But 17 segments
of 1 is greater than 9 segments of 2 and 1.
方法:使用的方法是动态编程。基本dp 0将为0,因为最初它没有任何分段。之后,从1迭代到n,并针对dp i + a ,dp i + b和dp i + c这3种状态的每一个,存储通过使用或不使用a,b或c获得的最大值部分。
要处理的三个州是:
dpi+a=max(dpi+1, dpi+a);
dpi+b=max(dpi+1, dpi+b);
dpi+c=max(dpi+1, dpi+c);
下面是上述想法的实现:
C++
// C++ implementation to divide N into
// maximum number of segments
// of length a, b and c
#include
using namespace std;
// function to find the maximum
// number of segments
int maximumSegments(int n, int a,
int b, int c)
{
// stores the maximum number of
// segments each index can have
int dp[n + 1];
// initialize with -1
memset(dp, -1, sizeof(dp));
// 0th index will have 0 segments
// base case
dp[0] = 0;
// traverse for all possible
// segments till n
for (int i = 0; i < n; i++)
{
if (dp[i] != -1) {
// conditions
if(i + a <= n ) //avoid buffer overflow
dp[i + a] = max(dp[i] + 1,
dp[i + a]);
if(i + b <= n ) //avoid buffer overflow
dp[i + b] = max(dp[i] + 1,
dp[i + b]);
if(i + c <= n ) //avoid buffer overflow
dp[i + c] = max(dp[i] + 1,
dp[i + c]);
}
}
return dp[n];
}
// Driver code
int main()
{
int n = 7, a = 5, b = 2, c = 5;
cout << maximumSegments(n, a, b, c);
return 0;
}
Java
// Java implementation to divide N into
// maximum number of segments
// of length a, b and c
import java.util.*;
class GFG
{
// function to find the maximum
// number of segments
static int maximumSegments(int n, int a,
int b, int c)
{
// stores the maximum number of
// segments each index can have
int dp[] = new int[n + 10];
// initialize with -1
Arrays.fill(dp, -1);
// 0th index will have 0 segments
// base case
dp[0] = 0;
// traverse for all possible
// segments till n
for (int i = 0; i < n; i++)
{
if (dp[i] != -1)
{
// conditions
if(i + a <= n ) //avoid buffer overflow
dp[i + a] = Math.max(dp[i] + 1,
dp[i + a]);
if(i + b <= n ) //avoid buffer overflow
dp[i + b] = Math.max(dp[i] + 1,
dp[i + b]);
if(i + c <= n ) //avoid buffer overflow
dp[i + c] = Math.max(dp[i] + 1,
dp[i + c]);
}
}
return dp[n];
}
// Driver code
public static void main(String arg[])
{
int n = 7, a = 5, b = 2, c = 5;
System.out.print(maximumSegments(n, a, b, c));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python implementation
# to divide N into maximum
# number of segments of
# length a, b and c
# function to find
# the maximum number
# of segments
def maximumSegments(n, a, b, c) :
# stores the maximum
# number of segments
# each index can have
dp = [-1] * (n + 10)
# 0th index will have
# 0 segments base case
dp[0] = 0
# traverse for all possible
# segments till n
for i in range(0, n) :
if (dp[i] != -1) :
# conditions
if(i + a <= n ): # avoid buffer overflow
dp[i + a] = max(dp[i] + 1,
dp[i + a])
if(i + b <= n ): # avoid buffer overflow
dp[i + b] = max(dp[i] + 1,
dp[i + b])
if(i + c <= n ): # avoid buffer overflow
dp[i + c] = max(dp[i] + 1,
dp[i + c])
return dp[n]
# Driver code
n = 7
a = 5
b = 2
c = 5
print (maximumSegments(n, a,
b, c))
# This code is contributed by
# Manish Shaw(manishshaw1)
C#
// C# implementation to divide N into
// maximum number of segments
// of length a, b and c
using System;
class GFG
{
// function to find the maximum
// number of segments
static int maximumSegments(int n, int a,
int b, int c)
{
// stores the maximum number of
// segments each index can have
int []dp = new int[n + 10];
// initialize with -1
for(int i = 0; i < n + 10; i++)
dp[i]= -1;
// 0th index will have 0 segments
// base case
dp[0] = 0;
// traverse for all possible
// segments till n
for (int i = 0; i < n; i++)
{
if (dp[i] != -1)
{
// conditions
if(i + a <= n ) // avoid buffer overflow
dp[i + a] = Math.Max(dp[i] + 1,
dp[i + a]);
if(i + b <= n ) // avoid buffer overflow
dp[i + b] = Math.Max(dp[i] + 1,
dp[i + b]);
if(i + c <= n ) // avoid buffer overflow
dp[i + c] = Math.Max(dp[i] + 1,
dp[i + c]);
}
}
return dp[n];
}
// Driver code
public static void Main()
{
int n = 7, a = 5, b = 2, c = 5;
Console.Write(maximumSegments(n, a, b, c));
}
}
// This code is contributed by nitin mittal
PHP
Javascript
输出 :
2
时间复杂度:O(n)