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name
stringlengths 2
112
| description
stringlengths 29
13k
| source
int64 1
7
| difficulty
int64 0
25
| solution
stringlengths 7
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| language
stringclasses 4
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1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | /*
import java.util.Arrays;
import java.util.Scanner;
public class KvassAndTheFairNut {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
long s = in.nextLong();
long[] kegs = new long[n];
long sum = 0;
for (int i = 0; i < n; i++) {
kegs[i] = in.nextLong();
sum+=kegs[i];
}
if (sum < s) {
System.out.println(-1);
return;
}
Arrays.sort(kegs);
long Min = kegs[0];
long cur = 0;
for (int i = n - 1; i >= 0; i--) {
if (kegs[i] == Min) {
break;
}
cur += kegs[i] - Min;
kegs[i] = Min;
if (cur >= s) {
System.out.println(Min);
return;
}
}
while (kegs[0] > 0) {
cur += n;
kegs[0]--;
if (cur >= s) {
System.out.println(kegs[0]);
return;
}
}
System.out.println(-1);
}
}
*/
import java.util.*;
import java.io.*;
public class KvassAndTheFairNut {
public static void main(String args[]) {
FastReader ss = new FastReader();
int n = ss.nextInt();
long s = ss.nextLong();
long sum = 0l;
int[] kegs = new int[n];
while (n-- > 0) {
int temp = ss.nextInt();
kegs[n] = temp;
sum += (long) temp;
}
if (s > sum) {
System.out.println("-1");
System.exit(0);
}
Arrays.sort(kegs);
int min = kegs[0];
n = kegs.length;
for (int i = 1; i < n; i++) {
int temp = kegs[i] - min;
s -= (long) temp;
}
while (s > 0l && min > 0) {
s -= n;
min--;
}
System.out.println(min);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def Prime_factors(n):
i = 2
factors = set()
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.add(i)
if n > 1:
factors.add(n)
print(factors)
if len(factors)==2:
return True
else:
return False
def Divisors(n) :
l = []
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
return l
def Sieve(n):
prime=[True for i in range(n+1)]
lst=[]
p=2
while p*p<=n:
if prime[p]:
for i in range(p*p,n+1,p):
prime[i]=False
p=p+1
for i in range(2,n+1):
if prime[i]:
lst.append(i)
return lst
def minimum_integer_not_in_list(a):
b=max(a)
if(b<1):
return 1
A=set(a)
B=set(range(1,b+1))
D=B-A
if len(D)==0:
return b+1
else:
return min(D)
# Driver Code
def solution():
n,s=Mint()
a=List()
if(sum(a)<s):
print(-1)
else:
print(min(min(a),(sum(a)-s)//n))
#Call the solve function
if __name__ == "__main__":
solution() | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
ll s;
cin >> n >> s;
vector<int> a(n);
ll sum = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
sum += a[i];
}
if (sum < s) {
cout << -1 << '\n';
return 0;
}
sort(a.begin(), a.end());
int lo = 0, hi = a[0] + 1;
while (lo + 1 < hi) {
int mid = (lo + hi) / 2;
ll t = 0;
for (int i = 0; i < n; i++) {
t += max(0, a[i] - mid);
}
if (t >= s) {
lo = mid;
} else {
hi = mid;
}
}
cout << lo << '\n';
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | first = input()
first = first.split(" ")
second = input()
second = second.split(" ")
n= int(first[0])
s = int(first[1])
v = []
# people_ = people_.split(" ")
for p in second:
v.append(int(p))
if sum(v) < s:
print(-1)
elif sum(v) == s:
print(0)
else:
min_volume = min(v)
sum = 0
v_ = []
for each in v:
sum += each - min_volume
v_.append(min_volume)
if sum>=s:
print(min_volume)
else:
s = s-sum
volume_r = s/n
if s%n !=0:
volume_r +=1
print(min_volume-int(volume_r)) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
public class problem2 {
public static void main(String[]args){
Scanner sc=new Scanner(System.in);
long n=sc.nextLong(),s=sc.nextLong();
long sum=0,min=Long.MAX_VALUE;
for(long i=0;i<n;i++){
long vi=sc.nextLong();
sum+=vi;
if(vi<min)
min=vi;
}
sc.close();
System.out.println(s>sum?-1:Math.min(min,(sum-s)/n));
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 |
import java.io.*;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.StringTokenizer;
public class d
{
public static void print(String str,long val){
System.out.println(str+" "+val);
}
public long gcd(long a, long b) {
if (b==0L) return a;
return gcd(b,a%b);
}
public static void debug(long[][] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.println(Arrays.toString(arr[i]));
}
}
public static void debug(int[][] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.println(Arrays.toString(arr[i]));
}
}
public static void debug(String[] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.println(arr[i]);
}
}
public static void print(int[] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.print(arr[i]+" ");
}
System.out.print('\n');
}
public static void print(Object[] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.print(arr[i]+" ");
}
System.out.print('\n');
}
public static void print(String[] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.print(arr[i]+" ");
}
System.out.print('\n');
}
public static void print(long[] arr){
int len = arr.length;
for(int i=0;i<len;i++){
System.out.print(arr[i]+" ");
}
System.out.print('\n');
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
public FastReader(String path) throws FileNotFoundException {
br = new BufferedReader(new FileReader(path));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
static void sort(long[] arr){
Long[] temp = new Long[arr.length];
for(int i=0;i<arr.length;i++){
temp[i] = arr[i];
}
Arrays.sort(temp);
for(int i=0;i<arr.length;i++){
arr[i] = temp[i];
}
}
public static void main(String[] args)
{
FastReader s=new FastReader();
int n = s.nextInt();
long vol = s.nextLong();
long[] arr = new long[n];
long sum =0;
for(int i=0;i<n;i++){
arr[i] = s.nextInt();
sum+=arr[i];
}
if(sum<vol){
System.out.println(-1);
return;
}
long start = 0;
long end = (long)(1e9);
while (end>start+1){
long mid = (end+start)/2;
if(search(mid,arr,vol)){
start =mid;
}
else {
end = mid-1;
}
}
if(search(end,arr,vol)){
System.out.println(end);
return;
}
System.out.println(start);
}
static boolean search(long k,long[] arr,long vol){
long sum =0;
for(int i=0;i<arr.length;i++){
if(arr[i]<k){
return false;
}
sum+=(arr[i]-k)*1l;
}
return (sum>=vol);
}
// OutputStream out = new BufferedOutputStream( System.out );
// for(int i=1;i<n;i++){
// out.write((arr[i]+" ").getBytes());
// }
// out.flush();
// long start_time = System.currentTimeMillis();
// long end_time = System.currentTimeMillis();
// System.out.println((end_time - start_time) + "ms");
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
long long int s, sum = 0;
cin >> n >> s;
long long int* v = new long long int[n];
for (int i = 0; i < n; i++) {
cin >> v[i];
sum += v[i];
}
if (s > sum)
cout << -1 << endl;
else {
sort(v, v + n);
if (s < (sum - v[0] * n))
cout << v[0] << endl;
else {
s -= sum - v[0] * n;
if (s % n == 0)
cout << v[0] - s / n << endl;
else
cout << v[0] - (s / n) - 1 << endl;
}
}
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
n,q=map(int,input().split())
h=list(map(int,input().split()))
t=0
for i in range (n):
t+=h[i]
if t<q:
print(-1)
exit()
low=min(h)
for i in range (n):
q-=(h[i]-low)
if q<=0:
print(low)
exit()
low-=math.ceil(q/n)
print(low)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.Arrays;
import java.util.Scanner;
public class Kvass526B {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int kegs = s.nextInt();
long glass = s.nextLong();
long sum =0l;
int arr[] = new int[kegs];
for(int i =0;i<kegs;i++) {
arr[i] = s.nextInt();
sum+=arr[i];
}
if(sum<glass) {
System.out.println(-1);
return;
}
Arrays.sort(arr);
long filled = 0l;
int min = arr[0];
for(int i = arr.length-1;i>=0;i--) {
if(arr[i]!=min) {
filled+=arr[i]-min;
arr[i] = min;
}
if(filled>=glass)
break;
}
// for(int o:arr) {
// System.out.print(o+" ");
// }
// System.out.println();
// System.out.println("Filled :"+filled);
if(filled>=glass) {
System.out.println(min);
return;
}
long remaining = glass-filled;
//System.out.println("Remaining : "+remaining);
sum = 0l;
for(int i=0;i<arr.length;i++)
sum+=arr[i];
long ans = (sum-remaining)/arr.length;
System.out.println(ans);
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
from decimal import Decimal
import heapq
def na():
n = int(input())
b = [int(x) for x in input().split()]
return n,b
def nab():
n = int(input())
b = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
return n,b,c
def dv():
n, m = map(int, input().split())
return n,m
def dva():
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
return n,m,b
def eratosthenes(n):
sieve = list(range(n + 1))
for i in sieve:
if i > 1:
for j in range(i + i, len(sieve), i):
sieve[j] = 0
return sorted(set(sieve))
def nm():
n = int(input())
b = [int(x) for x in input().split()]
m = int(input())
c = [int(x) for x in input().split()]
return n,b,m,c
def dvs():
n = int(input())
m = int(input())
return n, m
n,s = map(int, input().split())
a = list(map(int, input().split()))
if sum(a) < s:
print(-1)
exit()
d = min(a)
tc = 0
for i in a:
tc += i - d
if tc >= s:
print(d)
else:
tc = s - tc
if tc % n == 0:
print(d - tc // n)
else:
print(d - tc //n - 1)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Q2 {
public static void main(String[] args) throws IOException{
Reader.init(System.in);
int n = Reader.nextInt();
Long s = Reader.nextlong();
int[] values = new int[n];
int place = 0;
for (int i=0;i<n;i++) {
values[i] = Reader.nextInt();
if (values[i]<values[place]) {
place = i;
}
}
Long diff = (long) 0;
for (int i=0;i<n;i++) {
diff += values[i] - values[place];
}
if (diff >= s) {
System.out.println(values[place]);
}
else {
s = s - diff;
long hey = (long) n*values[place];
if (hey < s) {
System.out.println(-1);
}
else {
if (s%n==0)
System.out.println(values[place] - (s/n));
else {
System.out.println(values[place] - (s/n + 1));
}
}
}
}
}
class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
tokenizer = new StringTokenizer(
reader.readLine() );
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt( next() );
}
static double nextDouble() throws IOException {
return Double.parseDouble( next() );
}
