Problem:


Score : 500 points

Problem Statement

Takahashi has come to a party as a special guest.
There are N ordinary guests at the party. The i-th ordinary guest has a power of A_i.

Takahashi has decided to perform M handshakes to increase the happiness of the party (let the current happiness be 0).
A handshake will be performed as follows:

Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same).

Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y.

However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold:

Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q).

What is the maximum possible happiness after M handshakes?

Constraints

1 \leq N \leq 10^5

1 \leq M \leq N^2

1 \leq A_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_N

Output

Print the maximum possible happiness after M handshakes.

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Let us say that Takahashi performs the following handshakes:

In the first handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 4.

In the second handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 5.

In the third handshake, Takahashi shakes the left hand of Guest 5 and the right hand of Guest 4.

Then, we will have the happiness of (34+34)+(34+33)+(33+34)=202.

We cannot achieve the happiness of 203 or greater, so the answer is 202.

Sample Input 2

9 14
1 3 5 110 24 21 34 5 3

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170



Sample input:
5 3
10 14 19 34 33


Sample output:
202

Submission status: Accepted

Solution code:
#!/usr/bin/env python3

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))


#O(nlogn)
def f(x):
    cnt = 0
    for i in a:
        idx = bl(a,x-i)
        cnt += n - idx
        if cnt >= m:
            return True
    return False

n,m = LI()
a = LI()
a.sort()
ng,ok = 2*max(a)+1,2
#O(log(max(a)))
while ng - ok > 1:
    mid = (ng+ok) // 2
    if f(mid):
        ok = mid
    else:
        ng = mid

ans = 0
tmp = 0
b = [0] + list(accumulate(a))
#O(nlogn)
for i in a:
    idx = bl(a,ok-i)
    # i*(n-idx) := 左手で握手する人の幸福度(i)*左手で握手する回数(n-idx)
    # (b[-1] - b[idx]) := 右手で握手する人の幸福度の総和
    ans += i*(n - idx) + (b[-1] - b[idx])
    # tmp := 左手で握手する人を決めたときの右手で握手した人の数の総和 = ペアの総数
    tmp += n - idx

# ペアの数がm以上なら答えからペアの数がmになるよう最小のペアの幸福度(ok)を引く
ans = ans - (tmp - m)*ok
print(ans)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3

import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))


def f(x):
    cnt = 0
    for i in a:
        idx = bl(a,x-i)
        cnt += n - idx
        if cnt >= m:
            return True
    return False

n,m = LI()
a = LI()
a.sort()
ok,ng = 2*max(a)+1,2
while ok - ng > 1:
    mid = (ok+ng) // 2
    if f(mid):
        ng = mid
    else:
        ok = mid

ans = 0
tmp = 0
b = [0] + list(accumulate(a))
for i in a:
    idx = bl(a,ng-i)
    ans += i*(n - idx) + (b[-1] - b[idx])
    tmp += n - idx
ans = ans - (tmp - m)*ng
print(ans)

Submission status: Accepted

Solution code:
def count(pre, a, x):
  cnt = 0
  val = 0
  for i in reversed(range(n)):
    if 2*a[i] < x:
      break
    lo, hi = 0, n-1
    while lo < hi:
      mid = (lo + hi)//2
      if a[mid] + a[i] < x:
        lo = mid + 1
      else:
        hi = mid
    cnt += 1 + 2*(i - lo)
    val += 2*(i-lo+1)*a[i] + (pre[i] - pre[lo])*2
  return cnt, val

n, m = map(int, input().split())
a = list(sorted(map(int, input().split())))
pre = [0]
for i in range(n):
  pre.append(a[i] + pre[-1])
lo, hi = 0, 2*a[-1]
while lo < hi:
  mid = (lo + hi + 1) // 2
  cnt, _ = count(pre, a, mid)
  if cnt < m:
    hi = mid - 1
  else:
    lo = mid
cnt, val = count(pre, a, lo)
print(val - (cnt - m)*lo)

Submission status: Accepted

Solution code:
import bisect

n, m = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)

ruiseki = [0]*(n+1)
for i in range(n):
    ruiseki[i+1] = ruiseki[i] + a[i]

def solve(min_power):
    ans = 0
    for pos in range(n):
        tmp = min_power - a[pos]
        next_pos = bisect.bisect_left(a, tmp, pos) - pos
        times = n - pos - next_pos
        if times == n - pos:
            ans += 2 * (times - 1) + 1 
        else:
            ans += 2 * times
    if ans >= m:
        return True
    else:
        return False

ok = 0
ng = 10** 9
while abs(ok - ng) > 1:
    mid =  (ok + ng) // 2 
    if solve(mid):
        ok = mid
    else:
        ng = mid

cnt = 0
ans = 0

for pos in range(n):
    tmp = ok - a[pos]
    next_pos = bisect.bisect_left(a, tmp, pos) - pos
    times = n - pos - next_pos
    if times == n - pos:
        cnt += 2 * (times - 1) + 1 
        ans += 2 * (ruiseki[n] - ruiseki[n-times+1] + (times-1)*a[pos]) + 2*a[pos]
    else:
        cnt += 2 * times
        ans += 2 * (ruiseki[n] - ruiseki[n-times] + (times)*a[pos])

print(ans - (cnt-m)*ok)

Submission status: Accepted

Solution code:
from bisect import bisect_left


def check(x, m):
    # 幸福度が x 以上となる握手が m 通り以上か
    cnt = 0
    for i in range(N):
        y = x - A[i]
        p = bisect_left(A, y)
        cnt += N - p
    return cnt >= m


N, M = map(int, input().split())
A = list(map(int, input().split()))
A.sort()

S = [0] * (N + 1)
for i, a in enumerate(A):
    S[i + 1] = S[i] + a

ok = 0
ng = 2 * 10 ** 5 + 1
while abs(ok - ng) > 1:
    mid = (ok + ng) // 2
    if check(mid, M):
        ok = mid
    else:
        ng = mid

X = ok
ans = 0
cnt = 0
amari = 2 * 10 ** 5 + 1
for i in range(N):
    Aj = X - A[i]
    j = bisect_left(A, Aj)
    ans += S[N] - S[j] + A[i] * (N - j)
    cnt += N - j
    if j < N:
        amari = min(amari, A[i] + A[j])

print(ans - (cnt - M) * amari)

Submission status: Accepted

Solution code:
from itertools import accumulate
import bisect
N, M = map(int, input().split())
A = [int(i) for i in input().split()]
A.sort()

OK = 0
NG = max(A)*2+1
c = 0
# 幸福度がmid点以上上がる握手をM回できるか
while NG-OK > 1:
    mid = (OK+NG)//2
    cnt = 0

    for a in A:
        cnt += N-bisect.bisect_left(A, mid-a)

    if cnt >= M:
        OK = mid
    else:
        NG = mid

S = list(accumulate(A))

ans = 0
c = 0
for a in A:
    idx = bisect.bisect_right(A, OK-a)
    if idx >= 1:
        ans += a*(N-idx)+S[-1]-S[idx-1]
    else:
        ans += a*(N-idx)+S[-1]

    c += N-idx
ans += OK*(M-c)
print(ans)