text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
N,K = map(int,input().split())
R,S,P = map(int,input().split())
T = list(input())
d = {'r':P,'s':R,'p':S}
ans = 0
for i,c in enumerate(T):
if i < K or c != T[i-K]:
ans += d[c]
continue
else:
T[i] = '_'
print(ans)
```
| 8,200 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
from itertools import groupby
N, K = map(int, input().split())
R, S, P = map(int, input().split())
T = input()
d = {"r": P, "s": R, "p": S}
print(sum(sum(map(lambda p: d[p[0]]*((len(list(p[1]))+1)//2), groupby(T[k::K]))) for k in range(K)))
```
| 8,201 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
dict = {"r":p, "s":r, "p":s}
val = 0
for i in range(n):
if i < k:
val += dict[t[i]]
elif t[i-k] != t[i]:
val += dict[t[i]]
else:
t[i] = " "
print(val)
```
| 8,202 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
n,k = map(int,input().split())
r,s,p = map(int,input().split())
t = list(input())
for i in range(k,n):
if t[i]==t[i-k]:
t[i]=''
ans = t.count('r')*p+t.count('s')*r+t.count('p')*s
print(ans)
```
| 8,203 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
d = {'r': p, 's': r, 'p': s}
point = 0
for i in range(k):
point += d[t[i]]
for i in range(k, n):
if t[i] == t[i - k]:
t[i] = 'x'
continue
point += d[t[i]]
print(point)
```
| 8,204 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
d = {"r":p, "s":r, "p":s}
t = list(input())
dp = [0]*n
for i in range(n):
if i>=k and t[i]==t[i-k]:
dp[i] = 0
t[i] = "x"
else:
dp[i] = d[t[i]]
print(sum(dp))
```
| 8,205 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
N, K = map(int, input().split())
points = list(map(int, input().split()))
hand = {'r': 0, 's': 1, 'p': 2}
T = [hand[t] for t in input()]
win = [False] * N
ans = 0
for i in range(N):
if i < K or T[i] != T[i-K] or not win[i-K]:
win[i] = True
ans += points[(T[i]+2) % 3]
print(ans)
```
| 8,206 |
Provide a correct Python 3 solution for this coding contest problem.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
"Correct Solution:
```
N,K=map(int,input().split())
R,S,P=map(int,input().split())
T=list(input())
for i in range(N-K):
if T[K+i]==T[i]:
T[K+i]=""
ans=T.count("r")*P+T.count("s")*R+T.count("p")*S
print(ans)
```
| 8,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k=map(int,input().split())
r,s,p=map(int,input().split())
t=list(input())
cnt={'r':0,'s':0,'p':0}
for i in range(n):
if i<=k-1:
cnt[t[i]]+=1
elif t[i]==t[i-k]:
t[i]='o'
continue
else:
cnt[t[i]]+=1
ans=p*cnt['r']+r*cnt['s']+s*cnt['p']
print(ans)
```
Yes
| 8,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = list(input())
score = {"r":p, "s":r, "p":s}
ans = 0
for i in range(n):
if i<k:
ans += score[t[i]]
else :
if t[i] != t[i-k] :
ans += score[t[i]]
else :
t[i]="x"
print(ans)
```
Yes
| 8,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k = map(int,input().split())
r,s,p = map(int,input().split())
t = list(input())
ans = 0
for i in range(k,n) :
if t[i] == t[i-k] :
t[i] = 'x'
for i in range(n) :
if t[i] == 'r' :
ans += p
elif t[i] == 's' :
ans += r
elif t[i] == 'p' :
ans += s
print(ans)
```
Yes
| 8,210 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
N,K=map(int,input().split())
R,S,P=map(int,input().split())
T=input()
arr=[0]*N
d={'r':P,'s':R,'p':S}
for i,t in enumerate(T):
if i-K>=0:
if T[i-K]!=t:
arr[i]=d[t]
elif arr[i-K]==0:
arr[i]=d[t]
else:
arr[i]=d[t]
print(sum(arr))
```
Yes
| 8,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
from collections import deque
N, K = [int(x) for x in input().split()]
r, s, p = [int(x) for x in input().split()]
win_point = {
'r': r,
's': s,
'p': p,
}
next_hands = {
'r': ['s', 'p'],
's': ['r', 'p'],
'p': ['r', 's'],
}
enemy_hands = input()
def can_win(enemy_hand, my_hands):
# 勝てる場合にはTrueと勝てる手を教えてくれる
if enemy_hand == 'r' and 'p' in my_hands:
return True, 'p'
if enemy_hand == 's' and 'r' in my_hands:
return True, 'r'
if enemy_hand == 'p' and 's' in my_hands:
return True, 's'
# 勝てる手がない場合
return False, None
def choose_hand(enemy_hand, next_enemy_hand, now_hands):
# 次の手に勝てるようにその手以外を選択する
if next_enemy_hand == enemy_hand:
# 次回も同じ手ならどっちでも良い
return now_hands[0]
win, hand = can_win(next_enemy_hand, enemy_hand)
# 次の手と今の相手の手を戦わせる
# 後の手が勝ったら今の手と同じものを出す
# 今の手が勝ったら後の手と同じものと出す
if win:
return now_hands[enemy_hand]
else:
return now_hands[next_enemy_hand]
point = 0
for index in range(K):
target = index
now_hands = ['r', 'p', 's']
for i in range(index, N, K):
win, hand = can_win(enemy_hands[i], now_hands)
if win:
point += win_point[hand]
now_hands = next_hands[hand]
else:
if i + K < N:
next_enemy_hand = enemy_hands[i+K]
now_hands = next_hands[choose_hand(
enemy_hands[i], next_enemy_hand, now_hands)]
print(point)
```
No
| 8,212 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
r, s, p = map(int, input().split())
t = str(input())
```
No
| 8,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n, k = map(int, input().split())
R, S, P = map(int, input().split())
t = input()
ans = 0
hand = []
for i in t:
if i == "r":
hand.append("p")
if i == "s":
hand.append("r")
if i == "p":
hand.append("s")
for i, h in enumerate(t):
if i >= k:
if h == "r":
if hand[i-k] != "p":
ans += P
else:
hand[i] = "r"
if h == "s":
if hand[i-k] != "r":
ans += R
else:
hand[i] = "s"
if h == "p":
if hand[i-k] != "s":
ans += S
else:
hand[i] = "p"
else:
if h == "r":
ans += P
if h == "s":
ans += R
if h == "p":
ans += S
print(ans)
```
No
| 8,214 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:
* The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)
* Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):
* R points for winning with Rock;
* S points for winning with Scissors;
* P points for winning with Paper.