static long nextlong() throws IOException {
return Long.parseLong(next());
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
public class gr1 {
static class InputReader {
public BufferedReader br;
public StringTokenizer token;
public InputReader(InputStream stream)
{
br=new BufferedReader(new InputStreamReader(stream),32768);
token=null;
}
public String next()
{
while(token==null || !token.hasMoreTokens())
{
try
{
token=new StringTokenizer(br.readLine());
}
catch(IOException e)
{
throw new RuntimeException(e);
}
}
return token.nextToken();
}
public int nextInt()
{
return Integer.parseInt(next());
}
public long nextLong()
{
return Long.parseLong(next());
}
public double nextDouble()
{
return Double.parseDouble(next());
}
}
static class card{
long a;
int cnt;
int i;
public card(long a,int cnt,int i)
{
this.a=a;
this.cnt=cnt;
this.i=i;
}
}
static class ascend implements Comparator<pair>
{
public int compare(pair o1,pair o2)
{
if(o1.b!=o2.b)
return (int)(o1.b-o2.b);
else
return (int)(o1.a-o2.a);
}
}
/*static class descend implements Comparator<pair>
{
public int compare(pair o1,pair o2)
{
if(o1.a!=o2.a){
return (o1.a-o2.a)*-1;
} else {
return (o1.b-o2.b);
}
}
}*/
static class extra
{
static void shuffle(long a[])
{
List<Long> l=new ArrayList<>();
for(int i=0;i<a.length;i++)
l.add(a[i]);
Collections.shuffle(l);
for(int i=0;i<a.length;i++)
a[i]=l.get(i);
}
static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
static boolean valid(int i,int j,int r,int c)
{
if(i>=0 && i<r && j>=0 && j<c)
return true;
else
return false;
}
static boolean v[]=new boolean[100001];
static List<Integer> l=new ArrayList<>();
static int t;
static void seive()
{
for(int i=2;i<100001;i++)
{
if(!v[i])
{
t++;
l.add(i);
for(int j=2*i;j<100001;j+=i)
v[j]=true;
}
}
}
static int binary(pair a[],int val,int n)
{
int mid=0,l=0,r=n-1,ans=0;
while(l<=r)
{
mid=(l+r)>>1;
if(a[mid].a==val)
{
r=mid-1;
ans=mid;
}
else if(a[mid].a>val)
r=mid-1;
else
{
l=mid+1;
ans=l;
}
}
return (ans);
}
}
static class pair{
long a;
int b;
public pair(long a,int n)
{
this.a=a;
this.b=n;
}
}
static InputReader sc=new InputReader(System.in);
static PrintWriter out=new PrintWriter(System.out);
public static void main(String[] args) {
solver s=new solver();
int t=1;
while(t>0)
{
s.solve();
t--;
}
}
static class solver
{
void solve()
{
int n=sc.nextInt();
long s=sc.nextLong();
long v[]=new long[n];
long min=Long.MAX_VALUE;
long sum=0;
for(int i=0;i<n;i++)
{
v[i]=sc.nextLong();
sum+=v[i];
if(v[i]<min)
min=v[i];
}
if(s>sum)
System.out.println(-1);
else
{
new extra().shuffle(v);
Arrays.sort(v);
for(int i=n-1;i>=1;i--)
{
s-=Math.min(v[i]-min,s);
if(s==0)
{
System.out.println(min);
System.exit(0);
}
}
min-=(s/n)+((s%n==0)?0:1);
System.out.println(min);
}
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | x,s=list(map(int,input().split()))
l=list(map(int,input().split()))
sum=0
for i in l:
sum+=i
if sum<s:
print(-1)
elif sum==s:
print(0)
else:
k=sum-len(l)*min(l)
if k>=s:
print(min(l))
else:
sum-=k
s-=k
ans=s//len(l)
if s%len(l)!=0:
ans+=1
print(min(l)-ans) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | # -*- coding: utf-8 -*-
"""
Created on Sat Dec 15 09:06:15 2018
@author: Arsanuos
"""
def solve():
rd = lambda: list(map(int, input().split()))
n, s = rd()
arr = rd()
k = sum(arr)
if s > k:
print(-1)
return
if s == k:
print(0)
return
arr.sort(reverse=True)
mi = arr[-1]
tmp = 0
for item in arr:
tmp += (item - mi)
if tmp >= s:
print(mi)
else:
tmp = s - tmp
l = len(arr)
mi -= int(tmp/l)
tmp -= int(tmp/l) * l
if tmp > 0 and tmp % l != 0:
mi -= 1
if mi >= 0:
print(mi)
else:
print(-1)
solve() | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | def ifPoss(arr,v,h):
tot = 0
for i in arr:
tot += (i-h)
return tot>=v
def prog():
n,v = map(int,input().split())
arr = [int(x) for x in input().split()]
minLev = -1
l,h = 0,min(arr)
while(l<=h):
m = int((l+h)/2)
if(ifPoss(arr,v,m)):
minLev = max(minLev,m)
l = m+1
else:
h = m-1
return minLev
print(prog()) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
const long long INF = 1e18;
const int M = 1e9 + 7;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << " = " << a << '\n';
err(++it, args...);
}
int dx4[] = {0, 0, 1, -1};
int dy4[] = {1, -1, 0, 0};
int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[] = {0, 1, 1, 1, 0, -1, -1, -1};
int FX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int FY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long n, m, l, r, d, a, b, k, u, p = -1, q, x, y, z, mn, mx, rem, ans,
res = 0, c = 0;
long long aa[2000];
long long ch(long long NN) {
long long sm = 0;
for (long long i = 0; i < n; i++) {
if (aa[i] < NN)
return 0;
else
sm += aa[i] - NN;
}
if (sm >= b)
return sm;
else
return 0;
}
void _CODE() {
cin >> n >> b;
for (long long i = 0; i < n; i++) cin >> aa[i];
p = -1;
l = 0;
r = 1e9;
while (l <= r) {
m = (l + r) >> 1ll;
if (ch(m))
l = m + 1, p = m;
else
r = m - 1;
}
cout << p;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
_CODE();
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | R = lambda: map(int, input().split())
n, s = R()
v = [*R()]
print(max(-1, min((sum(v) - s) // n, *v)))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.awt.geom.*;
import java.io.*;
import java.math.*;
public class Main
{
public static void main(String[] args) throws Exception
{
long startTime = System.nanoTime();
int n = in.nextInt();
long s = in.nextLong();
long [] data = new long[n];
for(int i=0;i<n;i++)
data[i] = in.nextLong();
if(n==1){
if(data[0]>=s){
out.println((data[0]-s));
exit(0);
}else{
out.println(-1);
exit(0);
}
}
Arrays.sort(data);
long covered = process(data);
if(s<=covered){
out.println(data[0]);
exit(0);
}else{
long pending = s-covered;
if(pending > data[0]*n) {
out.println(-1);
exit(0);
}
else{
long temp = (long)Math.ceil((pending+0.0)/n);
out.println((data[0]-temp));
exit(0);
}
}
long endTime = System.nanoTime();
err.println("Execution Time : +" + (endTime-startTime)/1000000 + " ms");
exit(0);
}
static long process(long [] data){
long min = data[0];
long res = 0;
for(int i=1;i<data.length;i++){
if(data[i]>min){
res += (data[i]-min);
data[i] = min;
}
}
return res;
}
static class InputReader
{
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble(){
return Double.parseDouble(next());
}
}
static void exit(int a)
{
out.close();
err.close();
System.exit(a);
}
static InputStream inputStream = System.in;
static OutputStream outputStream = System.out;
static OutputStream errStream = System.err;
static InputReader in = new InputReader(inputStream);
static PrintWriter out = new PrintWriter(outputStream);
static PrintWriter err = new PrintWriter(errStream);
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.Writer;
import java.math.BigInteger;
import java.util.Scanner;
import java.util.StringTokenizer;
public class temp {
boolean possible(int v[],int min,long s)
{
for(int i=0;i<v.length;i++)
{
if(s==0)
{
if(v[i] < min)
return false;
}
else
{
if(v[i] >= min)
{
if(s >= (v[i]-min))
s-=(v[i]-min);
else
s = 0;
}
else
return false;
}
}
return s==0 ? true:false;
}
void solve() throws IOException
{
FastReader sc = new FastReader();
int n = sc.nextInt();
long s = sc.nextLong();
int v[] = new int[n];
int l = 0,r = -1;
for(int i=0;i<n;i++)
{
v[i] = sc.nextInt();
r = Math.max(r, v[i]);
}
int ans = -1;
while(l<=r)
{
int mid = (l+r)/2;
if(possible(v,mid,s))
{
ans = mid;
l = mid+1;
}
else
r = mid-1;
}
System.out.println(ans);
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
new temp().solve();
}
class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
n, s = (int(s2) for s2 in input().split(' '))
kegs = [int(s2) for s2 in input().split(' ')]
min_keg = min(kegs)
overflow = sum(keg - min_keg for keg in kegs)
if s <= overflow:
print(min_keg)
else:
remaining = s - overflow
n = int(math.ceil(remaining / len(kegs)))
if n > min_keg:
print(-1)
else:
print(min_keg - n)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,s=map(int,input().split())
lis=list(map(int,input().split()))
if s>sum(lis):
print(-1)
exit()
lis.sort(reverse=True)
i=0
while i<n and s>0:
if s<(lis[i]-lis[-1]):
print(lis[-1])
exit()
s-=(lis[i]-lis[-1])
lis[i]=lis[-1]
i+=1
if s==0:
print(lis[-1])
elif s%n==0:
print(lis[-1]-s//n)
else:
print(lis[-1]-s//n-1)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.io.*;
public class Main{
public static long primeFactors(long n)
{
// Print the number of 2s that divide n
while (n%2 == 0)
{
return 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (long i = 3; i <= Math.sqrt(n); i = i+2)
{
// While i divides n, print i and divide n
while (n%i == 0)
{
return i;
}
}
// This condition is to handle the case when n
// is a prime number greater than 2
if (n > 2)
return n;
return 0;
}
public static boolean isPrime(long n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0) return false;
for (long i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader inp = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
Solver solver = new Solver();
solver.solve(inp, out);
out.close();
}
static class Solver {
public boolean possible(long x,long arr[],long s)
{
long sum=0;
for(int i=0;i<arr.length;i++)
{
if(arr[i]-x<0)
return false;
sum+=arr[i]-x;
}
if(sum>=s)
return true;
return false;
}
private void solve(InputReader inp, PrintWriter out1) {
int n = inp.nextInt();
long s = inp.nextLong();
long arr[] = new long[n];
long sum=0;
for(int i=0;i<n;i++)
{
arr[i] = inp.nextLong();
sum+=arr[i];
}
if(sum<s)
{
out1.println(-1);
return;
}
Arrays.sort(arr);
long left =0;
long right = arr[n-1]+5;
long ans =0;
while(left<=right)
{
long mid = (left+right)/2;
// out1.println("mid "+mid);
if(possible(mid,arr,s))
{
left = mid+1;
ans = mid;
}
else
{
right=mid-1;
}
}
out1.println(ans);
}
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
}
}
class ele{
long value;
long i;
boolean flag;
public ele(long value,long i)
{
this.value = value;
this.i=i;
this.flag = false;
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
/**
* Created by Mach on 10.12.2018.
