C++ program: Calculate the number of operations required to reach n using coin payments-C++-php.cn

C++ program: Calculate the number of operations required to reach n using coin payments

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Release： 2023-09-14 20:53:04

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C++ program: Calculate the number of operations required to reach n using coin payments

Suppose we have five numbers, N, A, B, C, D. We start with the number 0 and end with N. We can change a number by using a certain number of coins. The specific operation is as follows:

Multiply the number by 2 and pay A coins
Multiply the number by 3 and pay B coins Coins
Multiply the number by 5, pay C coins
Increase or decrease the number by 1, pay D coins

We can do this any number of times and in any order these operations. We need to find the minimum number of coins required to reach N

So if the input is N = 11; A = 1; B = 2; C = 2; D = 8, then the output will be 19 because initially x is 0.

Use 8 coins to increase x by 1 (x=1).

Use 1 coin to multiply x by 2 (x=2).

Use 2 coins to multiply x by 5 (x=10).

Use 8 coins to increase it by 1 (x=11).

Steps

To solve this problem, we will follow the following steps:

Define one map f for integer type key and value
Define one map vis for integer type key and Boolean type value
Define a function calc, this will take n
if n is zero, then:
   return 0
if n is in vis, then:
   return f[n]
vis[n] := 1
res := calc(n / 2) + n mod 2 * d + a
if n mod 2 is non-zero, then:
   res := minimum of res and calc((n / 2 + 1) + (2 - n mod 2)) * d + a)
res := minimum of res and calc(n / 3) + n mod 3 * d + b
if n mod 3 is non-zero, then:
   res := minimum of res and calc((n / 3 + 1) + (3 - n mod 3)) * d + b)
res := minimum of res and calc(n / 5) + n mod 5 * d + c
if n mod 5 is non-zero, then:
   res := minimum of res and calc((n / 5 + 1) + (5 - n mod 5))
if (res - 1) / n + 1 > d, then:
   res := n * d
return f[n] = res
From the main method, set a, b, c and d, and call calc(n)

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Example

Let us see the following implementation for better understanding Understand −

#include <bits/stdc++.h>
using namespace std;

int a, b, c, d;
map<long, long> f;
map<long, bool> vis;

long calc(long n){
   if (!n)
      return 0;
   if (vis.find(n) != vis.end())
      return f[n];
   vis[n] = 1;
   long res = calc(n / 2) + n % 2 * d + a;
   if (n % 2)
      res = min(res, calc(n / 2 + 1) + (2 - n % 2) * d + a);
   res = min(res, calc(n / 3) + n % 3 * d + b);
   if (n % 3)
      res = min(res, calc(n / 3 + 1) + (3 - n % 3) * d + b);
   res = min(res, calc(n / 5) + n % 5 * d + c);
   if (n % 5)
      res = min(res, calc(n / 5 + 1) + (5 - n % 5) * d + c);
   if ((res - 1) / n + 1 > d)
      res = n * d;
   return f[n] = res;
}
int solve(int N, int A, int B, int C, int D){
   a = A;
   b = B;
   c = C;
   d = D;
   return calc(N);
}
int main(){
   int N = 11;
   int A = 1;
   int B = 2;
   int C = 2;
   int D = 8;
   cout << solve(N, A, B, C, D) << endl;
}

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Input

11, 1, 2, 2, 8

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Output

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