* However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)
Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.
The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is `r`, the machine will play Rock in the i-th round. Similarly, `p` and `s` stand for Paper and Scissors, respectively.
What is the maximum total score earned in the game by adequately choosing the hand to play in each round?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq K \leq N-1
* 1 \leq R,S,P \leq 10^4
* N,K,R,S, and P are all integers.
* |T| = N
* T consists of `r`, `p`, and `s`.
Input
Input is given from Standard Input in the following format:
N K
R S P
T
Output
Print the maximum total score earned in the game.
Examples
Input
5 2
8 7 6
rsrpr
Output
27
Input
7 1
100 10 1
ssssppr
Output
211
Input
30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr
Output
4996
Submitted Solution:
```
n,k=map(int,input().split())
p,r,s=map(int,input().split())
mc=input()
ans=0
for i in range(len(mc)):
if i>=k and mc[i-k]!=mc[i]:
if mc[i]=='r':
ans+=r
elif mc[i]=='p':
ans+=p
else:
ans+=s
```
No
| 8,215 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
### BIT binary
def init(bit, values):
for i,v in enumerate(values):
add(bit,i+1,v)
#a1 ~ aiまでの和 O(logn)
def query(bit,i):
res = 0
while i > 0:
res += bit[i]
i -= i&(-i)
return res
#ai += x(logN)
def add(bit,i,x):
if i==0:
raise RuntimeError
while i <= len(bit)-1:
bit[i] += x
i += i&(-i)
return
n = int(input())
xy = [tuple(map(int, input().split())) for _ in range(n)]
from bisect import bisect_left
def press(l):
# xs[inds[i]]==l[i]となる
xs = sorted(set(l))
inds = [None] * len(l)
for i,item in enumerate(l):
inds[i] = bisect_left(xs, item)
return xs, inds
_, indx = press([item[0] for item in xy])
_, indy = press([item[1] for item in xy])
xy = list(zip(indx, indy))
xy.sort()
M = 998244353
# ans = (pow(2, n, M) * n) % M
pows = [0]*(n+1)
v = 1
for i in range(n+1):
pows[i] = v-1
v *= 2
v %= M
ans = (pows[n]*n) % M
bit = [0]*(n+1)
for i,(x,y) in enumerate(xy):
ans -= pows[i] + pows[n-i-1]
ans += pows[query(bit, y+1)] + pows[i - query(bit, y+1)]
add(bit, y+1, 1)
ans %= M
from operator import itemgetter
xy.sort(key=itemgetter(1))
bit = [0]*(n+1)
for i,(x,y) in enumerate(xy):
ans -= pows[i] + pows[n-i-1]
ans += pows[i - query(bit, x+1)]
add(bit, x+1, 1)
ans %= M
bit = [0]*(n+1)
for i in range(n-1, -1, -1):
x,y = xy[i]
ans += pows[(n-1-i) - query(bit, x+1)]
add(bit, x+1, 1)
ans %= M
print(ans%M)
```
| 8,216 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
input = sys.stdin.readline
N=int(input())
POINT=[list(map(int,input().split())) for i in range(N)]
mod=998244353
PX=[p[0] for p in POINT]
PY=[p[1] for p in POINT]
compression_dict_x={a: ind for ind, a in enumerate(sorted(set(PX)))}
compression_dict_y={a: ind for ind, a in enumerate(sorted(set(PY)))}
for i in range(N):
POINT[i]=[compression_dict_x[POINT[i][0]]+1,compression_dict_y[POINT[i][1]]+1]
P_Y=sorted(POINT,key=lambda x:x[1])
# BIT(BIT-indexed tree)
LEN=len(compression_dict_x)# 必要なら座標圧縮する
BIT=[0]*(LEN+1)# 1-indexedなtree
def update(v,w):# vにwを加える
while v<=LEN:
BIT[v]+=w
v+=(v&(-v))# 自分を含む大きなノードへ. たとえばv=3→v=4
def getvalue(v):# [1,v]の区間の和を求める
ANS=0
while v!=0:
ANS+=BIT[v]
v-=(v&(-v))# 自分より小さい2ベキのノードへ. たとえばv=3→v=2へ
return ANS
ALL = pow(2,N,mod)-1
ANS= ALL*N%mod
for i in range(N):
ANS = (ANS - (pow(2,i,mod) -1)*4)%mod
for i in range(N):
x,y=P_Y[i]
up=getvalue(x)
ANS=(ANS+pow(2,up,mod)+pow(2,i-up,mod)-2)%mod
update(x,1)
P_Y2=P_Y[::-1]
BIT=[0]*(LEN+1)
for i in range(N):
x,y=P_Y2[i]
down=getvalue(x)
ANS=(ANS+pow(2,down,mod)+pow(2,i-down,mod)-2)%mod
update(x,1)
print(ANS)
```
| 8,217 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
import array
from operator import itemgetter
def main():
input = sys.stdin.readline
md = 998244353
n = int(input())
tl=n+1
ft=[0]*tl
xy = [[0]*2 for _ in range(n)]
for i in range(n):
xy[i] = [int(item) for item in input().split()]
xy.sort(key=itemgetter(0))
yxi = [y for x, y in xy]
*YS, = set(yxi)
YS.sort()
yxi= list(map({e: i for i, e in enumerate(YS)}.__getitem__, yxi))
ct=[0]*(n+1)
ct[0]=1
for i in range(1,n+1):
ct[i]=ct[i-1]*2%md
cnt=tuple(ct)
def upd(i):
i+=1
while(i<=n):
ft[i]+=1
i+=i&-i
def get(i):
i+=1
ret=0
while(i!=0):
ret+=ft[i]
i-=i&-i
return ret
def calc(get,upd):
for i, y in enumerate(yxi):
v = get(y); upd(y)
p1 = cnt[v]; p0 = cnt[y - v]
q1 = cnt[i - v]; q0 = cnt[(n - y - 1) - (i - v)]
yield (p0 + p1 + q0 + q1 - (p0 + q1) * (p1 + q0)) % md
print((sum(calc(get,upd))+n*cnt[n] - n)%md)
main()
```
| 8,218 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
class BitSum:
def __init__(self, n):
self.n = n + 1
self.table = [0] * self.n
def add(self, i, x):
i += 1
while i < self.n:
self.table[i] += x
i += i & -i
def sum(self, i):
