Problem:


(Please read problem A first. The maximum score you can get by solving this problem C is 1, which will have almost no effect on your ranking.)

Beginner's Guide

"Local search" is a powerful method for finding a high-quality solution.
In this method, instead of constructing a solution from scratch, we try to find a better solution by slightly modifying the already found solution.
If the solution gets better, update it, and if it gets worse, restore it.
By repeating this process, the quality of the solution is gradually improved over time.
The pseudo-code is as follows.

solution = compute an initial solution (by random generation, or by applying other methods such as greedy)
while the remaining time > 0:
    slightly modify the solution (randomly)
    if the solution gets worse:
        restore the solution

For example, in this problem, we can use the following modification: pick the date d and contest type q at random and change the type of contest to be held on day d to q.
The pseudo-code is as follows.

t[1..D] = compute an initial solution (by random generation, or by applying other methods such as greedy)
while the remaining time > 0:
    pick d and q at random
    old = t[d] # Remember the original value so that we can restore it later
    t[d] = q
    if the solution gets worse:
        t[d] = old

The most important thing when using the local search method is the design of how to modify solutions.

If the amount of modification is too small, we will soon fall into a dead-end (local optimum) and, conversely, if the amount of modification is too large, the probability of finding an improving move becomes extremely small.

In order to increase the number of iterations, it is desirable to be able to quickly calculate the score after applying a modification.

In this problem C, we focus on the second point.
The score after the modification can, of course, be obtained by calculating the score from scratch.
However, by focusing on only the parts that have been modified, it may be possible to quickly compute the difference between the scores before and after the modification.
From another viewpoint, the impossibility of such a fast incremental calculation implies that a small modification to the solution affects a majority of the score calculation.
In such a case, we may need to redesign how to modify solutions, or there is a high possibility that the problem is not suitable for local search.
Let's implement fast incremental score computation.
It's time to demonstrate the skills of algorithms and data structures you have developed in ABC and ARC!

In this kind of contest, where the objective is to find a better solution instead of the optimal one, a bug in a program does not result in a wrong answer, which may delay the discovery of the bug.
For early detection of bugs, it is a good idea to unit test functions you implemented complicated routines.
For example, if you implement fast incremental score calculation, it is a good idea to test that the scores computed by the fast implementation match the scores computed from scratch, as we will do in this problem C.

Problem Statement

You will be given a contest schedule for D days and M queries of schedule modification.
In the i-th query, given integers d_i and q_i, change the type of contest to be held on day d_i to q_i, and then output the final satisfaction at the end of day D on the updated schedule.
Note that we do not revert each query. That is, the i-th query is applied to the new schedule obtained by the (i-1)-th query.

Input

Input is given from Standard Input in the form of the input of Problem A followed by the output of Problem A and the queries.

D
c_1 c_2 \cdots c_{26}
s_{1,1} s_{1,2} \cdots s_{1,26}
\vdots
s_{D,1} s_{D,2} \cdots s_{D,26}
t_1
t_2
\vdots
t_D
M
d_1 q_1
d_2 q_2
\vdots
d_M q_M

The constraints and generation methods for the input part are the same as those for Problem A.

For each d=1,\ldots,D, t_d is an integer generated independently and uniformly at random from {1,2,\ldots,26}.

The number of queries M is an integer satisfying 1\leq M\leq 10^5.

For each i=1,\ldots,M, d_i is an integer generated independently and uniformly at random from {1,2,\ldots,D}.

For each i=1,\ldots,26, q_i is an integer satisfying 1\leq q_i\leq 26 generated uniformly at random from the 25 values that differ from the type of contest on day d_i.