*/
public class round526a {
public static void main(String[] args) {
Reader526a r=new Reader526a();
r.init(System.in);
int n=r.nextInt();
long s=r.nextLong();
int a[]=new int[n];
int min=Integer.MAX_VALUE;
long sum=0;
for(int i=0; i<n; i++) {
a[i]=r.nextInt();
if (a[i]<min) min=a[i];
sum=sum+a[i];
}
long res=-1;
if (s<=(sum-(long)min*n)) {System.out.println(min); return;}
if (s>(sum)) {System.out.println(res); return;}
res=(sum-s)/(long)n;
System.out.println(res);
// 980'103'855'476
}
}
class Reader526a {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(
new InputStreamReader(input) );
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() /* throws IOException */ {
while ( ! tokenizer.hasMoreTokens() ) {
//TODO add check for eof if necessary
try {
tokenizer = new StringTokenizer(
reader.readLine());
} catch(Exception e) {e.printStackTrace();}
}
return tokenizer.nextToken();
}
static int nextInt() {
return Integer.parseInt( next() );
}
static long nextLong(){ return Long.parseLong(next());}
static double nextDouble() {
return Double.parseDouble( next() );
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
long long int power(long long int x, long long int b) {
long long int p = 1;
while (b > 0) {
if (b & 1) {
p = p * x;
p %= 1000000007;
}
b >>= 1;
x *= x;
x %= 1000000007;
}
return p % 1000000007;
}
using namespace std;
struct lex_compare {
bool operator()(
const pair<long long int, pair<long long int, long long int> > p1,
const pair<long long int, pair<long long int, long long int> > p2) {
return (p1.first == p2.first) ? p1.second.first < p2.second.first
: p1.first > p2.first;
}
};
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
;
long long int n, s;
cin >> n >> s;
long long int arr[n];
long long int sum = 0;
long long int mint = INT_MAX;
for (long long int i = 0; i < n; i++) {
cin >> arr[i];
sum += arr[i];
mint = min(arr[i], mint);
}
if (sum < s) {
cout << -1 << "\n";
return 0;
} else if (sum == s) {
cout << 0 << "\n";
return 0;
}
long long int sum2 = sum - n * mint;
if (sum2 >= s) {
cout << mint << "\n";
} else {
s -= sum2;
cout << (n * mint - s) / n << "\n";
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, s, v[1010], t = 0;
cin >> n >> s;
for (long long i = 0; i < n; ++i) {
cin >> v[i];
t += v[i];
}
if (t < s) {
puts("-1");
return 0;
}
sort(v, v + n);
t = 0;
for (int i = 1; i < n; ++i) {
t += v[i] - v[0];
}
if (t >= s) {
cout << v[0] << endl;
return 0;
}
long long k = (s - t) / n;
t += k * n;
if (t < s) ++k;
cout << v[0] - k << endl;
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | // package PCS;
// Working program with FastReader
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Kvass {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
long glass = reader.nextLong();
long min = Long.MAX_VALUE;
long sum = 0;
long poured = 0;
long[] kvass = new long[n];
for (int i=0;i<n;i++) {
long next = reader.nextLong();
sum += next;
if (next < min) min = next;
kvass[i] = next;
}
if (sum < glass) {
System.out.println("-1");
return;
}
// long inEach = (sum - glass) / n;
// System.out.println(inEach);
for (int i=0;i<n;i++) {
poured += kvass[i] - min;
kvass[i] = min;
}
if (poured >= glass) {
System.out.println(min);
} else {
long low = 0;
long high = min;
long mid = 0;
while (low < high) {
mid = (low + high) /2;
// System.out.println(low + "\t" + mid + "\t" + high);
if (poured + (min-mid) * n >= glass) {
low = mid;
if (poured + (min-mid-1) * n < glass) {
break;
}
} else {
high = mid;
}
}
System.out.println(mid);
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | # -*- coding: utf-8 -*-
N, S = map(int, input().split())
aN = list(map(int, input().split()))
if (sum(aN) - S) / N < 0:
print(-1)
else:
print(min((sum(aN) - S) // N, min(aN)))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
BKvassAndTheFairNut solver = new BKvassAndTheFairNut();
solver.solve(1, in, out);
out.close();
}
static class BKvassAndTheFairNut {
long[] v;
long s;
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
s = in.nextLong();
v = in.nextLongArray(n);
long sum = 0;
long min = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
sum += v[i];
min = Math.min(v[i], min);
}
if (sum < s) {
out.println(-1);
} else {
long l = -1;
long r = min + 1;
while (r - l > 1) {
long mid = (l + r) / 2;
if (f(mid)) l = mid;
else r = mid;
}
out.println(l);
}
}
boolean f(long check) {
long fill = 0;
for (int i = 0; i < v.length; i++) {
fill += (v[i] - check);
}
if (fill >= s) return true;
return false;
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public long[] nextLongArray(int size) {
long[] array = new long[size];
for (int i = 0; i < size; i++)
array[i] = nextLong();
return array;
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long a[5000];
int main() {
int n;
long long m;
cin >> n >> m;
long long sum = 0ll;
for (int i = 1; i <= n; i++) cin >> a[i], sum += a[i];
sort(a + 1, a + 1 + n);
long long fuck = 0ll;
for (int i = 2; i <= n; i++) fuck += a[i] - a[1];
m -= fuck;
if (m <= 0) return cout << a[1], 0;
int flag = 0;
if (m % n != 0) flag = 1;
m /= n;
m += flag;
cout << max(-1ll, a[1] - m);
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, s;
cin >> n >> s;
vector<long long> v(n);
long long sum = 0, minimum = 1e9;
for (int i = 0; i < n; i++) {
cin >> v[i];
sum += v[i];
if (v[i] < minimum) minimum = v[i];
}
sum -= s;
if (sum < 0)
cout << -1 << endl;
else
cout << min(minimum, sum / n) << endl;
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, s = map(int, input().split())
k = list(map(int, input().split()))
sumk = sum(k)
mink = min(k)
if sumk < s:
print(-1)
exit()
if sumk-n*mink >= s:
print(mink)
else:
s -= sumk-n*mink
if s%n == 0:
print(mink-s//n)
else:
print(mink-s//n-1) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,s=map(int,input().split())
l=list(map(int,input().split()))
sa=sum(l)
sa-=s
if sa<0:
print(-1)
else:
sa//=n
if min(l)<sa:
print(min(l))
else:
print(sa)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
const char dl = '\n';
const long double eps = 0.00000001;
const long long MOD = 1e9 + 7;
const double PI = 3.141592653589793238463;
using namespace std;
void debug() { cout << endl; }
template <typename H, typename... T>
void debug(H p, T... t) {
std::cout << p << " ";
debug(t...);
}
long long ans;
int n, m, k;
long long second, sum;
int main() {
fflush(stdin);
cout << fixed, cout.precision(18);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int i, j;
cin >> n >> second;
long long a[n];
for (i = 0; i < n; ++i) cin >> a[i], sum += a[i];
if (sum < second) return cout << -1, 0;
long long up = 0;
sort(a, a + n);
long long mi = a[0];
for (i = 0; i < n; ++i) up += (a[i] - mi);
second -= up;
if (second <= 0) return cout << mi, 0;
long long need = second / n;
second %= n;
mi -= need;
if (second) mi--;
cout << mi;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #!/usr/bin/env python2
"""
This file is part of https://github.com/cheran-senthil/PyRival
Copyright 2019 Cheran Senthilkumar <[email protected]>
"""
from __future__ import division, print_function
import math
import itertools
import os
import sys
from atexit import register
from io import BytesIO
class dict(dict):
""" dict() -> new empty dictionary """
def items(self):
""" D.items() -> a set-like object providing a view on D's items """
return dict.iteritems(self)
def keys(self):
""" D.keys() -> a set-like object providing a view on D's keys """
return dict.iterkeys(self)
def values(self):
""" D.values() -> an object providing a view on D's values """
return dict.itervalues(self)
def gcd(x, y):
""" greatest common divisor of x and y """
while y:
x, y = y, x % y
return x
range = xrange
filter = itertools.ifilter
map = itertools.imap
zip = itertools.izip
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))
sys.stdout = BytesIO()
register(lambda: os.write(1, sys.stdout.getvalue()))
input = lambda: sys.stdin.readline().rstrip('\r\n')
def main():
n, s = map(int, input().split())
v = sorted(int(i) for i in input().split())
sum_v = sum(v)
if sum_v < s:
print(-1)
return
min_v = min(v)
max_drink = sum_v - n * min_v
if max_drink >= s:
print(min_v)
return
print(min_v - int(math.ceil((s - max_drink) / n)))
if __name__ == '__main__':
main()
| PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
long long int s;
cin >> s;
long long int arr[n];
long long int sum = 0;
for (int i = 0; i < n; i++) {
cin >> arr[i];
sum += arr[i];
}
sort(arr, arr + n);
