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C++ program to count number of operations needed to reach n by paying coins
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
Updated on: 03-Mar-2022
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