Output

Let v_i be the final satisfaction at the end of day D on the schedule after applying the i-th query.
Print M integers v_i to Standard Output in the following format:

v_1
v_2
\vdots
v_M

Sample Input 1

5
86 90 69 51 2 96 71 47 88 34 45 46 89 34 31 38 97 84 41 80 14 4 50 83 7 82
19771 12979 18912 10432 10544 12928 13403 3047 10527 9740 8100 92 2856 14730 1396 15905 6534 4650 11469 3628 8433 2994 10899 16396 18355 11424
6674 17707 13855 16407 12232 2886 11908 1705 5000 1537 10440 10711 4917 10770 17272 15364 19277 18094 3929 3705 7169 6159 18683 15410 9092 4570
6878 4239 19925 1799 375 9563 3445 5658 19857 11401 6997 6498 19933 3848 2426 2146 19745 16880 17773 18359 3921 14172 16730 11157 5439 256
8633 15862 15303 10749 18499 7792 10317 5901 9395 11433 3514 3959 5202 19850 19469 9790 5653 784 18500 10552 17975 16615 7852 197 8471 7452
19855 17918 7990 10572 4333 438 9140 9104 12622 4985 12319 4028 19922 12132 16259 17476 2976 547 19195 19830 16285 4806 4471 9457 2864 2192
1
17
13
14
13
5
1 7
4 11
3 4
5 24
4 19

Sample Output 1

72882
56634
38425
27930
42884

Note that this example is a small one for checking the problem specification. It does not satisfy the constraint D=365 and is never actually given as a test case.

Next Step

Let's go back to Problem A and implement the local search algorithm by utilizing the incremental score calculator you just implemented!
For this problem, the current modification "pick the date d and contest type q at random and change the type of contest to be held on day d to q" is actually not so good. By considering why it is not good, let's improve the modification operation.
One of the most powerful and widely used variant of the local search method is "Simulated Annealing (SA)", which makes it easier to reach a better solution by stochastically accepting worsening moves.
For more information about SA and other local search techniques, please refer to the editorial that will be published after the contest.



Sample input:
5
86 90 69 51 2 96 71 47 88 34 45 46 89 34 31 38 97 84 41 80 14 4 50 83 7 82
19771 12979 18912 10432 10544 12928 13403 3047 10527 9740 8100 92 2856 14730 1396 15905 6534 4650 11469 3628 8433 2994 10899 16396 18355 11424
6674 17707 13855 16407 12232 2886 11908 1705 5000 1537 10440 10711 4917 10770 17272 15364 19277 18094 3929 3705 7169 6159 18683 15410 9092 4570
6878 4239 19925 1799 375 9563 3445 5658 19857 11401 6997 6498 19933 3848 2426 2146 19745 16880 17773 18359 3921 14172 16730 11157 5439 256
8633 15862 15303 10749 18499 7792 10317 5901 9395 11433 3514 3959 5202 19850 19469 9790 5653 784 18500 10552 17975 16615 7852 197 8471 7452
19855 17918 7990 10572 4333 438 9140 9104 12622 4985 12319 4028 19922 12132 16259 17476 2976 547 19195 19830 16285 4806 4471 9457 2864 2192
1
17
13
14
13
5
1 7
4 11
3 4
5 24
4 19


Sample output:
72882
56634
38425
27930
42884

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache

#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)

#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n): return kaijo_memo[n]
  if(len(kaijo_memo) == 0): kaijo_memo.append(1)
  while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if n == r: return 1
  if n < r or r < 0: return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1: arr.append([temp, 1])
  if arr==[]: arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b: a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n-1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n: return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

def base_10_to_n_without_0(X, n):
  X -= 1
  if X//n: return base_10_to_n_without_0(X//n, n)+[X%n]
  return [X%n]

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

D = I()
C = IL()
S = ILL(D)

ans = 0
score = 0
last = [[-1] for i in range(26)]
schedule = []

for d in range(D):
  t = I()-1
  last[t].append(d)
  score += S[d][t]
  schedule.append(t)

for i in range(26):
  last[i].append(D)
  for k in range(len(last[i])-1):
    g = last[i][k+1]-last[i][k]-1
    score -= C[i]*g*(g+1)//2

M = I()

for _ in range(M):
  d,q = IL()
  d,q =  d-1,q-1
  p = schedule[d]
  schedule[d] = q
  score -= S[d][p]
  score += S[d][q]
  
  pos = last[p].index(d)
  score -= C[p]*(last[p][pos]-last[p][pos-1])*(last[p][pos+1]-last[p][pos])
  last[p].remove(d)
  
  pos = bisect.bisect(last[q],d)
  score += C[q]*(d-last[q][pos-1])*(last[q][pos]-d)
  bisect.insort(last[q],d)
  
  print(score)