i += 1
res = 0
while i > 0:
res += self.table[i]
i -= i & -i
return res
def main():
md=998244353
n=II()
xy=LLI(n)
yy=set()
for x,y in xy:yy.add(y)
ytoj={y:j for j,y in enumerate(sorted(yy))}
ans=n*(pow(2,n,md)-1)-4*(pow(2,n,md)-1-n)
ans%=md
xy.sort()
bit=BitSum(200005)
for i,[x,y] in enumerate(xy):
j=ytoj[y]
c=bit.sum(j-1)
ans+=pow(2,c,md)-1
ans+=pow(2,i-c,md)-1
ans%=md
bit.add(j,1)
bit=BitSum(200005)
for i,[x,y] in enumerate(xy[::-1]):
j=ytoj[y]
c=bit.sum(j-1)
ans+=pow(2,c,md)-1
ans+=pow(2,i-c,md)-1
ans%=md
bit.add(j,1)
print(ans)
main()
```
| 8,219 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
input=sys.stdin.readline
N=int(input())
p=[]
for i in range(N):
x,y=map(int,input().split())
p.append([x,y])
mod=998244353
pow=[1]
for i in range(N):
pow.append((pow[-1]*2)%mod)
p.sort(key=lambda x:x[1])
for i in range(N):
p[i][1]=i+1
data=[[0 for i in range(4)] for i in range(N)]
#A1 ... AnのBIT(1-indexed)
n=N
BIT = [0]*(n+1)
#A1 ~ Aiまでの和 O(logN)
def BIT_query(idx):
if idx==0:
return 0
res_sum = 0
while idx > 0:
res_sum += BIT[idx]
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def BIT_update(idx,x):
while idx <= n:
BIT[idx] += x
idx += idx&(-idx)
return
p.sort()
for i in range(N):
x,y=p[i]
a,b=i,BIT_query(y)
data[i][0]=a-b
data[i][3]=b
BIT_update(y,1)
BIT=[0]*(n+1)
for i in range(N-1,-1,-1):
x,y=p[i]
a,b=N-1-i,BIT_query(y)
data[i][1]=a-b
data[i][2]=b
BIT_update(y,1)
ans=0
for i in range(N):
ban=0
for j in range(4):
ban+=(pow[data[i][j]]-1)*pow[data[i][j-1]]
ban%=mod
ans+=pow[N]-ban-1
ans%=mod
print(ans)
```
| 8,220 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
input = sys.stdin.readline
N=int(input())
POINT=[list(map(int,input().split())) for i in range(N)]
mod=998244353
compression_dict_x={a: ind for ind, a in enumerate(sorted(set([p[0] for p in POINT])))}
POINT=[[compression_dict_x[x]+1,y] for x,y in POINT]
P_Y=sorted(POINT,key=lambda x:x[1])
# BIT(BIT-indexed tree)
LEN=len(compression_dict_x)
BIT=[0]*(LEN+1)# 1-indexedなtree
def update(v,w):# vにwを加える
while v<=LEN:
BIT[v]+=w
v+=(v&(-v))# 自分を含む大きなノードへ. たとえばv=3→v=4
def getvalue(v):# [1,v]の区間の和を求める
ANS=0
while v!=0:
ANS+=BIT[v]
v-=(v&(-v))# 自分より小さい2ベキのノードへ. たとえばv=3→v=2へ
return ANS
ANS= 4*N+(pow(2,N,mod)-1)*(N-4)%mod
for i in range(N):
x,y=P_Y[i]
left=getvalue(x)
ANS=(ANS+pow(2,left,mod)+pow(2,i-left,mod)-2)%mod
update(x,1)
P_Y.reverse()
BIT=[0]*(LEN+1)
for i in range(N):
x,y=P_Y[i]
left=getvalue(x)
ANS=(ANS+pow(2,left,mod)+pow(2,i-left,mod)-2)%mod
update(x,1)
print(ANS)
```
| 8,221 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
import sys
input = sys.stdin.readline
mod = 998244353
n = int(input())
XY = sorted(tuple(map(int, input().split())) for _ in range(n))
X, Y = zip(*XY)
L = {y:i+1 for i, y in enumerate(sorted(Y))}
Y = tuple(L[y] for y in Y)
BIT = [0]*(n+1)
def add(i):
while i<=n:
BIT[i] += 1
i += i&-i
def acc(i):
res = 0
while i:
res += BIT[i]
i -= i&-i
return res
A = [0]*n
B = [0]*n
C = [0]*n
D = [0]*n
for i, y in enumerate(Y):
A[i] = acc(y)
B[i] = i - A[i]
add(y)
BIT = [0]*(n+1)
for i in range(n-1, -1, -1):
y = Y[i]
C[i] = acc(y)
D[i] = n-1-i - C[i]
add(y)
T = [1]*n
for i in range(n-1):
T[i+1] = T[i]*2%mod
ans = T[n-1] * n % mod
for a, b, c, d in zip(A, B, C, D):
a = T[a]
b = T[b]
c = T[c]
d = T[d]
temp = (a-1)*b*c*(d-1) + a*(b-1)*(c-1)*d - (a-1)*(b-1)*(c-1)*(d-1)
ans += temp
ans %= mod
print(ans)
```
| 8,222 |
Provide a correct Python 3 solution for this coding contest problem.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
"Correct Solution:
```
from bisect import bisect_right, bisect_left
# instead of AVLTree
class BITbisect():
def __init__(self, InputProbNumbers):
# 座圧
self.ind_to_co = [-10**18]
self.co_to_ind = {}
for ind, num in enumerate(sorted(list(set(InputProbNumbers)))):
self.ind_to_co.append(num)
self.co_to_ind[num] = ind+1
self.max = len(self.co_to_ind)
self.data = [0]*(self.max+1)
def __str__(self):
retList = []
for i in range(1, self.max+1):
x = self.ind_to_co[i]
if self.count(x):
c = self.count(x)
for _ in range(c):
retList.append(x)
return "[" + ", ".join([str(a) for a in retList]) + "]"
def __getitem__(self, key):
key += 1
s = 0
ind = 0
l = self.max.bit_length()
for i in reversed(range(l)):
if ind + (1<<i) <= self.max:
if s + self.data[ind+(1<<i)] < key:
s += self.data[ind+(1<<i)]
ind += (1<<i)
if ind == self.max or key < 0:
raise IndexError("BIT index out of range")
return self.ind_to_co[ind+1]
def __len__(self):
return self._query_sum(self.max)
def __contains__(self, num):
if not num in self.co_to_ind:
return False
return self.count(num) > 0
# 0からiまでの区間和
# 左に進んでいく
def _query_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
# i番目の要素にxを足す
# 上に登っていく
def _add(self, i, x):
while i <= self.max:
self.data[i] += x
i += i & -i