if (s > sum) {
cout << "-1" << endl;
} else {
long long int m = 0;
int id = n;
for (int i = 1; i < n; i++) {
if (arr[i] != arr[0]) {
id = i;
break;
}
}
for (int i = id; i < n; i++) {
m = m + arr[i] - arr[0];
}
s = s - m;
if (s <= 0) {
cout << arr[0] << endl;
} else {
long long int ans;
ans = arr[0] - ceil(s / (n + 0.0));
cout << ans << endl;
}
}
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import sys
input = sys.stdin.readline
import math
def main():
n,s = map(int,input().split())
tab = list(map(int,input().split()))
cost = 0
if sum(tab) < s:
print(-1)
return
m = min(tab)
for x in range(n):
cost += tab[x] - m
if cost >= s:
print(m)
return
s -= cost
nb_couches = s//n
if cost == 0:
nb_couches = s//n
if nb_couches == 0 or s == 1:
print(m-1)
else:
if s%nb_couches==0:
print(m-nb_couches)
else:
print(m-nb_couches-1)
main() | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, s = tuple(map(int,input().split()))
v = list(map(int,input().split()))
mn = min(v)
total = 0
possible = 0
for vi in v:
possible += vi - mn
total += vi
if total < s:
print(-1)
elif possible >= s:
print(mn)
else:
remain = s-possible
print( mn - (remain+n-1)//n ) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | def kvas(s, lst):
ss1, ss2 = sum(lst), 0
if ss1 < s:
return -1
a = sorted(lst)
for i in range(len(a)):
ss2 += a[i] - a[0]
if ss2 > s:
return a[0]
return a[0] - (s - ss2 + len(lst) - 1) // len(lst)
N, S = [int(j) for j in input().split()]
b = [int(z) for z in input().split()]
print(kvas(S, b))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,s=map(int,input().split())
a=sorted(list(map(int,input().split())))
k=0
mi=a[0]
for i in range(n):
k+=(a[n-i-1]-a[0])
a[n-i-1]-=(a[n-i-1]-a[0])
if k>=s:
exit(print(a[0]))
else:
if sum(a)<(s-k):
exit(print(-1))
else:
r=(-(-(s-k)//n))
print(a[0]-r) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
public class Main {
static class TaskSolver {
public static void solve(MyScanner in, MyWriter out) throws IOException {
int n = in.nextInt();
long s = in.nextLong();
long[] inp = new long[n];
long low = 0, high = (long) 1e15, ans = -1;
for(int i = 0; i < n; i++) {
inp[i] = in.nextLong();
high = Math.min(high, inp[i]);
}
while(low <= high) {
long mid = (low + high) / 2;
long hand = 0;
for(int i = 0; i < n; i++) {
if(inp[i] >= mid) {
hand += inp[i] - mid;
}
}
if(hand >= s) {
ans = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
out.println(ans);
}
}
public static void main(String[] arg) {
try {
MyScanner in = new MyScanner();
MyWriter out = new MyWriter();
TaskSolver.solve(in, out);
out.close();
} catch (IOException e) {
e.printStackTrace();
}
}
static class MyWriter extends PrintWriter {
MyWriter() throws FileNotFoundException {
super(System.out);
}
void printArrayInOneLine(int[] a) {
print('[');
for (int i = 0; i < a.length; i++) {
print(a[i] + " ");
}
println(']');
}
}
static class MyScanner {
private BufferedReader br;
private StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
return null;
}
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
try {
return br.readLine();
} catch (Exception e) {
e.printStackTrace();
}
return null;
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.*;
public class Main {
static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
public static void main(String[] args) throws IOException {
int n = readInt();
long s = readLong();
int minVolume = Integer.MAX_VALUE;
long total = 0;
for (int index = 0; index < n; index++) {
int volume = readInt();
total += volume;
minVolume = Math.min(minVolume, volume);
}
System.out.println(total - s >= 0 ? (total - s) / n < minVolume ? (total - s) / n : minVolume : "-1");
}
static long readLong() throws IOException {
in.nextToken();
return (long) in.nval;
}
static int readInt() throws IOException {
in.nextToken();
return (int) in.nval;
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #!/usr/bin/env python
# -*- coding: utf-8 -*-
# File : b.py
# Author : recurze
# Date : 22:14 10.12.2018
# Last Modified Date: 22:18 10.12.2018
n, s = [int(x) for x in raw_input().split()]
a = list(map(int, raw_input().split()))
su = sum(a)
if su < s:
print -1
else:
m = min(a)
if su - m*n >= s:
print m
else:
s -= su - m*n
from math import ceil
print int(m-ceil(float(s)/n))
| PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
long long int v[1001];
int main() {
int i, j, k, n, t, l;
long long int min = 1000000001;
long long int s, r, d, total = 0;
scanf("%d %lld", &n, &s);
for (int a = 0; a < n; a++) {
scanf("%lld", &v[a]);
if (min > v[a]) min = v[a];
total += v[a];
}
if (s > total)
printf("-1\n");
else if (s == total)
printf("0\n");
else {
if (((total - s) / n) >= min)
printf("%lld\n", min);
else
printf("%lld\n", ((total - s) / n));
}
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long n, s;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
t = 1;
while (t--) {
cin >> n >> s;
vector<long long> v(n);
long long sum = 0;
long long mini = INT_MAX;
for (long long i = 0; i < n; i++) {
cin >> v[i];
sum += v[i];
mini = min(mini, v[i]);
}
if (sum < s) {
cout << -1;
continue;
}
sum = 0;
for (long long i = 0; i < n; i++) {
sum += (v[i] - mini);
}
if (sum >= s) {
cout << mini;
continue;
}
s -= sum;
long long ans = mini;
while (s > 0) {
s -= n;
ans--;
}
cout << ans;
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long n, m, mn, res, sum;
cin >> n >> m;
long long arr[n];
sum = 0;
mn = INT_MAX;
for (int i = 0; i < n; i++) {
cin >> arr[i];
mn = min(arr[i], mn);
sum += arr[i];
}
res = mn;
if (sum < m) {
cout << "-1";
return 0;
}
if (sum == 0) {
cout << "0";
return 0;
}
for (int i = 0; i < n; i++) {
m = m - (arr[i] - mn);
if (m <= 0) break;
}
if (m > 0) {
{ mn = mn - ((m + n - 1) / n); }
}
cout << mn << "\n";
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
bool comp(pair<int, int> &a, pair<int, int> &b) {
if (a.second == b.second)
return a.first < b.first;
else
return a.second < b.second;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int N;
long long S;
cin >> N >> S;
vector<long long> a(N);
long long sum1 = 0ll, sum2 = 0ll;
for (int i = 0; i < N; ++i) {
cin >> a[i];
sum1 += a[i];
}
if (sum1 < S) {
cout << -1;
return 0;
}
sort(a.begin(), a.end());
for (int i = 0; i < N; ++i) {
sum2 += a[i] - a[0];
}
if (sum2 >= S) {
cout << a[0];
return 0;
}
S -= sum2;
cout << a[0] - ((S - 1ll) / (long long)N + 1ll);
return 0;
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,s = map(int,input().split())
l = list(map(int,input().split()))
l.sort(reverse=True)
if sum(l)<s:
print(-1)
exit()
cnt = 0
z = l[-1]
for i in range(n):
if s == 0:
break
k = l[i]-z
if s>=k:
s-=k
else:
print(z)
exit()
ka = s//n
ans = z-ka
if s%n == 0:
print(ans)
else:
print(ans-1) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,k = [int(j) for j in input().split()]
a = [int(j) for j in input().split()]
t = min(a)
summ = 0
for i in range(n):
summ += a[i]
psumm = summ-t*n
ram = (k-psumm)
if ram%n == 0:
ans = t-(ram//n)
else:
ans = t-1-(ram//n)
if k>summ:
print(-1)
elif k<=psumm:
print(t)
else:
print(ans) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | '''input
3 4
5 3 4
'''
import math
from fractions import gcd
n,s=map(int,raw_input().split())
v=map(int,raw_input().split())
v.sort()
if sum(v)<s:
print -1
else:
for i in range(n):
if v[i]-v[0]>=s:
print v[0]
exit(0)
else:
s-=(v[i]-v[0])
v[i]=v[0]
k=s/n
z=v[0]
z-=k
s-=k*n
if(s==0):
print z
else:
print z-1
| PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
n,s = map(int, input().split(" "))
ar = list(map(int, input().split(" ")))
sm = sum(ar)
if s > sm:
print(-1)
else:
mn = min(ar)
ex = (sm - mn) - ((n-1)*mn)
if ex >= s:
print(mn)
else:
print(mn - math.ceil((s - ex)/n)) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
long long int n, s;
vector<long long int> v;
bool check(long long int val) {
long long int bebe = 0;
for (auto x : v) {
if (x < val) return false;
bebe += abs(x - val);
}
return bebe >= s;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> s;
long long int tot = 0;
for (int i = 1; i <= n; i++) {
long long int x;
cin >> x;
v.push_back(x);
tot += x;
}
if (tot < s) {
cout << -1 << endl;
return 0;
}
sort(v.begin(), v.end());
long long int ini = 0;
long long int end = 1e9;
long long int best = -INF;
while (ini <= end) {
long long int mid = (ini + end) >> 1;
if (check(mid)) {
best = max(best, mid);
ini = mid + 1;
} else
end = mid - 1;
}