Submission status: Accepted

Solution code:
import heapq
D=int(input())
c=list(map(int,input().split()))
s=[0]
last=[0]*26
for i in range(D):
  s.append(list(map(int,input().split())))
contests=[]
for i in range(26):
  contests.append([])
  contests[-1].append(0)
  contests[-1].append(D+1)
schedule=[0]*(D+1)
score=0
for i in range(1,D+1):
  t=int(input())-1
  score+=s[i][t]
  last[t]=i
  contests[t].append(i)
  schedule[i]=t+1
  for j in range(26):
    score-=c[j]*(i-last[j])
for i in range(26):
  contests[i].sort()
m=int(input())
for i in range(m):
  d,q=map(int,input().split())
  changebefore=schedule[d]-1
  changeafter=q-1
  r=contests[changebefore]
  score-=s[d][changebefore]
  score+=s[d][changeafter]
  for j in range(len(r)-1):
    if r[j]==d:
      score-=c[changebefore]*((r[j+1]-r[j-1]-1)*(r[j+1]-r[j-1]))//2
      score+=c[changebefore]*((r[j+1]-d-1)*(r[j+1]-d))//2
      score+=c[changebefore]*((d-r[j-1]-1)*(d-r[j-1]))//2
      break
  contests[changebefore].remove(d)
  contests[changebefore].sort()
  r=contests[changeafter]
  for j in range(len(r)-1):
    if r[j+1]>d:
      score+=c[changeafter]*((r[j+1]-r[j]-1)*(r[j+1]-r[j]))//2
      score-=c[changeafter]*((r[j+1]-d-1)*(r[j+1]-d))//2
      score-=c[changeafter]*((d-r[j]-1)*(d-r[j]))//2
      break
  contests[changeafter].append(d)
  contests[changeafter].sort()
  schedule[d]=q
  print(score)

Submission status: Accepted

Solution code:
from collections import deque, Counter, defaultdict
from itertools import chain, combinations
import json
# import numpy as np
import bisect
import sys
import math
import bisect
from functools import lru_cache
import itertools

sys.setrecursionlimit(10 ** 8)

M = 10 ** 9 + 7
INF = 10 ** 17


def main():
    D = int(input())
    c = [int(a) for a in input().split()]
    s = [
        [int(a) for a in input().split()]
        for _ in range(D)
    ]
    t = [
        int(input())
        for _ in range(D)
    ]
    M = int(input())
    dq = [
        [int(a) for a in input().split()]
        for _ in range(M)
    ]

    def calc_score(t):
        score = 0
        last = [0] * len(c)
        res = []

        for d, ti in enumerate(t):
            sdi = s[d][ti - 1]
            score += sdi
            last[ti - 1] = d + 1
            for i, ci in enumerate(c):
                score -= ci * ((d + 1) - last[i])
            res.append(score)
        return res

    def calc_score_diff(t, di, qi):
        old_q = t[di - 1]
        last_old_q_date = 0
        for i in range(di - 2, -1, -1):
            if t[i] == old_q:
                last_old_q_date = i + 1
                break
        next_old_q_date = D + 1
        for i in range(di, D):
            if t[i] == old_q:
                next_old_q_date = i + 1
                break

        old_d_last = di - last_old_q_date
        old_d_next = next_old_q_date - di
        old_diff = old_d_next * old_d_last * c[old_q - 1]

        new_q = qi
        last_new_q_date = 0
        for i in range(di - 2, -1, -1):
            if t[i] == new_q:
                last_new_q_date = i + 1
                break
        next_new_q_date = D + 1
        for i in range(di, D):
            if t[i] == new_q:
                next_new_q_date = i + 1
                break

        new_d_last = di - last_new_q_date
        new_d_next = next_new_q_date - di
        new_diff = new_d_next * new_d_last * c[new_q - 1]

        old_score = s[di - 1][old_q - 1]
        new_score = s[di - 1][new_q - 1]
        return new_diff - old_diff + new_score - old_score

    score = calc_score(t)[-1]
    for di, qi in dq:
        diff = calc_score_diff(t, di, qi)
        score += diff
        print(score)
        t[di - 1] = qi