# 値xを挿入
def push(self, x):
if not x in self.co_to_ind:
raise KeyError("The pushing number didnt initialized")
self._add(self.co_to_ind[x], 1)
# 値xを削除
def delete(self, x):
if not x in self.co_to_ind:
raise KeyError("The deleting number didnt initialized")
if self.count(x) <= 0:
raise ValueError("The deleting number doesnt exist")
self._add(self.co_to_ind[x], -1)
# 要素xの個数
def count(self, x):
return self._query_sum(self.co_to_ind[x]) - self._query_sum(self.co_to_ind[x]-1)
# 値xを超える最低ind
def bisect_right(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_right(self.ind_to_co, x) - 1
return self._query_sum(i)
# 値xを下回る最低ind
def bisect_left(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_left(self.ind_to_co, x)
if i == 1:
return 0
return self._query_sum(i-1)
mod = 998244353
import sys
input = sys.stdin.readline
N = int(input())
XY = [list(map(int, input().split())) for _ in range(N)]
Ys = []
for x, y in XY:
Ys.append(y)
XY.sort()
D = [[0, 0, 0, 0] for _ in range(N)]
B1 = BITbisect(Ys)
for i, (x, y) in enumerate(XY):
k = B1.bisect_left(y)
D[i][0] = k
D[i][1] = i-k
B1.push(y)
B2 = BITbisect(Ys)
for i in reversed(range(N)):
x, y = XY[i]
k = B2.bisect_left(y)
D[i][2] = k
D[i][3] = (N-1-i)-k
B2.push(y)
N2 = [1]
for _ in range(N):
N2.append(N2[-1]*2%mod)
ans = 0
for i in range(N):
a, b, c, d = D[i]
p = N2[N] - 1 - (N2[a+b]+N2[b+d]+N2[d+c]+N2[c+a])%mod + (N2[a]+N2[b]+N2[c]+N2[d])%mod
ans = (ans + p) % mod
print(ans)
```
| 8,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
from operator import itemgetter
class BIT:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
i += 1
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
i += 1
while i <= self.size:
self.tree[i] += x
i += i & -i
mod = 998244353
N = int(input())
xy = [list(map(int,input().split())) for _ in range(N)]
xy.sort(key=itemgetter(1))
for i in range(N):
xy[i][1] = i
xy.sort(key=itemgetter(0))
pow2 = [1]*(N+1)
for i in range(N):
pow2[i+1] = 2*pow2[i]%mod
ans = pow2[N-1]*N%mod
bit = BIT(N)
for i in range(N):
l1 = bit.sum(xy[i][1])
l2 = i-l1
r1 = xy[i][1]-l1
r2 = N-xy[i][1]-l2-1
bit.add(xy[i][1],1)
ans += (pow2[l2]-1)*(pow2[r1]-1)*pow2[r2]*pow2[l1]
ans += (pow2[r2]-1)*(pow2[l1]-1)*(pow2[l2]+pow2[r1]-1)
ans %= mod
print(ans)
```
Yes
| 8,224 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
import sys,queue,math,copy,itertools,bisect,collections,heapq
def main():
MOD = 998244353
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())
N = NI()
dat = [LI() for _ in range(N)]
dat.sort()
yo = [y for _,y in dat]
yo.sort()
up = [0] * N
dw = [0] * N
bit = [0] * (N+1)
def bit_add(i):
while i <= N:
bit[i] += 1
i += i & -i
def bit_sum(i):
ret = 0
while i > 0:
ret += bit[i]
i -= i & -i
return ret
for i in range(N):
x,y = dat[i]
y = bisect.bisect_left(yo,y) + 1
up[i] = bit_sum(y)
bit_add(y)
bit = [0] * (N+1)
for i in range(N-1,-1,-1):
x,y = dat[i]
y = bisect.bisect_left(yo,y) + 1
dw[i] = bit_sum(y)
bit_add(y)
ans = N * pow(2,N-1,MOD)
for i in range(N):
a = i - up[i]
b = up[i]
c = N-(i+1) - dw[i]
d = dw[i]
p = (pow(2,a+d,MOD)-1) - (pow(2,a,MOD)-1) - (pow(2,d,MOD)-1)
q = (pow(2,b+c,MOD)-1) - (pow(2,b,MOD)-1) - (pow(2,c,MOD)-1)
x = p * (pow(2,b,MOD) + pow(2,c,MOD)-1)
y = q * (pow(2,a,MOD) + pow(2,d,MOD)-1)
z = p * q
ans = (ans + x + y + z) % MOD
print(ans)
if __name__ == '__main__':
main()
```
Yes
| 8,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 998244353
class BIT:
def __init__(self, size):
self.bit = [0] * size
self.size = size
def add(self, i, w):
x = i + 1
while x <= self.size:
self.bit[x - 1] += w
x += x & -x
return
def sum(self, i):
res = 0
x = i + 1
while x:
res += self.bit[x - 1]
x -= x & -x
return res
n = I()
X = []
Y = []
XY = []
for x, y in LIR(n):
Y += [y]
X += [x]
co_to_ind = {e: i for i, e in enumerate(sorted(Y))}
Y = [co_to_ind[k] for k in Y]
Y = [Y[i] for i in sorted(range(n), key=lambda j:X[j])]
pow2 = [1]
for i in range(n):
pow2 += [pow2[-1] * 2 % mod]
ans = pow2[n - 1] * n % mod
bit = BIT(n)
for i, y in enumerate(Y):
bit.add(y, 1)
ld = bit.sum(y - 1)
lu = i - ld
rd = y - ld
ru = n - y - 1 - lu
ans = ans + (pow2[ld] - 1) * (pow2[ru] - 1) % mod * pow2[lu] % mod * pow2[rd] % mod
ans = ans + (pow2[lu] - 1) * (pow2[rd] - 1) % mod * pow2[ld] % mod * pow2[ru] % mod
ans = ans - (pow2[ld] - 1) * (pow2[ru] - 1) % mod * (pow2[rd] - 1) % mod * (pow2[lu] - 1) % mod
print(ans % mod)
```
Yes
| 8,226 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
### BIT binary
def init(bit, values):
for i,v in enumerate(values):
add(bit,i+1,v)
#a1 ~ aiまでの和 O(logn)
def query(bit,i):
res = 0
while i > 0:
res += bit[i]
i -= i&(-i)
return res
#ai += x(logN)
def add(bit,i,x):
if i==0:
raise RuntimeError
while i <= len(bit)-1:
bit[i] += x
i += i&(-i)
return
n = int(input())