cout << best << endl;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | [n, s] = [int(x) for x in input().split(' ')]
v = [int(x) for x in input().split(' ')]
v.sort()
minV = v[0]
totalV = sum(v)
if totalV < s:
print(-1)
else:
if totalV - minV * n >= s:
print(minV)
else:
remain = totalV - s
print(remain//n) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long n, ans = 0, acum = 0, s, tonel;
cin >> n >> s >> ans;
acum = ans;
for (int i = 1; i < n; ++i) {
cin >> tonel;
acum += tonel;
ans = min(ans, tonel);
}
if (acum < s)
ans = -1;
else if (acum == s)
ans = 0;
else {
s -= (acum - ans * n);
while (s > 0) {
ans--;
s -= n;
}
}
cout << ans << "\n";
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class codeforces
{
static class Student{
int a,s,gd,gs;
Student(int a,int b,int c,int d){
this.a=a;
this.s=b;
gd=c;
gs=d;
}
}
static int sieveOfEratosthenes(int n)
{
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
int pos=0;
boolean prime[] = new boolean[n+1];
for(int i=0;i<=n;i++)
prime[i] = true;
for(int p = 2; p*p <=n; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == true)
{
// Update all multiples of p
for(int i = p*p; i <= n; i += p)
prime[i] = false;
}
}
// Print all prime numbers
for(int i = 2; i <= n; i++)
{
if(prime[i] == true)
++pos;
}
return pos;
}
static class Sortbyroll implements Comparator<Student>
{
// Used for sorting in ascending order of
// roll number
public int compare(Student c, Student b)
{
if(c.s>b.s)
return -1;
else if(c.s==b.s){
if(c.gd>b.gd)
return -1;
else if(c.gd==b.gd){
if(c.gs>b.gs)
return -1;
else if(c.gs==b.gs){
if(c.a>b.a)
return -1;
else
return 1;
}
}
else
return 1;
}
return 1;
}
}
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args){
//long sum=0;
int t,n,d,i,q,p,r,pos=0,j,c_0=-1,c_1=-1,si=0,ans2=0,e,x,mid,min,m,h,c_2,c,a1,b;
long ans=0,ans1=0,s;
String s1;
FastReader sc=new FastReader();
n=sc.nextInt();
s=sc.nextLong();
long a[]=new long[n];
for(i=0;i<n;i++)
a[i]=sc.nextLong();
Arrays.sort(a);
for(i=n-1;i>=0;i--){
s-=a[i]-a[0];
}
if(s<=0)
System.out.println(a[0]);
else{
if(s>n*a[0])
System.out.println(-1);
else
{
if(s%(long)n==0)
System.out.println(a[0]-s/(long)n);
else
System.out.println(a[0]-s/(long)n-1);
}
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
public class KVASS {
public static void main(String[] args)throws IOException {
BufferedReader ob=new BufferedReader(new InputStreamReader(System.in));
String s[]=ob.readLine().split(" ");
int n=Integer.parseInt(s[0]);
long k=Long.parseLong(s[1]);
int v[]=new int[n];
String s1[]=ob.readLine().split(" ");
int min=Integer.MAX_VALUE;
long total=0;
for(int i=0;i<n;i++) {
v[i]=Integer.parseInt(s1[i]);
total+=v[i];
if(v[i]<min)
min=v[i];
}
// System.out.println("total = "+total);
if(total<k) {
System.out.println("-1");
return;
}
long temp=0;
for(int i=0;i<n;i++)
temp+=(long)(v[i]-min);
// System.out.println("temp = "+temp);
if(temp>=k) {
System.out.println(min);
return;
}
k=k-temp;
// System.out.println(" k after temp = "+k);
long x=k/n;
if(k%n!=0)
x++;
// System.out.println("x = "+x);
total=min-x;
System.out.println(total);
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int n;
int v[1005];
long long s, sum;
bool cmp(int x, int y) { return x > y; }
int main() {
scanf("%d %lld", &n, &s);
for (int i = 1; i <= n; i++) scanf("%d", &v[i]), sum += v[i];
if (sum < s) {
printf("-1\n");
return 0;
}
sort(v + 1, v + n + 1, cmp);
for (int i = 1; i <= n; i++) {
long long vi = (long long)i * (v[i] - v[i + 1]);
if (s > vi)
s -= vi;
else {
if (i == n)
printf("%d\n", v[i] - (s + i - 1) / i);
else
printf("%d\n", v[n]);
return 0;
}
}
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
n,s=map(int,input().split())
li=input().split()
for i in range(n):
li[i]=int(li[i])
su=sum(li)
if su<s:
print(-1)
exit(0)
if su-s<n:
print(0)
exit(0)
li.sort()
k=(su-s)/n
k=int(math.floor(k))
if k>li[0]:
k=li[0]
print(k) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, s = map(int, input().split())
l = [int(i) for i in input().split()]
'''
for i in range(n):
x = max(l) - s // n
l.remove(max(l))
l.append(x)
for i in range(s % n):
x = max(l) - 1
l.remove(max(l))
l.append(x)
if min(l) < 0:
print(-1)
else:
print(min(l))
'''
if sum(l) < s:
print(-1)
else:
print(min( [(sum(l)-s)//n, min(l)] )) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | l = raw_input().split()
n, s = int(l[0]), int(l[1])
l = raw_input().split()
add = 0
m = 10**18
for i in range(n):
l[i] = int(l[i])
add += l[i]
m = min(m, l[i])
if add < s:
print -1
elif add == s:
print 0
else:
excess = 0
for i in range(n):
excess += (l[i]-m)
if excess >= s:
print m
else:
ans = m-(s-excess)/n
if (s-excess)%n == 0:
print ans
else:
print ans-1 | PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.io.*;
public class B{
public static void main(String[] args)
{
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int n = fs.nextInt(); long s = fs.nextLong();
long[] arr = new long[n];
long sum = 0;
long min = Long.MAX_VALUE;
for(int i=0;i<n;i++)
{
arr[i] = fs.nextLong();
sum += arr[i];
min = Math.min(min,arr[i]);
}
if(sum < s)
{
out.println("-1");
}
else
{
long my = 0;
for(long i:arr)
{
my += i-min;
}
long more = s-my;
long times = (more+n-1)/n;
long ans = min - Math.max(0,times);
out.println(ans);
}
out.close();
}
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
public static int[] sort(int[] arr)
{
List<Integer> temp = new ArrayList();
for(int i:arr)temp.add(i);
Collections.sort(temp,Collections.reverseOrder());
int start = 0;
for(int i:temp)arr[start++]=i;
return arr;
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*
* @author Mufaddal Naya
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
BKvassAndTheFairNut solver = new BKvassAndTheFairNut();
solver.solve(1, in, out);
out.close();
}
static class BKvassAndTheFairNut {
public void solve(int testNumber, InputReader in, PrintWriter out) {
int n = in.nextInt();
long s = in.nextLong();
int a[] = new int[n];
for (int i = 0; i < n; i++) a[i] = in.nextInt();
int min = a[0];
for (int i = 0; i < n; i++) if (a[i] < min) min = a[i];
long k = 0;
for (int i = 0; i < n; i++) {
k += a[i] - min;
}
if (k >= s) {
out.println(min);
return;
}
double g = s - k;
g /= n;
long kt = (long) Math.ceil(g);
if (kt <= min) {
out.println((min - kt));
} else {
out.println("-1");
}
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long ar[1001];
int main() {
long long int n, s;
cin >> n >> s;
for (int i = 1; i <= n; i++) cin >> ar[i];
long long sum = 0;
for (int i = 1; i <= n; i++) sum += ar[i];
if (sum < s) {
cout << -1 << endl;
exit(0);
}
long long m = *min_element(ar + 1, ar + n + 1);
long long diff = sum - m * n;
if (diff > s)
cout << m << endl;
else {
s -= diff;
int h;
if (s % n == 0)
h = s / n;
else
h = s / n + 1;
cout << m - h << endl;
}
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long n, s;
vector<long long> v(1001);
long long check(long long x) {
long long sum = 0;
for (long long i = 0; i < n; i++) {
if (v[i] < x) return 0;
sum += max((long long)0, v[i] - x);
}
return (sum >= s);
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
cin >> n >> s;
for (long long i = 0; i < n; i++) cin >> v[i];
long long l = 0;
long long r = 1e15;
while (l < r) {
long long mid = (l + r + 1) / 2;
if (check(mid))
l = mid;
else
r = mid - 1;
}
if (check(l))
cout << l;
else
cout << -1;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import sys
input = sys.stdin.readline
n,s=map(int,input().split())
V=list(map(int,input().split()))
MIN=min(V)
SUM=sum(V)
if s>SUM:
print(-1)
sys.exit()
if SUM-MIN*n>=s:
print(MIN)
sys.exit()
print((SUM-s)//n)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class KvassAndTheFairNut {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader sc = new FastReader();
int n = sc.nextInt();
long s = sc.nextLong();
long sum = 0;
long[] a = new long[n];
for(int i=0; i<n; i++)
{
a[i] = sc.nextLong();
sum = sum + a[i];
}
if(sum<s)
{
System.out.println("-1");
}
else
{
Arrays.sort(a);
sum = 0;
for(int i=1; i<n; i++)
{
sum = sum + a[i]-a[0];
}
if(sum>=s)
{
System.out.println(a[0]);
}
else
{
s = s-sum;
long rem = (long)Math.ceil((double)s/(double)n);
System.out.println(a[0]-rem);
}
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | /*
_oo0oo_
o8888888o
88" . "88
(| -_- |)
0\ = /0
___/`---'\___
.' \\| |// '.
/ \\||| : |||// \
/ _||||| -:- |||||- \
| | \\\ - /// | |
| \_| ''\---/'' |_/ |
\ .-\__ '-' ___/-. /
___'. .' /--.--\ `. .'___
."" '< `.___\_<|>_/___.' >' "".
