Problem:
Score : 200 points

Problem Statement

Takahashi loves gold coins. He gains 1000 happiness points for each 500-yen coin he has and gains 5 happiness points for each 5-yen coin he has. (Yen is the currency of Japan.)

Takahashi has X yen. If he exchanges his money so that he will gain the most happiness points, how many happiness points will he earn?

(We assume that there are six kinds of coins available: 500-yen, 100-yen, 50-yen, 10-yen, 5-yen, and 1-yen coins.)

Constraints

0 \leq X \leq 10^9

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the maximum number of happiness points that can be earned.

Sample Input 1

1024

Sample Output 1

2020

By exchanging his money so that he gets two 500-yen coins and four 5-yen coins, he gains 2020 happiness points, which is the maximum number of happiness points that can be earned.

Sample Input 2

0

Sample Output 2

0

He is penniless - or yenless.

Sample Input 3

1000000000

Sample Output 3

2000000000

He is a billionaire - in yen.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using intpair=pair<int,int>;
using llpair=pair<ll,ll>;
using llvec=vector<ll>;
using llmat=vector<vector<ll>>;
#define PI 3.141592653589793
#define llmattp(name,a,b,num) name(a,vector<ll>(b,num))
#define INTINF 1<<30
#define LLINF 1LL<<60
#define ABS(x) ( (x)>0 ? (x) : -(x) )

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}

template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

ll gcd(ll a, ll b) { //最大公約数
	if(a==0||b==0) return 0;
	if(a<b) swap(a,b);
	ll tmp = a%b;
	while(tmp!=0) {
		a = b;
		b = tmp;
		tmp = a%b;
	}
	return b;
}

ll factorial(ll x) { //階乗
	ll f=1;
	for(ll i=2; i<x; i++) {
		f*=i;
	}
	return f;
}

ll nPr(ll n, ll r) {
	ll result=1;
	for(ll i=r+1; i<=n; i++) result*=i;
	return result;
}

ll nCr(ll n, ll r) {
	if (n == r) { return 1; }
	if (r > n) { return 0; }

	if (r > n / 2) { r = n - r; }

	if (n == 0) { return 0; }
	if (r == 0) { return 1; }
	if (r == 1) { return n; }

	double result = 1;
	for (double i = 1; i <= r; i++) {
		result *= (n - i + 1) / i;
	}

	return (ll)result;
}

bool IsPrime(ll num) { //素数判定
	if (num < 2) return false;
	else if (num == 2) return true;
	else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

	double sqrtNum = sqrt(num);
	for (ll i = 3; i <= sqrtNum; i += 2)
	{
		if (num % i == 0)
		{
			// 素数ではない
			return false;
		}
	}

	// 素数である
	return true;
}

ll GreaterBinarySearch(ll *array, ll key, ll max, ll min) {
	if(array[max]<array[min]) {
		return NULL;
	} else {
		ll mid = max + (min-max)/2;
		if(array[mid]<key) {
			return GreaterBinarySearch(array,key,max,mid-1);
		} if(array[mid]>key) {
			return GreaterBinarySearch(array,key,mid+1,min);
		} else {
			return mid;
		}
	}
}

int DigitNum(ll n) { //桁数
	int digit=0;
	ll wari=1LL;
	while(n/wari) {
		digit++;
		wari*=10;
	}
	return digit;
}

void PrimeFact(ll x) { // 素因数分解 出力部を適切に書き換え
	int i=2;
	while(x%2&&i*i<=x) {
		cout<<2<<endl;
		x/=2;
	}
	i++;
	while(i*i<=x) {
		if(x%i) {
			i+=2;
			continue;
		}
		cout<<i<<endl;
		x/=i;
	}
}

bool Palind(string s) { //回文判定
	return s == string(s.rbegin(), s.rend());
}

string stcut[4][4]={
	{"AA","AB","AX","AY"},
	{"BA","BB","BX","BY"},
	{"XA","XB","XX","XY"},
	{"YA","YB","YX","YY"}
};

int main() {
	int N; cin>>N;
	ll result=0LL;
	result+=N/500*1000;
	N%=500;
	result+=N/5*5;
	cout<<result<<endl;
}