xy = [tuple(map(int, input().split())) for _ in range(n)]
from bisect import bisect_left
def press(l):
# xs[inds[i]]==l[i]となる
xs = sorted(set(l))
inds = [None] * len(l)
for i,item in enumerate(l):
inds[i] = bisect_left(xs, item)
return xs, inds
_, indx = press([item[0] for item in xy])
_, indy = press([item[1] for item in xy])
xy = []
for i in range(n):
xy.append(indy[i]+indx[i]*n)
xy.sort()
M = 998244353
# ans = (pow(2, n, M) * n) % M
pows = [0]*(n+1)
v = 1
for i in range(n+1):
pows[i] = v-1
v *= 2
v %= M
ans = (pows[n]*n) % M
bit = [0]*(n+1)
for i in range(n):
x,y = divmod(xy[i],n)
ans -= pows[i] + pows[n-i-1]
ans += pows[query(bit, y+1)] + pows[i - query(bit, y+1)]
add(bit, y+1, 1)
ans %= M
from operator import itemgetter
xy.sort(key=lambda item: item%n)
bit = [0]*(n+1)
for i in range(n):
x,y = divmod(xy[i],n)
ans -= pows[i] + pows[n-i-1]
ans += pows[i - query(bit, x+1)]
add(bit, x+1, 1)
ans %= M
bit = [0]*(n+1)
for i in range(n-1, -1, -1):
x,y = divmod(xy[i], n)
ans += pows[(n-1-i) - query(bit, x+1)]
add(bit, x+1, 1)
ans %= M
print(ans%M)
```
Yes
| 8,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
N = int(input())
p = 998244353
S = [[0, 0] for i in range(N)]
for i in range(N):
S[i] = [int(x) for x in input().split()]
S.sort()
Y = sorted([i for i in range(N)], key=lambda y:S[y][1])
D = [[0, 0, 0, 0] for i in range(N)]
BIT = [0] * N
def query(y):
_y = y + 1
ans = 0
while _y > 0:
ans += BIT[_y - 1]
_y -= _y & -_y
return ans
def update(y):
_y = y + 1
while _y <= N:
BIT[_y - 1] += 1
_y += _y & -_y
return 0
for i, y in enumerate(Y):
D[y][0] = query(y)
D[y][1] = i - D[y][0]
update(y)
BIT = [0] * N
for i, y in enumerate(Y[::-1]):
D[y][2] = query(y)
D[y][3] = i - D[y][2]
update(y)
cnt = 0
for i in range(N):
a, b, c, d = [pow(2, d, p) - 1 for d in D[i]]
cnt += pow(2, N - 1, p)
cnt += a * d * (b + c + 1)
cnt += b * c * (a + d + 1)
cnt += a * b * c * d
cnt %= p
print(int(cnt))
```
No
| 8,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
N = int(input())
XY = []
for _ in range(N):
x, y = map(int, input().split())
XY.append((x, y))
# BITの定義
import math
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
# index 1からiの値の和を返す
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
# index iの値にxを加算する
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
#######################
# 開始
p_list = XY[:]
# y座標の圧縮
from operator import itemgetter
p_list = sorted(p_list, key=itemgetter(1))
p_list = [ (x, y, i+1) for i, (x, y) in enumerate(p_list)]
# xでソート
p_list= sorted(p_list)
a = [0] * N # 左下
b = [0] * N# 左上
c = [0] * N# 右下
d = [0] * N# 右上
# a,bを求める
bit = Bit(N)
for i, (x, y, r) in enumerate(p_list):
tmp = bit.sum(r)
a[i] = tmp
b[i] = i-tmp
bit.add(r, 1)
# c,dを求める
bit = Bit(N)
for i, (x, y, r) in enumerate(reversed(p_list)):
tmp = bit.sum(r)
c[i] = tmp
d[i] = i-tmp
bit.add(r, 1)
c = c[::-1]
d = d[::-1]
# ある点が四角に含まれる場合の総数を求めるメソッド
power = [1] * N
for i in range(1, N):
power[i] = power[i-1] * 2 % 998244353
ans = 0
for _a,_b,_c,_d in zip(a,b,c,d):
_a = power[_a] - 1
_b = power[_b] - 1
_c = power[_c] - 1
_d = power[_d] - 1
ret = (_a+1) * (_d+1) * _c * _b
ret += (_b+1) * (_c+1) * _a * _d
ret -= _a*_b*_c*_d
ret += power[N-1]
ans += ret % 998244353
print(ans % 998244353)
```
No
| 8,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline()[:-1]
MOD = 998244353
class BIT():#1-indexed
def __init__(self, size):
self.table = [0 for _ in range(size+2)]
self.size = size
def Sum(self, i):#1からiまでの和
s = 0
while i > 0:
s += self.table[i]
i -= (i & -i)
return s
def PointAdd(self, i, x):#
while i <= self.size:
self.table[i] += x
i += (i & -i)
return
n = int(input())
dd = [list(map(int, input().split())) + [i] for i in range(n)]
d = [[-1, -1] for _ in range(n)]
dd.sort()
for i, x in enumerate(dd):
d[x[2]][0] = i+1
dd.sort(key=lambda x: x[1])
for i, x in enumerate(dd):
d[x[2]][1] = i+1
d.sort()
d = [(x[0], x[1]) for x in d]
lu = [0 for _ in range(n)]
ld = [0 for _ in range(n)]
rd = [0 for _ in range(n)]
ru = [0 for _ in range(n)]
bitl, bitr = BIT(n), BIT(n)
for i in range(n):
x = d[i][1]
s = bitl.Sum(x)
lu[i] = i - s
ld[i] = s
bitl.PointAdd(x, 1)
for i in range(n-1, -1, -1):
x = d[i][1]
s = bitr.Sum(x)
ru[i] = n-1-i - s
rd[i] = s
bitr.PointAdd(x, 1)
#print(lu, ld, rd, ru, sep="\n")
ans = (n*pow(2, n, MOD) - n) % MOD
for i in range(n):
ans -= pow(2, lu[i]+ld[i], MOD) + pow(2, ld[i]+rd[i], MOD) + pow(2, rd[i]+ru[i], MOD) + pow(2, ru[i]+lu[i], MOD)
ans += pow(2, lu[i], MOD) + pow(2, ld[i], MOD) + pow(2, rd[i], MOD) + pow(2, ru[i], MOD)
ans %= MOD
print(ans)
```
No
| 8,230 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a set S of N points in a two-dimensional plane. The coordinates of the i-th point are (x_i, y_i). The N points have distinct x-coordinates and distinct y-coordinates.