| | : `- \`.;`\ _ /`;.`/ - ` : | |
\ \ `_. \_ __\ /__ _/ .-` / /
=====`-.____`.___ \_____/___.-`___.-'=====
`=---='
*/
import java.util.function.Consumer;
import java.util.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.io.*;
import java.lang.Math.*;
public class KickStart2020{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){br = new BufferedReader(
new InputStreamReader(System.in));}
String next(){
while (st == null || !st.hasMoreElements()) {
try { st = new StringTokenizer(br.readLine()); }
catch (IOException e) {e.printStackTrace();}}
return st.nextToken();}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next());}
float nextFloat() {return Float.parseFloat(next());}
String nextLine() {
String str = "";
try {str = br.readLine();}
catch (IOException e) { e.printStackTrace();}
return str; }}
static boolean isBracketSequence(String s, int a, int b) {
Stack<Character> ss = new Stack<>();
boolean hachu = true;
for(int i = a; i <= b; i++) {
if(s.charAt(i) == ')' && ss.isEmpty()) {hachu = false; break;}
if(s.charAt(i) == '(') ss.add('(');
else ss.pop();
}
return ss.empty() && hachu;
}
static String reverseOfString(String a) {
StringBuilder ssd = new StringBuilder();
for(int i = a.length() - 1; i >= 0; i--) {
ssd.append(a.charAt(i));
}
return ssd.toString();
}
static char[] reverseOfChar(char a[]) {
char b[] = new char[a.length];
int j = 0;
for(int i = a.length - 1; i >= 0; i--) {
b[i] = a[j];
j++;
}
return b;
}
static boolean isPalindrome(char a[]) {
boolean hachu = true;
for(int i = 0; i <= a.length / 2; i++) {
if(a[i] != a[a.length - 1 - i]) {
hachu = false;
break;
}
}
return hachu;
}
static long gcd(long a, long b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long powermod(long x, long y, long mod){
long ans = 1;
x = x % mod;
if (x == 0)
return 0;
int i = 1;
while (y > 0){
if ((y & 1) != 0)
ans = (ans * x) % mod;
y = y >> 1;
x = (x * x) % mod;
}
return ans;
}
static long power(long x, long y){
long ans = 1;
if (x == 0)
return 0;
int i = 1;
while (y > 0){
if ((y & 1) != 0)
ans = (ans * x);
y = y >> 1;
x = (x * x);
}
return ans;
}
static boolean check(String a) {
boolean hachu = true;
for(int i = 0; i < a.length(); i++) {
if(a.charAt(0) != a.charAt(i)) {hachu = false; break;}
}
return hachu;
}
public static class Pair implements Comparable<Pair> {
public final int index;
public final int value;
public Pair(int index, int value) {
this.index = index;
this.value = value;
}
@Override
public int compareTo(Pair other) {
//multiplied to -1 as the author need descending sort order
return -1 * Integer.valueOf(this.value).compareTo(other.value);
}
}
static boolean equalString(int i, int j, int arr[], String b) {
int brr[] = new int[26];
for(int k = i; k <= j; k++) brr[b.charAt(k) - 'a']++;
for(int k = 0; k < 26; k++) {
if(arr[k] != brr[k]) return false;
}
return true;
}
static boolean cequalArray(String a, String b) {
int count[] = new int[26];
int count1[] = new int[26];
for(int i = 0; i < a.length(); i++) count[a.charAt(i) - 'a']++;
for(int i = 0; i < a.length(); i++) count1[b.charAt(i) - 'a']++;
for(int i = 0; i < 26; i++) if(count[i] != count1[i]) return false;
return true;
}
public static void main(String[] args) throws Exception{
FastReader sc = new FastReader();
PrintWriter out = new PrintWriter(System.out);
int n = sc.nextInt();
long s = sc.nextLong();
int min = Integer.MAX_VALUE;
int arr[] = new int[n];
long sum = 0;
for(int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
min = Math.min(arr[i], min);
sum += arr[i];
}
if(sum < s)out.println(-1);
else {
if(sum - (long)min * n >= s) out.println(min);
else {
long z = s - sum + (long) min * n;
long x = z / n;
long d = z % (long)n;
if(d == 0) out.println(min - x);
else out.println(min - x - 1);
}
}
out.close();
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | # -*- coding: utf-8 -*-
"""
Created on Wed Dec 12 11:09:39 2018
@author: pc
"""
from math import ceil
n,s=map(int,input().split())
x=[int(v) for v in input().split()]
t=sum(x)
m=min(x)
if t<s:
print(-1)
else:
if t-n*m>=s:
print(m)
else:
l=ceil((s-t+n*m)/n)
print(m-l) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import sys
def leastKeg(kegs,s):
if sum(kegs)<s:
return(-1)
kegs.sort()
kegs.reverse()
mini=kegs[-1]
# print(kegs)
for i in xrange(len(kegs)):
if s>0 :
if kegs[i]-mini>=s:
s=0
kegs[i]=kegs[i]-s
return(mini)
else:
s-=(kegs[i]-mini)
kegs[i]=mini
kegs[0]-=s/len(kegs)
s-=len(kegs)*(s/len(kegs))
if s!=0:
mini=kegs[0]-1
else:
mini=kegs[0]
return(mini)
[n,s]=list(map(int,raw_input().split()))
kegs=list(map(int,raw_input().split()))
print(leastKeg(kegs,s)) | PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.Scanner;
public class Codeforces2{
private static int n;
private static long s;
private static int[] arr;
private static Scanner in;
public static long findDoable(){
int min=Integer.MAX_VALUE;
for(int i=0;i<n;i++){
if(arr[i] < min){
min=arr[i];
}
}
long got=0;
long temp=0;
for(int i=0;i<n;i++){
if(arr[i]==min){
continue;
}
temp=(long)arr[i]-min;
got+=temp;
if(got >= s){
return (long)min;
}
}
//System.out.println("min "+min);
/*for(int k=min;k>0;k--){
got+=n;
//System.out.println("got "+got);
if(got >= s){
return k-1;
}
}*/
long minL=(long) min;
long r,q;
long left=s-got;
r=left%n;
q=left/n;
if(q > minL ){
return -1;
}
else if(r==0){
return minL-q;
}
else if(r > 0){
if(minL-q == 0)
return -1;
return minL-q-1;
}
return -1;
}
public static void main(String[] args){
in=new Scanner(System.in);
n=in.nextInt();
s=in.nextLong();
arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=in.nextInt();
}
System.out.println(findDoable());
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long n, s, a[1001];
int main() {
while (cin >> n >> s) {
long long maxa = 0, sum = 0, mina = 0x3f3f3f3f;
for (int i = 0; i < n; i++) {
cin >> a[i];
maxa = max(a[i], maxa);
mina = min(mina, a[i]);
sum += a[i];
}
long long l = 0, r = maxa;
while (l + 1 < r) {
long long mid = (l + r) / 2;
if (sum - mid * n >= s)
l = mid;
else
r = mid;
}
while (r >= 0 && sum - r * n < s) {
r--;
}
cout << min(r, mina) << endl;
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
bool P(long long r, vector<long long> &v, long long &s) {
int i;
long long sum = 0;
for (i = 0; i < v.size(); i++)
if (v[i] - r >= 0) sum += v[i] - r;
if (sum >= s) return true;
return false;
}
int main() {
long long n, s, i, sum = 0;
cin >> n >> s;
vector<long long> v(n);
for (i = 0; i < n; i++) {
cin >> v[i];
sum += v[i];
}
if (sum < s) {
cout << -1;
return 0;
}
long long l = 0, h = *min_element(v.begin(), v.end()), ans, mid;
while (l <= h) {
mid = (l + h) / 2;
if (P(mid, v, s)) {
ans = mid;
l = mid + 1;
} else
h = mid - 1;
}
cout << ans;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
inline char gch() {
static char buf[100010], *h = buf, *t = buf;
return h == t && (t = (h = buf) + fread(buf, 1, 100000, stdin), h == t)
? EOF
: *h++;
}
inline void re(long long &x) {
x = 0;
char a;
bool b = 0;
while (!isdigit(a = getchar())) b = a == '-';
while (isdigit(a)) x = x * 10 + a - '0', a = getchar();
if (b == 1) x = -x;
}
long long n, s, ans = -1, a[1010];
inline bool check(long long x) {
long long rt = 0;
for (register int i = 1; i <= n; i++) {
if (a[i] < x) return 0;
rt += (a[i] - x);
}
return rt >= s;
}
int main() {
re(n), re(s);
long long l = 0, r = 0;
for (register int i = 1; i <= n; i++) re(a[i]), r += a[i];
while (l <= r) {
long long mid = (l + r) >> 1;
if (check(mid))
ans = mid, l = mid + 1;
else
r = mid - 1;
}
cout << ans;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
long long int t, n, m, j, ans, k, a, b, c, d, e, f, sum, i, sz;
string s, s2, s3, s4;
vector<long long int> v;
int ar[(int)(1e6 + 10)], ar2[(int)(1e6 + 10)];
void brainfuck();
int main() {
ios_base::sync_with_stdio(NULL);
cin.tie(NULL);
cout.tie(NULL);
brainfuck();
return 0;
}
void brainfuck() {
cin >> n >> m;
b = INT_MAX;
for (i = 1; i <= n; i++) {
cin >> a;
v.push_back(a);
b = min(a, b);
sum += a;
}
if (sum < m) {
cout << "-1";
return;
}
a = 0;
for (i = 1; i <= n; i++) {
a += abs(v[i - 1] - b);
v[i - 1] = b;
}
if (m <= a) {
cout << b;
return;
}
c = m - a;
d = c / n;
if (!d) d = 1;
if (d * n >= c) {
cout << v[0] - d;
return;
} else
cout << v[0] - d - 1;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, size = list(map(int, input().split()))
array = list(map(int, input().split()))
if (sum(array) - size) < 0:
print(-1)
else:
print(min(min(array), (sum(array) - size) // n))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.*;
import java.util.*;
import java.math.*;
public
class Main{
//static variable
static final int mod = (int) 1e9 + 7;
static final double eps = 1e-6;
static final double pi = Math.PI;
static final long inf = Long.MAX_VALUE / 2;
// .......static class
static class Pair{
int key,value;
Pair(int key,int value){
this.key=key;
this.value=value;
}
int Key(){
return key;
}
int Value(){
return value;
}
}
static class mycomparator implements Comparator<Pair>{
@Override
public int compare(Pair o1,Pair o2){
Integer key1=o1.Key(),key2=o2.Key();
return key1.compareTo(key2);
}
}
//.............staic class end.
BufferedReader br;
PrintWriter out;
public static void main(String[] args) {
new Main().main1();
}
void main1()
{
try{
br=new BufferedReader(new InputStreamReader(System.in));
out=new PrintWriter(System.out);
int t=1;
// t=ii();
while(t-->0){
//........solution start
long ns[] = ilong();
long n = ns[0] , s = ns[1];
long a[] = ilong();
long min = getmin(a);
long sum = 0;
for(int i=0;i<n;i++){
sum+=a[i];
}
if(sum<s){
System.out.println("-1");
continue;
}
long ans = findSol(0,min,sum,n,s);
System.out.println(ans);
//..........solution end.
}
out.flush();
out.close();
}
catch(Exception e){
e.printStackTrace();}
}
// ...............required method.
long findSol(long l , long r ,long sum,long n , long s){
long mid = 0;
long ans = -1;
while(l<=r){
mid = l+(r-l)/2;
long exp = calcExpire(mid,sum,n);
if(exp >= s){
ans = Math.max(ans,mid);
l=mid+1;
}
else {
r=mid-1;
}
}
if(ans<0) ans=0;
return ans;
}
int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b%a, a);
}
long calcExpire(long mid , long sum , long n){
long temp = n*mid;
return (sum - temp);
}
//................end.