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
D=int(input())
c = list(map(int,input().split()))

s=[None]*D
for i in range(D):
    s[i] = list(map(int,input().split()))
t=[0]*D
for i in range(D):
    t[i]=int(input())
    
m=int(input())
d=[None]*m
q=[None]*m
for i in range(m):
    d[i],q[i] = map(int,input().split())
sumc=sum(c)



def calc_score(t):
    last=[0]*26
    satis=0
    pena=0
    delta_pena=0
    for j in range(D):
        ss=s[j]
        tt=t[j]
        satis += ss[tt-1]
        delta_pena += sumc - c[tt-1]*(j+1-last[tt-1])

        last[tt-1]=j+1   
        pena += delta_pena
    return satis-pena

for i in range(m):
    t[d[i]-1] = q[i]
    print(calc_score(t))

Submission status: Accepted

Solution code:
from bisect import bisect_left, bisect_right

class Set(): #BinaryIndexedTree
    def __init__(self, valuelist, INF=10**18+1):
        self.n = len(valuelist) + 1
        self.value = sorted(valuelist) + [INF]
        self.comp = {v : k for k, v in enumerate(self.value, 1)}
        self.bit = BinaryIndexedTree(self.n)

    def insert(self, x, k=1):
        self.bit.add(self.comp[x], k)

    def delete(self, x, k=1):
        self.bit.add(self.comp[x], -k)

    def count(self, x):
        return self.bit.get(self.comp[x])

    def sum(self, x):
        return self.bit.sum(self.comp[x])

    def get(self, k): #k個以下のときはINFを返す
        return self.value[self.bit.bisect_left(k) - 1]

class BinaryIndexedTree(): #1-indexed
    def __init__(self, n):
        self.n = n
        self.tree = [0 for _ in range(n + 1)]

    def sum(self, idx):
        res = 0
        while idx:
            res += self.tree[idx]
            idx -= idx & -idx
        return res

    def get(self, idx):
        return self.sum(idx) - self.sum(idx - 1)

    def add(self, idx, x):
        while idx <= self.n:
            self.tree[idx] += x
            idx += idx & -idx

    def bisect_left(self, x):
        if x <= 0: return 0
        res, tmp = 0, 2**((self.n).bit_length() - 1)
        while tmp:
            if res + tmp <= self.n and self.tree[res + tmp] < x:
                x -= self.tree[res + tmp]
                res += tmp
            tmp >>= 1
        if res >= self.n:
            return self.n
        return res + 1

D = int(input())
C = list(map(int, input().split()))
S = [list(map(int, input().split())) for _ in range(D)]
T = [int(input()) for _ in range(D)]

M = int(input())
Q = [tuple(map(int, input().split())) for _ in range(M)]

res = 0
last = [0 for _ in range(26)]
used = [Set(list(range(D))) for i in range(26)]

for d in range(D):
    res += S[d][T[d] - 1]
    last[T[d] - 1] = d + 1
    used[T[d] - 1].insert(d)
    for j in range(26):
        res -= C[j] * ((d + 1) - last[j])

for i in range(M):
    d, q = Q[i]
    if T[d - 1] == q:
        print(res)
        continue
    res -= S[d - 1][T[d - 1] - 1]
    res += S[d - 1][q - 1]
    k = used[T[d - 1] - 1].sum(d - 1)
    l = used[T[d - 1] - 1].get(k - 1) if k != 1 else -1
    r = used[T[d - 1] - 1].get(k + 1)
    if r == 1000000000000000001:
        r = D
    res -= (d - 1 - l) * C[T[d - 1] - 1] * (r - d + 1)
    used[T[d - 1] - 1].delete(d - 1)
    used[q - 1].insert(d - 1)
    k = used[q - 1].sum(d - 1)
    l = used[q - 1].get(k - 1) if k != 1 else -1
    r = used[q - 1].get(k + 1)
    if r == 1000000000000000001:
        r = D
    res += (d - 1 - l) * C[q - 1] * (r - d + 1)
    T[d - 1] = q
    print(res)