For a non-empty subset T of S, let f(T) be the number of points contained in the smallest rectangle, whose sides are parallel to the coordinate axes, that contains all the points in T. More formally, we define f(T) as follows:
* f(T) := (the number of integers i (1 \leq i \leq N) such that a \leq x_i \leq b and c \leq y_i \leq d, where a, b, c, and d are the minimum x-coordinate, the maximum x-coordinate, the minimum y-coordinate, and the maximum y-coordinate of the points in T)
Find the sum of f(T) over all non-empty subset T of S. Since it can be enormous, print the sum modulo 998244353.
Constraints
* 1 \leq N \leq 2 \times 10^5
* -10^9 \leq x_i, y_i \leq 10^9
* x_i \neq x_j (i \neq j)
* y_i \neq y_j (i \neq j)
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
:
x_N y_N
Output
Print the sum of f(T) over all non-empty subset T of S, modulo 998244353.
Examples
Input
3
-1 3
2 1
3 -2
Output
13
Input
4
1 4
2 1
3 3
4 2
Output
34
Input
10
19 -11
-3 -12
5 3
3 -15
8 -14
-9 -20
10 -9
0 2
-7 17
6 -6
Output
7222
Submitted Solution:
```
def main():
import sys
from operator import itemgetter
input = sys.stdin.readline
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
mod = 998244353
N = int(input())
X_raw = [0] * N
Y_raw = [0] * N
for i in range(N):
x, y = map(int, input().split())
X_raw[i] = x
Y_raw[i] = y
val2idx_X = {}
val2idx_Y = {}
for i in range(N):
val2idx_X[X_raw[i]] = i
val2idx_Y[Y_raw[i]] = i
X_raw.sort()
Y_raw.sort()
X = [0] * N
Y = [0] * N
for i in range(N):
X[val2idx_X[X_raw[i]]] = i+1
Y[val2idx_Y[Y_raw[i]]] = i+1
XY = [(x, y) for x, y in zip(X, Y)]
XY.sort(key=itemgetter(0))
bit_ul = Bit(N + 2)
bit_ur = Bit(N+2)
ul = []
ur = []
for x, y in XY:
ul.append(bit_ul.sum(y))
ur.append(bit_ur.sum(N+1) - bit_ur.sum(y))
bit_ul.add(y, 1)
bit_ur.add(y, 1)
bit_dl = Bit(N + 2)
bit_dr = Bit(N + 2)
dl = []
dr = []
for x, y in reversed(XY):
dl.append(bit_dl.sum(y))
dr.append(bit_dr.sum(N + 1) - bit_dr.sum(y))
bit_dl.add(y, 1)
bit_dr.add(y, 1)
dl.reverse()
dr.reverse()
ans = 0
for i in range(N):
ans = (ans + pow(2, N, mod) - 1
- pow(2, ul[i]+ur[i], mod)
- pow(2, dl[i]+dr[i], mod)
- pow(2, ul[i]+dl[i], mod)
- pow(2, ur[i]+dr[i], mod)
+ pow(2, ul[i], mod)
+ pow(2, ur[i], mod)
+ pow(2, dl[i], mod)
+ pow(2, dr[i], mod))%mod
print(ans)
if __name__ == '__main__':
main()
```
No
| 8,231 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
import sys
readline = sys.stdin.readline
N, A, B = map(int, readline().split())
*P, = map(int, readline().split())
Q = [0]*N
for i, p in enumerate(P):
Q[p-1] = i
INF = 10**18
dp = [[INF]*(N+1) for i in range(N+1)]
dp[0][0] = 0
for i in range(N):
qi = Q[i]
for j in range(N):
v = dp[i][j]
dp[i+1][j] = min(dp[i+1][j], v + (A if qi < j else B))
dp[i][j+1] = min(dp[i][j+1], v)
if Q[i] == j:
dp[i+1][j+1] = min(dp[i+1][j+1], v)
for i in range(N):
dp[i+1][N] = min(dp[i+1][N], dp[i][N] + A)
for j in range(N):
dp[N][j+1] = min(dp[N][j+1], dp[N][j])
sys.stdout.write("%d\n" % dp[N][N])
```
| 8,232 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
n, a, b = map(int, input().split())
dp = [2 ** 60] * (n + 1)
dp[0] = 0
p = list(map(int, input().split()))
for i in range(n):
dp[i + 1] = dp[0]
dp[0] += a
for j in range(i):
if p[j] < p[i]:
dp[i + 1] = min(dp[i + 1], dp[j + 1])
dp[j + 1] += a
else:
dp[j + 1] += b
print(min(dp))
```
| 8,233 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
N, A, B = map(int, input().split())
P = [int(a) - 1 for a in input().split()]
X = [0] * N
for i in range(N):
pre = X
X = [0] * N
mi = pre[0]
for j in range(N):
mi = min(mi, pre[j])
X[j] = mi + (A if j < P[i] else B if j > P[i] else 0)
print(min(X))
```
| 8,234 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
import sys
input=sys.stdin.readline
N,A,B=map(int,input().split())
P=list(map(int,input().split()))
data=[0]*(N+1)
flag=[0]*(N+1)
for i in range(N):
p=P[i]
flag[p]=1
cnt=0
for j in range(p+1,N+1):
if flag[j]==0:
data[j]=data[j-1]
else:
cnt+=1
data[j]=min(cnt*A+data[p-1],data[j]+B)
print(data[-1])
```
| 8,235 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
n, a, b = map(int, input().split())
p = list(map(int, input().split()))
INF = 10**13
dp = [[INF for _ in range(n+1)] for _ in range(n+1)]
dp[0][0] = 0
for i in range(n):
for j in range(n+1):
if j < p[i]:
dp[i+1][p[i]] = min(dp[i+1][p[i]], dp[i][j])
dp[i+1][j] = min(dp[i+1][j], dp[i][j] + a)
else:
dp[i+1][j] = min(dp[i+1][j], dp[i][j] + b)
print(min(dp[n]))
```
| 8,236 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
from bisect import bisect
n, a, b = map(int, input().split())
ppp = map(int, input().split())
qqq = [0] * n
for i, p in enumerate(ppp, start=1):
qqq[p - 1] = i