//..............input method start.
int getmax(int a[]){
int n=a.length;
int max=a[0];
for (int i=1;i<n ;i++ ) {
max=Math.max(a[i],max);
}
return max;
}
long getmin(long a[]){
int n=a.length;
long min=a[0];
for (int i=1;i<n ;i++ ) {
min=Math.min(a[i],min);
}
return min;
}
int[] iint() throws IOException{
String line[]=br.readLine().split(" ");
int[] a=new int[line.length];
for (int i=0;i<line.length ;i++ ) {
a[i]=Integer.parseInt(line[i]);
}
return a;
}
long[] ilong() throws IOException{
String line[]=br.readLine().split(" ");
long[] a=new long[line.length];
for (int i=0;i<line.length ;i++ ) {
a[i]=Long.parseLong(line[i]);
}
return a;
}
double[] idouble() throws IOException{
String line[]=br.readLine().split(" ");
double[] a=new double[line.length];
for (int i=0;i<line.length ;i++ ) {
a[i]=Long.parseLong(line[i]);
}
return a;
}
long li() throws IOException{
return Long.parseLong(br.readLine());
}
int ii() throws IOException{
return Integer.parseInt(br.readLine());
}
double di() throws IOException{
return Double.parseDouble(br.readLine());
}
char ci() throws IOException{
return (char)br.read();
}
String si() throws IOException{
return br.readLine();
}
String[] isa(int n) throws IOException{
String at =si();
return at.split(" ");
}
double[][] idm(int n, int m) throws IOException{
double a[][] = new double[n][m];
for (int i = 0; i < n; i++) {
double[] temp=idouble();
for (int j = 0; j < m; j++) a[i][j] = temp[j];
}
return a;
}
int[][] iim(int n, int m) throws IOException{
int a[][] = new int[n][m];
for (int i = 0; i < n; i++) {
int[] temp=iint();
for (int j = 0; j < m; j++) a[i][j] =temp[j];
}
return a;
}
long[][] ilm(int n, int m) throws IOException{
long a[][] = new long[n][m];
for (int i = 0; i < n; i++) {
long[] temp=ilong();
for (int j = 0; j < m; j++) a[i][j] =temp[j];
}
return a;
}
//..............input methods end;
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
def main():
n,s = map(int,input().split())
volume = list(map(int,input().split()))
if s > sum(volume):
print(-1)
return
min_val = min(volume)
for i in range(n):
diff = volume[i]-min_val
if s >= diff:
volume[i] = min_val
s -= diff
else:
volume[i] -= s
s = 0
if s == 0:
break
if s > 0:
remove1 = s//n
remove2 = s%n
#print(remove1,remove2)
print(min_val-remove1-math.ceil(remove2/n))
else:
print(min_val)
main()
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long inf = (long long)1e18;
long long mod = 1e9 + 7;
long long max1 = (long long)1e9;
int main() {
long long i, j, k, n, m, ct = 0, t, ans = 0;
cin >> n >> k;
long long sum1 = 0;
long long a[n], min1 = 1e18;
for (i = 0; i < n; i++) {
cin >> a[i];
sum1 += a[i];
if (min1 > a[i]) {
min1 = a[i];
j = i;
}
}
if (sum1 < k)
cout << "-1";
else {
long long sum = 0;
for (i = 0; i < n; i++) {
if (a[i] != min1) {
sum += (a[i] - min1);
a[i] = min1;
}
}
long long low = 0, high = min1;
while (low <= high) {
long long mid = (low + high) / 2;
if (sum + mid * n >= k) {
ans = mid;
high = mid - 1;
} else
low = mid + 1;
}
cout << a[0] - ans << endl;
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, s = map(int, input().split())
kvas = list(map(int, input().split()))
if s > sum(kvas):
print(-1)
else:
if s <= sum(kvas) - n*min(kvas):
print(min(kvas))
else:
print(min(kvas) - 1 - (s - sum(kvas) + n*min(kvas) - 1)//n) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 |
import java.util.Scanner;
public class Fairnut2
{
public int findmin(int v[],int n)
{
int min = 0;
for(int i=1;i<n;i++)
{
if (v[i]<v[min])
{
min=i;
}
}
return min;
}
public static void main(String[] args)
{
Fairnut2 obj=new Fairnut2();
Scanner sc = new Scanner(System.in);
int n =sc.nextInt();
long s =sc.nextLong();
int []v= new int[n];
long total=0;
for(int i=0;i<n;i++)
{
v[i]= sc.nextInt();
total=total+v[i];
}
int vmin=v[obj.findmin(v,n)];
if(total-s<0)
{
System.out.println(-1);
}
else if(((total-s)/n)>=vmin)
{
System.out.println(vmin);
}
else
{
vmin= (int) ((total-s)/n);
System.out.println(vmin);
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,s = map(int, raw_input().split())
l = map(int, raw_input().split())
if sum(l) < s:
print -1
else:
if s <= sum(l)-min(l)*n:
print min(l)
else:
print (sum(l)-s)/n | PYTHON |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 |
import java.util.Scanner;
public class KvassAndTheFairNut
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int numKegs = in.nextInt();
long vol = in.nextLong();
int min = Integer.MAX_VALUE;
for(int i = 0; i < numKegs; i++)
{
int tmp = in.nextInt();
vol -= tmp;
min = min > tmp ? tmp : min;
}
if(vol > 0)
{
System.out.println("-1");
}
else
{
System.out.println(Math.min(vol/numKegs*-1, min));
}
}
}
| JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t;
t = 1;
while (t--) solve();
return 0;
}
void solve() {
long long i, j, n, s;
cin >> n >> s;
vector<long long> a(n);
long long small, tot = 0;
for (i = 0; i < n; ++i) {
cin >> a[i];
tot += a[i];
}
small = *min_element(a.begin(), a.end());
if (s > tot) {
cout << -1;
return;
}
for (i = 0; i < n; ++i) s -= a[i] - small;
if (s <= 0) {
cout << small;
return;
}
if (s % n)
cout << small - (s / n) - 1;
else
cout << small - s / n;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long n, s;
cin >> n >> s;
long long a[n];
long long sumofa = 0;
long long minofa = INT_MAX;
for (long long i = 0; i < n; i++) {
cin >> a[i];
sumofa += a[i];
if (minofa > a[i]) {
minofa = a[i];
}
}
if (sumofa < s) {
cout << -1 << endl;
} else if (sumofa == s) {
cout << 0 << endl;
} else {
long long temp = sumofa - (n * minofa);
if (temp >= s) {
cout << minofa << endl;
} else {
long long cnt = s - temp;
long long res = ((n * minofa) - cnt) / n;
cout << res << endl;
}
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | a,b=map(int,input().split())
z=list(map(int,input().split()))
s=sum(z);r=min(z)
if b>s:print(-1)
else:print(min(r,(s-b)//a)) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
template <typename X>
inline X sqr(const X& a) {
return a * a;
}
int nxt() {
int x;
cin >> x;
return x;
}
int main() {
int n = nxt();
long long s;
cin >> s;
vector<int> a(n);
generate(a.begin(), a.end(), nxt);
if (accumulate(a.begin(), a.end(), 0ll) < s) {
cout << -1 << '\n';
return 0;
}
int mn = *min_element(a.begin(), a.end());
for (int i = 0; i < int(n); ++i) {
s -= a[i] - mn;
}
s = max(s, 0ll);
cout << mn - (s + n - 1) / n << '\n';
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long num[1005];
int main() {
long long n, k;
while (~scanf("%lld%lld", &n, &k)) {
long long minn = 0x3f3f3f3f, sum = 0;
for (int i = 0; i < n; i++) {
scanf("%lld", &num[i]);
minn = min(minn, num[i]);
sum += num[i];
}
if (sum < k)
printf("-1\n");
else if (sum == k)
printf("0\n");
else {
long long ans = 0;
for (int i = 0; i < n; i++) ans += num[i] - minn;
if (ans >= k)
printf("%lld\n", minn);
else {
long long tmp = ceil((k - ans) * 1.0 / n);
printf("%lld\n", minn - tmp);
}
}
}
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.util.*;
import java.io.*;
public class a{
static Scanner sc=new Scanner(System.in);
public static void main(String [] args){
int n=sc.nextInt();
long s=sc.nextLong();
long [] v=new long[n];
long sum=0;
long min=Integer.MAX_VALUE;
long max=Integer.MIN_VALUE;
for(int i=0;i<n;i++){
v[i]=sc.nextLong();
sum+=v[i];
min=Math.min(min,v[i]);
max=Math.max(max,v[i]);
}
if(sum<s){
System.out.println(-1);
}
else{
long diff=0;
for(int i=0;i<n;i++){
diff+=v[i]-min;
}
long ans=1;
if(diff>=s){
ans=min;
}
else{
s=s-diff;
long x=s/n;
if(x<=min){
ans=min-x;
min-=x;
if(s%n!=0){
if(min>0){
ans-=1;
}else{
ans=-1;
}
}
}
else{
ans=-1;
}
}
System.out.println(ans);
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
long long N, s, v[1010];
int main() {
cin >> N >> s;
long long minval = numeric_limits<long long>::max();
for (int i = 0; i < N; i++) {
cin >> v[i];
minval = min(minval, v[i]);
}
long long starttotal = 0;
for (int i = 0; i < N; i++) {
starttotal += v[i] - minval;
}
if (starttotal + minval * N < s) {
cout << -1 << endl;
return 0;
}
s -= starttotal;
if (s <= 0) {
cout << minval << endl;
return 0;
}
cout << minval - (long long)ceil((double)s / (double)N) << endl;
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, V = map(int, input().split())
s = list(map(int, input().split()))
if sum(s) < V:
print(-1)
else:
s.sort()
summ = 0
m = min(s)
F = False
for el in s[1:]:
summ += el - m
if summ >= V:
F = True
break
if F:
print(m)
else:
ost = V - summ
print(int(m - ost/len(s)))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import sys
import math,bisect
sys.setrecursionlimit(10 ** 5)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
def I(): return int(sys.stdin.readline())
def neo(): return map(int, sys.stdin.readline().split())
def Neo(): return list(map(int, sys.stdin.readline().split()))
n,s = neo()
A = Neo()
m = min(A)
if sum(A) < s:
print(-1)
exit()
for i in range(n):
s -= (A[i]-m)
if s <= 0:
print(m)
else:
t = math.ceil(s/n)
print(m-t)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.*;
import java.util.*;
public class Main {
static Scanner sc = new Scanner();
static PrintWriter out = new PrintWriter(System.out);
public static void main(String[] args) {
int n = sc.nextInt();
long s = sc.nextLong();
int[] v = new int[n];
int min = Integer.MAX_VALUE;
for(int i = 0; i < n; i++)
min = Math.min(min, v[i] = sc.nextInt());
int lo = 0, hi = min, best = -1;
while(lo <= hi) {
int mid = lo + hi >> 1;
long sumOfDifferences = 0;
for(int i = 0; i < n; i++)
sumOfDifferences += v[i] - mid;
if(sumOfDifferences >= s){
best = mid;
lo = mid + 1;
}else hi = mid - 1;
}
out.println(best);
out.flush();
out.close();
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
public Scanner(String s) throws FileNotFoundException {
br = new BufferedReader(new FileReader(new File(s)));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,m=map(int,input().split())
a=list(map(int,input().split()))
s=sum(a)
k=min(a)
if s<m:
print(-1)
else:
p=s-m
print(min(k,p//n)) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n, s = map(int, input().split())
arrv = list(map(int, input().split()))
sumv, minv = sum(arrv), min(arrv)
if sumv < s:
print(-1)
else:
k = (sumv - s) // n
print (min(k, minv)) | PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
const int N = 1123;
long long v[N];
int main(void) {
ios_base::sync_with_stdio(false);
cin.tie(0);
long long n;
cin >> n;
long long s;
cin >> s;
for (int i = 0; i < n; i++) cin >> v[i];
long long sum = 0, mn = LLONG_MAX;
for (int i = 0; i < n; i++) {
mn = min(mn, v[i]);
sum += v[i];
}
if (sum < s) {
cout << -1 << endl;
return 0;
}
long long cur = 0;
for (int i = 0; i < n; i++) {
cur += v[i] - mn;
v[i] = mn;
}
if (cur >= s) {
cout << mn << endl;
return 0;
}
cout << mn - (s - cur + n - 1) / n << endl;
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | # Lets goto the next level