Submission status: Accepted

Solution code:

def main():
    D = int(input())
    c_list = list(map(int, input().split()))
    s_list = [list(map(int, input().split())) for _ in range(D)]
    t_list = [int(input()) for _ in range(D)]

    basic_bad = sum(c_list)
    #res_list = []
    M = int(input())
    for m in range(M):
        d,q = map(int, input().split())
        t_list[d-1] = q
        res = 0
        
        bad = 0
        last_open = [0] * 26
        for i in range(D):
            t = t_list[i]
            res += s_list[i][t-1]

            # TODO: ここは毎回26ループ回す必要がないところになりうるので改善可能かも(なおすだけで20回分の処理が消せる)
            # bad = 0
            # for ci in range(26):
            #     bad += c_list[ci]*(i+1 - last_open[ci])
            bad += basic_bad - (i+1 - last_open[t-1]) * c_list[t-1]
            last_open[t-1] = i+1

            res -= bad

        print(res)

if __name__ == "__main__":
    main()

Submission status: Accepted

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

import sys
from typing import Any, Callable, Deque, Dict, List, Mapping, Optional, Sequence, Set, Tuple, TypeVar, Union
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random                                # random, uniform, randint, randrange, shuffle, sample
# import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict          # subclass of dict. defaultdict(facroty)
# from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce                 # reduce(f, iter[, init])
# from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate             # accumulate(iter[, f])
# from operator import itemgetter              # itemgetter(1), itemgetter('key')
# from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)



def main():
    Num = Union[int, float]
    mod = 1000000007                # 10^9+7
    inf = float('inf')              # sys.float_info.max = 1.79e+308
    # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
    sys.setrecursionlimit(10**6)    # 1000 -> 1000000
    def input(): return sys.stdin.readline().rstrip()
    def ii():    return int(input())
    def isp():   return input().split()
    def mi():    return map(int, input().split())
    def mi_0():  return map(lambda x: int(x)-1, input().split())
    def lmi():   return list(map(int, input().split()))
    def lmi_0(): return list(map(lambda x: int(x)-1, input().split()))
    def li():    return list(input())
    
    
    def calc_total_score(schedule):
        ans = []
        last = [-1] * 26
        satisfy = 0
        for day in range(D):
            contest_id = schedule[day]
            last[contest_id] = day
            satisfy += score[day][contest_id]
            for i in range(26):
                satisfy -= c[i] * (day - last[i])
            ans.append(satisfy)
        return ans
    

    def change_contest(schedule, target_day, new_contest):
        # print(schedule, target_day, new_contest)
        prev_contest = schedule[target_day]
        diff = 0
        diff += (score[target_day][new_contest] - score[target_day][prev_contest])
        i = target_day - 1
        while i >= 0:
            if schedule[i] == prev_contest:
                break
            i -= 1
        j = target_day + 1
        while j <= len(schedule) - 1:
            if schedule[j] == prev_contest:
                break
            j += 1
        diff -= (j - target_day) * (target_day - i) * c[prev_contest]
        i = target_day - 1
        while i >= 0:
            if schedule[i] == new_contest:
                break
            i -= 1
        j = target_day + 1
        while j <= len(schedule) - 1:
            if schedule[j] == new_contest:
                break
            j += 1
        diff += (j - target_day) * (target_day - i) * c[new_contest]
        schedule[target_day] = new_contest
        return diff

    

    D = ii()
    c = lmi()
    score = [lmi() for _ in range(D)]

    schedule = [ii() - 1 for _ in range(D)]
    satisfy_list = calc_total_score(schedule)
    satisfy = satisfy_list[-1]

    # print(*satisfy_list, sep='\n')

    m = ii()
    query = [lmi_0() for _ in range(m)]
    for target_day, new_contest in query:
        diff = change_contest(schedule, target_day, new_contest)
        satisfy += diff
        print(satisfy)




if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
def examA():
    N = 26
    D = I()
    C = LI()
    S = [LI()for _ in range(N)]
    ans = [0]*D

    for v in ans:
        print(v)
    return

def examB():
    N = 26
    D = I()
    C = LI()
    S = [LI()for _ in range(D)]
    T = [I()for _ in range(D)]
    ans = [0]*D
    used = [0]*N
    for i,t in enumerate(T):
        ans[i] += ans[i-1]
        t -= 1
        ans[i] += S[i][t]
        used[t] = i+1
        for j in range(N):
            ans[i] -= C[j]*(i+1-used[j])
    for v in ans:
        print(v)
    return