dp = [(0, 0)]
for i in qqq:
s = bisect(dp, (i,))
ndp = [(j, cost + b) for j, cost in dp[:s]]
stay_cost = dp[s - 1][1]
ndp.append((i, stay_cost))
remain = iter(dp[s:])
for j, cost in remain:
if stay_cost > cost + a:
ndp.append((j, cost + a))
break
ndp.extend((j, cost + a) for j, cost in remain)
dp = ndp
print(dp[-1][1])
```
| 8,237 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
N, R, L = map(int, input().split())
P = list(map(int, input().split()))
val2idx = [0] * (N+1)
for i, p in enumerate(P):
val2idx[p] = i + 1
bigger = [[0] * (N+1) for _ in range(N+1)]
for p in range(1, N+1):
for j in range(N):
if P[j] > p:
bigger[p][j+1] = bigger[p][j] + 1
else:
bigger[p][j + 1] = bigger[p][j]
#print(bigger)
inf = 10**14
dp = [[inf] * (N+1) for _ in range(N+1)]
dp[0][0] = 0
for p in range(1, N+1):
i = val2idx[p]
for j in range(N+1):
if i > j:
dp[p][j] = min(dp[p][j], dp[p-1][j] + L)
dp[p][i] = min(dp[p][i], dp[p-1][j] + R * (bigger[p][i] - bigger[p][j]))
else:
dp[p][j] = min(dp[p][j], dp[p - 1][j])
print(min(dp[-1]))
#[print(i, dp[i]) for i in range(N+1)]
if __name__ == '__main__':
main()
```
| 8,238 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
"Correct Solution:
```
n, a, b = map(int, input().split())
dp = [0] + [2 ** 60] * n
p = list(map(int, input().split()))
for i in range(n):
dp[i + 1] = dp[0]
dp[0] += a
for j in range(i):
if p[j] < p[i]: dp[i + 1] = min(dp[i + 1], dp[j + 1])
dp[j + 1] += [b, a][p[j] < p[i]]
print(min(dp))
```
| 8,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
N,A,B=map(int,input().split())
p=list(map(int,input().split()))
p=[p[i]-1 for i in range(N)]
dp=[[0 for i in range(N+1)] for j in range(N)]
for j in range(N+1):
if p[0]<=j:
dp[0][j]=0
else:
dp[0][j]=A
for i in range(1,N):
for j in range(N+1):
if p[i]<=j:
dp[i][j]=min(dp[i-1][p[i]],dp[i-1][j]+B)
else:
dp[i][j]=dp[i-1][j]+A
print(dp[-1][N])
```
Yes
| 8,240 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
N,A,B=map(int,input().split())
P=[int(i) for i in input().split()]
s=1e19
d=[s]*(N+1)
d[0]=0
for i in range(N):
e=[s]*(N+1)
for j in range(N+1):
if j < P[i]:
e[P[i]]=min(e[P[i]],d[j])
e[j]=min(e[j],d[j]+A)
else:
e[j]=min(e[j],d[j]+B)
d=e
print(min(d))
```
Yes
| 8,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
"""
Writer: SPD_9X2
https://atcoder.jp/contests/agc032/tasks/agc032_d
圧倒的挿入ソート
https://atcoder.jp/contests/abc006/tasks/abc006_4
これの、左右のコストが違うVer
最長部分増加列を計算すれば、A=Bの時の答えは分かる
その時の答えは, A * (N - 最長部分増加列の長さ)
残りのカードが昇順になるように抜き出し、適切な場所に戻す。
最長部分列として採用する or しないで dp?
dp[i][部分増加列の最後の数字] = i番目まで見た時のコストの最小値 かな？
増加列の最後に足せば、コストはかからない
足さないとき、最後の数字より大きければ、A払う
小さければ、B払う
O(N**2)で解けた？
"""
N,A,B = map(int,input().split())
p = list(map(int,input().split()))
dp = [float("inf")] * (N+1)
dp[0] = 0
for i in range(N):
ndp = [float("inf")] * (N+1)
for j in range(N+1):
if p[i] >= j: #p[i]よりjがおおきいので、Aコスト & 更新ができる
ndp[j] = min(ndp[j] , dp[j] + A)
ndp[p[i]] = min(ndp[p[i]] , dp[j])
else: #Bコストでの更新
ndp[j] = min(ndp[j] , dp[j] + B)
dp = ndp
print (min(dp))
```
Yes
| 8,242 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
from bisect import bisect
from operator import itemgetter
n, a, b = map(int, input().split())
ppp = map(int, input().split())
qqq = [i for i, p in sorted(enumerate(ppp, start=1), key=itemgetter(1))]
INF = 10 ** 15
dp = [(0, 0)]
for i in qqq:
ndp = []
s = bisect(dp, (i,))
tmp_min = INF
for j, cost in dp[:s]:
cost += b
ndp.append((j, cost))
tmp_min = cost
stay_cost = dp[s - 1][1]
if tmp_min > stay_cost:
ndp.append((i, stay_cost))
tmp_min = stay_cost
for j, cost in dp[s:]:
cost += a
if tmp_min > cost:
ndp.append((j, cost))
tmp_min = cost
dp = ndp
print(dp[-1][1])
```
Yes
| 8,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
from collections import deque
class Dinic:
def __init__(self, n):
self.n = n
self.links = [[] for _ in range(n)]
self.depth = None
self.progress = None
def add_link(self, _from, to, cap):
self.links[_from].append([cap, to, len(self.links[to])])
self.links[to].append([0, _from, len(self.links[_from]) - 1])
def bfs(self, s):
depth = [-1] * self.n
depth[s] = 0
q = deque([s])
while q:
v = q.popleft()
for cap, to, rev in self.links[v]:
if cap > 0 and depth[to] < 0:
depth[to] = depth[v] + 1
q.append(to)
self.depth = depth
def dfs(self, v, t, flow):
if v == t:
return flow
links_v = self.links[v]
for i in range(self.progress[v], len(links_v)):
self.progress[v] = i
cap, to, rev = link = links_v[i]
if cap == 0 or self.depth[v] >= self.depth[to]:
continue
d = self.dfs(to, t, min(flow, cap))
if d == 0:
continue
link[0] -= d
self.links[to][rev][0] += d
return d
return 0
def max_flow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.depth[t] < 0:
