所需的最小个位数素数，其总和等于 N

求和等于 N 的最小个位数素数。
例子：

Input: 11
Output: 3
Explanation: 5 + 3 + 3.
Another possibility is 3 + 3 + 3 + 2, but it is not
the minimal

Input: 12
Output: 2
Explanation: 7 + 5

方法：动态规划可用于解决上述问题。观察结果是：

• 只有 4 个个位数的素数 {2, 3, 5, 7}。
• 如果可以通过对一位数的素数求和得到 N，那么 N-2、N-3、N-5 或 N-7 中的至少一个也是可达的。
• 所需的最小个位数素数数将比构成{N-2, N-3, N-5, N-7}之一所需的最小素数位数多 1 。

使用这些观察结果，建立了一个循环来解决这个问题。重复将是：

dp[i] = 1 + min(dp[i-2], dp[i-3], dp[i-5], dp[i-7])

对于 {2, 3, 5, 7}，答案将是 1。对于每个其他数字，如果可能，使用观察 3 尝试找到可能的最小值。
下面是上述方法的实现。

C++

// CPP program to find the minimum number of single
// digit prime numbers required which when summed
// equals to a given number N.
#include 
using namespace std;
 
// function to check if i-th
// index is valid or not
bool check(int i, int val)
{
    if (i - val < 0)
        return false;
    return true;
}
 
// function to find the minimum number of single
// digit prime numbers required which when summed up
// equals to a given number N.
int MinimumPrimes(int n)
{
    int dp[n + 1];
 
    for (int i = 1; i <= n; i++)
        dp[i] = 1e9;
 
    dp[0] = dp[2] = dp[3] = dp[5] = dp[7] = 1;
    for (int i = 1; i <= n; i++) {
 
        if (check(i, 2))
            dp[i] = min(dp[i], 1 + dp[i - 2]);
 
        if (check(i, 3))
            dp[i] = min(dp[i], 1 + dp[i - 3]);
 
        if (check(i, 5))
            dp[i] = min(dp[i], 1 + dp[i - 5]);
 
        if (check(i, 7))
            dp[i] = min(dp[i], 1 + dp[i - 7]);
    }
 
    // Not possible
    if (dp[n] == (1e9))
        return -1;
    else
        return dp[n];
}
 
// Driver Code
int main()
{
 
    int n = 12;
 
    int minimal = MinimumPrimes(n);
    if (minimal != -1)
        cout << "Minimum number of single"
             << " digit primes required : "
             << minimal << endl;
 
    else
        cout << "Not possible";
 
    return 0;
}

Java

// Java program to find the minimum number
// of single digit prime numbers required
// which when summed equals to a given
// number N.
 
class Geeks {
     
// function to check if i-th
// index is valid or not
static boolean check(int i, int val)
{
    if (i - val < 0)
        return false;
    else
        return true;
}
 
// function to find the minimum number
// of single digit prime numbers required
// which when summed up equals to a given
// number N.
static double MinimumPrimes(int n)
{
    double[] dp;
    dp = new double[n+1];
 
    for (int i = 1; i <= n; i++)
        dp[i] = 1e9;
 
    dp[0] = dp[2] = dp[3] = dp[5] = dp[7] = 1;
    for (int i = 1; i <= n; i++) {
 
        if (check(i, 2))
            dp[i] = Math.min(dp[i], 1 + dp[i - 2]);
 
        if (check(i, 3))
            dp[i] = Math.min(dp[i], 1 + dp[i - 3]);
 
        if (check(i, 5))
            dp[i] = Math.min(dp[i], 1 + dp[i - 5]);
 
        if (check(i, 7))
            dp[i] = Math.min(dp[i], 1 + dp[i - 7]);
    }
 
    // Not possible
    if (dp[n] == (1e9))
        return -1;
    else
        return dp[n];
}
 
    // Driver Code
    public static void main(String args[])
    {
     
        int n = 12;
        int minimal = (int)MinimumPrimes(n);
         
        if (minimal != -1)
            System.out.println("Minimum number of single "+
                        "digit primes required: "+minimal);
     
        else
            System.out.println("Not Possible");
     
    }
}
 
// This code is contributed ankita_saini

Python 3

# Python3 program to find the minimum number
# of single digit prime numbers required 
# which when summed equals to a given 
# number N.
 