def examC():
    def calc_start():
        rep = [0] * D
        used = [0] * N
        for i, t in enumerate(T):
            rep[i] += rep[i - 1]
            rep[i] += S[i][t]
            used[t] = i + 1
            for j in range(N):
                rep[i] -= C[j] * (i + 1 - used[j])
        return rep

    def search(d,k):
        l = 0; r = D
        for i in range(d+1,D):
            if T[i]==k:
                r = i
                break
        for i in range(0,d)[::-1]:
            if T[i]==k:
                l = i + 1
                break
        return l, r

    def calc_cost(d,k):
        return C[k]*(d+1)*d//2

    N = 26
    D = I()
    C = LI()
    S = [LI()for _ in range(D)]
    T = [I()-1 for _ in range(D)]

    ans = calc_start()[D-1]

    #print(ans)

    M = I()
    for i in range(M):
        #d, q = random.randint(0,D-1), random.randint(0,N-1)
        d, q = LI()
        d -= 1; q -= 1
        old = deepcopy(T[d])
        T[d] = q
        #print(q,old)

        old_l, old_r = search(d,old)
        new_l, new_r = search(d,q)

        #print(old_l, old_r)
        #print(new_l, new_r)

        change = S[d][q] - S[d][old] \
                 - (calc_cost(old_r - old_l, old) - calc_cost(d-old_l, old) - calc_cost(old_r-d - 1, old))\
                 + (calc_cost(new_r - new_l, q) - calc_cost(d - new_l, q) - calc_cost(new_r - d - 1, q))

        ans += change

        print(ans)

    return


from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examC()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""

Submission status: Accepted

Solution code:
d = int(input())
c = list(map(int,input().split()))
s = []
for i in range(d):
  si = list(map(int,input().split()))
  s.append(si)
lastd = [-1 for _ in range(26)]
ans = 0
t = [0 for _ in range(d)]
for i in range(d):
  di = int(input())
  t[i] = di-1
  lastd[di-1] = i
  ans += s[i][di-1]
  for j in range(26):
    ans -= (i-lastd[j])*c[j]
m = int(input())
for _ in range(m):
  da,qa = map(int,input().split())
  da -= 1
  qa -= 1
  qb = t[da]
  t[da] = qa
  qb1 = -1
  qa1 = -1
  ans -= s[da][qb] 
  ans += s[da][qa]
  for i in range(da-1,-1,-1):
    if t[i] == qb:
      qb1 = i
      break
  for i in range(da-1,-1,-1):
    if t[i] == qa:
      qa1 = i
      break
  qb2 = d
  qa2 = d
  for i in range(da+1,d):
    if t[i] == qb:
      qb2 = i
      break
  for i in range(da+1,d):
    if t[i] == qa:
      qa2 = i
      break
  ans -= c[qb]*(qb2-da)*(da-qb1)
  ans += c[qa]*(qa2-da)*(da-qa1)
  print(ans)

Submission status: Accepted

Solution code:
d = int(input())
c = list(map(int,input().split()))
s = [list(map(int,input().split())) for i in range(d)]
t = [int(input()) for i in range(d)]
now = [0]*26

score = [0]*26
for i in range(d):
    
    now[t[i]-1] = i+1
    ind = t[i]-1
    for j in range(26):
        if j == ind:
            score[j] += s[i][ind]
        else:
            x = (i+1-now[j])*c[j]
            score[j]-= x
m = int(input())
l = [list(map(int,input().split())) for i in range(m)]
ans = sum(score)
for i in range(m):
    x,y = l[i]
    be = t[x-1]
    be -= 1
    now1 =  0
    now2 = 0
    t[x-1] = y
    y -= 1
    count1 = 0
    count2 = 0
    for j in range(d):
        if be == t[j]-1:
            now1 = j+1
            count1 +=s[j][be]
        else:
            count1 -= (j+1-now1)*c[be]
        if y == t[j]-1:
            now2 = j+1
            count2 +=s[j][y]
        else:
            count2 -= (j+1-now2)*c[y]

    ans += count1+count2-score[be]-score[y]
    print(ans)
    score[be] = count1
    score[y] = count2