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

Solution (Python):
from bisect import bisect_left


def check(x, m):
    
    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)

Solution (Python):
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

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)