# C++ program to count number of operations needed to reach n by paying coins

C++Server Side ProgrammingProgramming

Suppose we have five numbers, N, A, B, C, D. We start with a number 0 and end at N. We can change a number by certain number of coins with the following operations −

• Multiply the number by 2, paying A coins
• Multiply the number by 3, paying B coins
• Multiply the number by 5, paying C coins
• Increase or decrease the number by 1, paying D coins.

We can perform these operations any number of times in any order. We have to find the minimum number of coins we need to reach N

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

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

For 1 coin, multiply x by 2 (x=2).

For 2 coins, multiply x by 5 (x=10).

For 8 coins, increase it by 1 (x=11).

## Steps

To solve this, we will follow these 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)

## Example

Let us see the following implementation to get better understanding −

#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;
}

## Input

11, 1, 2, 2, 8

## Output

19