# AIM Specialist at CF *__* asap
# template taken from chaudhary_19
# Remember you were also a novice when you started,
# hence never be rude to anyone who wants to learn something
# Never open a ranklist untill and unless you are done with solving problems, wastes 3/4 minuts
# Donot treat CP as a placement thing, love it and enjoy it, you will succeed for sure.
# Any doubts or want to have a talk, contact https://www.facebook.com/chaudhary.mayank
# ///==========Libraries, Constants and Functions=============///
import sys
from bisect import bisect_left,bisect_right,insort
from collections import deque,Counter
from math import gcd,sqrt,factorial,ceil,log10,log2,floor
from itertools import permutations
from heapq import heappush,heappop,heapify
inf = float("inf")
mod = 10**9+7
#sys.setrecursionlimit(10**9)
def factorial_p(n, p):
ans = 1
if n <= p // 2:
for i in range(1, n + 1):
ans = (ans * i) % p
else:
for i in range(1, p - n):
ans = (ans * i) % p
ans = pow(ans, p - 2, p)
if n % 2 == 0:
ans = p - ans
return ans
def nCr_p(n, r, p):
ans = 1
while (n != 0) or (r != 0):
a, b = n % p, r % p
if a < b:
return 0
ans = (ans * factorial_p(a, p) * pow(factorial_p(b, p), p - 2, p) * pow(factorial_p(a - b, p), p - 2, p)) % p
n //= p
r //= p
return ans
def prime_sieve(n):
"""returns a sieve of primes >= 5 and < n"""
flag = n % 6 == 2
sieve = bytearray((n // 3 + flag >> 3) + 1)
for i in range(1, int(n**0.5) // 3 + 1):
if not (sieve[i >> 3] >> (i & 7)) & 1:
k = (3 * i + 1) | 1
for j in range(k * k // 3, n // 3 + flag, 2 * k):
sieve[j >> 3] |= 1 << (j & 7)
for j in range(k * (k - 2 * (i & 1) + 4) // 3, n // 3 + flag, 2 * k):
sieve[j >> 3] |= 1 << (j & 7)
return sieve
def prime_list(n): #<----- You have to call me if you wanna get list of primes upto range n
"""returns a list of primes <= n"""
res = []
if n > 1:
res.append(2)
if n > 2:
res.append(3)
if n > 4:
sieve = prime_sieve(n + 1)
res.extend(3 * i + 1 | 1 for i in range(1, (n + 1) // 3 + (n % 6 == 1)) if not (sieve[i >> 3] >> (i & 7)) & 1)
return res
def is_prime(n):
"""returns True if n is prime else False"""
if n < 5 or n & 1 == 0 or n % 3 == 0:
return 2 <= n <= 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
p = (p * p) % n
if p == n - 1:
break
else:
return False
return True
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [],[]
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n//i)
large.reverse()
small.extend(large)
return small
def count(n):
return ((n*(n+1))//2)
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
# ///===========MAIN=============///
n,s=get_ints()
Arr=get_array()
total=sum(Arr)
if total<s:
print(-1)
exit()
left=0;mini=min(Arr)
for i in range(n):
left+=(Arr[i]-mini)
if left>=s:
print(mini)
exit()
else:
s=s-left
think=mini-(s//n)
if s%n!=0:
think-=1
print(think)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, s;
cin >> n >> s;
long long int min = INT_MAX;
unsigned long long int sum = 0;
long long int i;
long long int arr[n];
for (i = 0; i < n; i++) {
cin >> arr[i];
sum += arr[i];
if (arr[i] < min) min = arr[i];
}
if (sum < s) {
cout << "-1";
return 0;
}
if (sum == s) {
cout << "0";
return 0;
}
long long int count = 0;
for (i = 0; i < n; i++) {
if (arr[i] > min) count++;
}
for (i = 0; i < n; i++) {
if (count <= 0) break;
if (arr[i] > min) {
count -= 1;
if (s - (arr[i] - min) <= 0) {
cout << min;
return 0;
}
s = s - (arr[i] - min);
arr[i] = min;
}
}
while (s != 0 && min >= 0) {
if (s - (n) <= 0) {
cout << (min - 1);
return 0;
} else {
min--;
s = s - n;
}
}
cout << "-1";
return 0;
}
| CPP |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;
import static java.lang.Math.*;
public class Cf182 implements Runnable
{
static class InputReader
{
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public InputReader(InputStream stream)
{
this.stream = stream;
}
public int read()
{
if (numChars==-1)
throw new InputMismatchException();
if (curChar >= numChars)
{
curChar = 0;
try
{
numChars = stream.read(buf);
}
catch (IOException e)
{
throw new InputMismatchException();
}
if(numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
public int nextInt()
{
int c = read();
while(isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
int res = 0;
do
{
if(c<'0'||c>'9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
long res = 0;
do
{
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble()
{
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-')
{
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.')
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
}
if (c == '.')
{
c = read();
double m = 1;
while (!isSpaceChar(c))
{
if (c == 'e' || c == 'E')
return res * Math.pow(10, nextInt());
if (c < '0' || c > '9')
throw new InputMismatchException();
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public String readString()
{
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do
{
res.appendCodePoint(c);
c = read();
}
while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c)
{
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next()
{
return readString();
}
public interface SpaceCharFilter
{
public boolean isSpaceChar(int ch);
}
}
public static void main(String args[]) throws Exception
{
new Thread(null, new Cf182(),"Main",1<<27).start();
}
public static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// array sorting by colm
public static void sortbyColumn(int arr[][], int col)
{
Arrays.sort(arr, new Comparator<int[]>() {
@Override
public int compare(final int[] entry1,
final int[] entry2) {
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
});
}
// gcd
public static long findGCD(long arr[], int n)
{
long result = arr[0];
for (int i = 1; i < n; i++)
result = gcd(arr[i], result);
return result;
}
// fibonaci
static int fib(int n)
{
int a = 0, b = 1, c;
if (n == 0)
return a;
for (int i = 2; i <= n; i++)
{
c = a + b;
a = b;
b = c;
}
return b;
}
// sort a string
public static String sortString(String inputString)
{
char tempArray[] = inputString.toCharArray();
Arrays.sort(tempArray);
return new String(tempArray);
}
static long power(long x, long y)
{
if (y == 0)
return 1;
long temp = power(x, y / 2);
if (y % 2 == 0)
return temp*temp;
else
return x * temp * temp;
}
// pair function
// list.add(new Pair<>(sc.nextInt(), i + 1));
// Collections.sort(list, (a, b) -> Integer.compare(b.first, a.first));
private static class Pair<F, S> {
private F first;
private S second;
public Pair() {}
public Pair(F first, S second) {
this.first = first;
this.second = second;
}
}
public void run()
{
InputReader sc = new InputReader(System.in);
PrintWriter w = new PrintWriter(System.out);
int n = sc.nextInt();
long s = sc.nextLong();
int a[] = new int[n];
long sum = 0;
int min = Integer.MAX_VALUE;
for(int i=0;i<n;i++)
{
a[i] = sc.nextInt();
sum += a[i];
if(a[i]<min)
min = a[i];
}
if(sum<s)
System.out.println("-1");
else
{
long target = 0;
for(int i=0;i<n;i++)
target += (a[i] - min);
if(s<=target)
System.out.println(min);
else
{
long x = s - target;
x = (long)Math.ceil((double)x/n);
System.out.println(min-x);
}
}
}
} | JAVA |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | #codeforces _1084B_live
gi = lambda : list(map(int,input().split()))
n,s = gi()
l = gi()
l.sort(reverse=True)
mii = min(l)
cur = 0
while s and cur < n:
s -= (l[cur]-mii)
cur += 1
cur = 0
if s < 0:
print(mii)
exit()
if s > mii * n:
print(-1)
exit()
temp = s//n
if s%n:
temp += 1
print(mii-temp)
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | n,k = map(int,input().split())
l = list(map(int,input().split()))
m , s = min(l) ,sum(l)
if s < k:
print(-1)
else:
print(min(m,(s-k)//n))
| PYTHON3 |
1084_B. Kvass and the Fair Nut | The Fair Nut likes kvass very much. On his birthday parents presented him n kegs of kvass. There are v_i liters of kvass in the i-th keg. Each keg has a lever. You can pour your glass by exactly 1 liter pulling this lever. The Fair Nut likes this drink very much, so he wants to pour his glass by s liters of kvass. But he wants to do it, so kvass level in the least keg is as much as possible.
Help him find out how much kvass can be in the least keg or define it's not possible to pour his glass by s liters of kvass.
Input
The first line contains two integers n and s (1 ≤ n ≤ 10^3, 1 ≤ s ≤ 10^{12}) — the number of kegs and glass volume.
The second line contains n integers v_1, v_2, …, v_n (1 ≤ v_i ≤ 10^9) — the volume of i-th keg.
Output
If the Fair Nut cannot pour his glass by s liters of kvass, print -1. Otherwise, print a single integer — how much kvass in the least keg can be.
Examples
Input
3 3
4 3 5
Output
3
Input
3 4
5 3 4
Output
2
Input
3 7
1 2 3
Output
-1
Note
In the first example, the answer is 3, the Fair Nut can take 1 liter from the first keg and 2 liters from the third keg. There are 3 liters of kvass in each keg.
In the second example, the answer is 2, the Fair Nut can take 3 liters from the first keg and 1 liter from the second keg.
In the third example, the Fair Nut can't pour his cup by 7 liters, so the answer is -1. | 2 | 8 | import math
m,n=map(int,input().split())
li=sorted(list(map(int,input().split())))
l=li[0]
bli=[]
if sum(li)<n:
print(-1)
exit()
for i in range(len(li)):
bli.append(li[i]-li[0])
if sum(bli)>=n:
print(l)
else:
print(l-math.ceil((n-sum(bli))/m))
| PYTHON3 |
Subsets and Splits