# function to check if i-th
# index is valid or not
 
def check(i,val):
    if i-val<0:
        return False
    return True
 
# function to find the minimum number of single
# digit prime numbers required which when summed up
# equals to a given number N.
 
def MinimumPrimes(n):
    dp=[10**9]*(n+1)
    dp[0]=dp[2]=dp[3]=dp[5]=dp[7]=1
    for i in range(1,n+1):
        if check(i,2):
            dp[i]=min(dp[i],1+dp[i-2])
        if check(i,3):
            dp[i]=min(dp[i],1+dp[i-3])
        if check(i,5):
            dp[i]=min(dp[i],1+dp[i-5])
        if check(i,7):
            dp[i]=min(dp[i],1+dp[i-7])
 
    # Not possible
    if dp[n]==10**9:
        return -1
    else:
        return dp[n]
 
 
# Driver Code
if __name__ == "__main__":
    n=12
    minimal=MinimumPrimes(n)
    if minimal!=-1:
        print("Minimum number of single digit primes required : ",minimal)
    else:
        print("Not possible")
#This code is contributed Saurabh Shukla

C#

// C# program to find the
// minimum number of single
// digit prime numbers required
// which when summed equals to
// a given number N.
using System;
 
class GFG
{
     
// function to check if i-th
// index is valid or not
static Boolean check(int i,
                     int val)
{
    if (i - val < 0)
        return false;
    else
        return true;
}
 
// function to find the
// minimum number of single
// digit prime numbers
// required which when summed
// up equals to a given
// number N.
static double MinimumPrimes(int n)
{
    double[] dp;
    dp = new double[n + 1];
 
    for (int i = 1; i <= n; i++)
        dp[i] = 1e9;
 
    dp[0] = dp[2] = dp[3] = dp[5] = dp[7] = 1;
    for (int i = 1; i <= n; i++)
    {
        if (check(i, 2))
            dp[i] = Math.Min(dp[i], 1 +
                             dp[i - 2]);
 
        if (check(i, 3))
            dp[i] = Math.Min(dp[i], 1 +
                             dp[i - 3]);
 
        if (check(i, 5))
            dp[i] = Math.Min(dp[i], 1 +
                             dp[i - 5]);
 
        if (check(i, 7))
            dp[i] = Math.Min(dp[i], 1 +
                             dp[i - 7]);
    }
 
    // Not possible
    if (dp[n] == (1e9))
        return -1;
    else
        return dp[n];
}
 
// Driver Code
public static void Main(String []args)
{
    int n = 12;
    int minimal = (int)MinimumPrimes(n);
     
    if (minimal != -1)
        Console.WriteLine("Minimum number of single " +
                            "digit primes required: " +
                                              minimal);
    else
        Console.WriteLine("Not Possible");
 
}
}
 
// This code is contributed
// by Ankita_Saini

Javascript

// Javascript program to find the minimum
// number of single digit prime
// numbers required which when summed
// equals to a given number N.
 
// function to check if i-th
// index is valid or not
function check(i, val)
{
    if (i - val < 0)
        return false;
    return true;
}
 
// function to find the minimum
// number of single digit prime
// numbers required which when
// summed up equals to a given
// number N.
function MinimumPrimes(n)
{
    let dp = new Array(n + 1)
 
    for (let i = 1; i<= n; i++)
        dp[i] = 1e9;
 
    dp[0] = dp[2] = dp[3] = dp[5] = dp[7] = 1;
    for (let i = 1; i <= n; i++)
    {
 
        if (check(i, 2))
            dp[i] = Math.min(dp[i], 1 +
                          dp[i - 2]);
 
        if (check(i, 3))
            dp[i] = Math.min(dp[i], 1 +
                          dp[i - 3]);
 
        if (check(i, 5))
            dp[i] = Math.min(dp[i], 1 +
                          dp[i - 5]);
 
        if (check(i, 7))
            dp[i] = Math.min(dp[i], 1 +
                          dp[i - 7]);
    }
 
    // Not possible
    if (dp[n] == (1e9))
        return -1;
    else
        return dp[n];
}
 
// Driver Code
let n = 12;
 
let minimal = MinimumPrimes(n);
if (minimal != -1)
{
    document.write("Minimum number of single "  + "digit primes required :");
    document.write( minimal );
}
else
{
    document.write("Not possible");
}
 
// This code is contributed
// by gfgking
 

输出：

Minimum number of single digit primes required : 2

时间复杂度： O(N)
注意：在多个查询的情况下，可以预先计算 dp[] 数组，我们可以在 O(1) 中回答每个查询。

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