return flow
self.progress = [0] * self.n
current_flow = self.dfs(s, t, float('inf'))
while current_flow > 0:
flow += current_flow
current_flow = self.dfs(s, t, float('inf'))
n, a, b = map(int, input().split())
aaa = list(map(int, input().split()))
mf = Dinic(2 * n + 2)
t = 2 * n + 1
for i in range(1, n + 1):
mf.add_link(0, i, b)
mf.add_link(n + i, t, a)
INF = 10 ** 18
completed = [False] * (n + 1)
for i in range(n):
r = aaa[i]
for l in reversed(aaa[:i]):
if l > r:
mf.add_link(r, l + n, INF)
if l + 1 == r:
mf.add_link(r, l, INF)
break
print('Time for Construct Graph') # Test
exit()
print(mf.max_flow(0, t))
```
No
| 8,244 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
import sys
sys.setrecursionlimit(10000)
n, a, b = map(int, input().split())
p = list(map(int, input().split()))
p.append(n+1)
p.insert(0,0)
#'''
#print (p)
INF = (a+b)*n
min_ans =INF # 仮のmin値
# l-shift で 右端を整列する
def ptr_l_shift(p, cost, l, r):
for i in range(l, r+1):
if p[i] == r :
if r == i:
return (0 + solve(p, cost, l, r-1))
# l_shift(l, r)
p = p[0:i] +p[i+1:r+1] +p[i:i+1] +p[r+1:]
ret = solve(p, cost, l, r-1)
p = p[0:i] +p[r:r+1] +p[i:r] +p[r+1:]
return (a + ret)
# print("error l_shift", p,l,r)
# r-shift で 左端を整列する
def ptr_r_shift(p, cost, l, r) :
for i in range(l, r+1):
if p[i] == l :
if i == l :
return (0 + solve(p, cost, l+1, r))
# r_shift(i, j)
p = p[0:l] +p[i:i+1] +p[l:i] +p[i+1:]
ret = solve(p, cost, l+1, r)
p = p[0:l] +p[l+1:i+1] +p[l:l+1] +p[i+1:]
return (b + ret)
# print("error r_shift", p,l,r)
def solve(p, cost, ptr_l, ptr_r):
global min_ans
if cost >= min_ans :
return (INF)
if ptr_l == ptr_r :
return (0)
if a < b : # コストの小さい方を優先する
ll = ptr_l_shift(p, cost, ptr_l, ptr_r)
rr = ptr_r_shift(p, cost, ptr_l, ptr_r)
else:
rr = ptr_r_shift(p, cost, ptr_l, ptr_r)
ll = ptr_l_shift(p, cost, ptr_l, ptr_r)
# print ("return ll, rr =", ll, rr)
if ll > rr :
if rr < min_ans :
min_ans = rr
return (rr)
else :
if ll < min_ans :
min_ans = ll
return (ll)
cost = 0
print (solve(p, cost, 1, n))
```
No
| 8,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
n, a, b = map(int, input().split())
p = list(map(int, input().split()))
q = [0] * n
for i, x in enumerate(p):
q[x - 1] = i
dp = [[0] * (n + 1) for _ in range(n + 1)]
mi = [[float('inf')] * (n + 1) for _ in range(n + 1)]
ma = [[0] * (n + 1) for _ in range(n + 1)]
for i in range(n + 1):
for j in range(i + 1, n + 1):
mi[i][j] = min(mi[i][j - 1], q[j - 1])
ma[i][j] = max(ma[i][j - 1], q[j - 1])
for i in range(1, n + 1):
for j in range(i + 1):
k = i - j
s, t = 1, 1
if j > 0 and j == p[mi[j - 1][n - k]]:
s = 0
if k > 0 and n - k + 1 == p[ma[j][n - k + 1]]:
t = 0
if j == 0:
dp[j][k] = dp[j][k - 1] + a * t
elif k == 0:
dp[j][k] = dp[j - 1][k] + b * s
else:
dp[j][k] = min(dp[j - 1][k] + b * s, dp[j][k - 1] + a * t)
ans = float('inf')
for i in range(n + 1):
ans = min(ans, dp[i][n - i])
print(ans)
```
No
| 8,246 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p = (p_1, \ldots, p_N) of \\{ 1, \ldots, N \\}. You can perform the following two kinds of operations repeatedly in any order:
* Pay a cost A. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the left by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_{l + 1}, p_{l + 2}, \ldots, p_r, p_l, respectively.
* Pay a cost B. Choose integers l and r (1 \leq l < r \leq N), and shift (p_l, \ldots, p_r) to the right by one. That is, replace p_l, p_{l + 1}, \ldots, p_{r - 1}, p_r with p_r, p_l, \ldots, p_{r - 2}, p_{r - 1}, respectively.
Find the minimum total cost required to sort p in ascending order.
Constraints
* All values in input are integers.
* 1 \leq N \leq 5000
* 1 \leq A, B \leq 10^9
* (p_1 \ldots, p_N) is a permutation of \\{ 1, \ldots, N \\}.
Input
Input is given from Standard Input in the following format:
N A B
p_1 \cdots p_N
Output
Print the minimum total cost required to sort p in ascending order.
Examples
Input
3 20 30
3 1 2
Output
20
Input
4 20 30
4 2 3 1
Output
50
Input
1 10 10
1
Output
0
Input
4 1000000000 1000000000
4 3 2 1
Output
3000000000
Input
9 40 50
5 3 4 7 6 1 2 9 8
Output
220
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Feb 16 20:52:46 2019
@author: Owner
"""
import collections
import scipy.misc
import sys
import numpy as np
import math
from operator import itemgetter
import itertools
import copy
import bisect
#素因数を並べる
def prime_decomposition(n):
i = 2
table = []
while i * i <= n:
while n % i == 0:
n /= i
table.append(int(i))
i += 1
if n > 1:
table.append(int(n))
return table
# 桁数を吐く
def digit(i):
if i > 0:
return digit(i//10) + [i%10]
else:
return []
def getNearestValueIndex(list, num):
"""
概要: リストからある値に最も近い値のインデックスを取得する関数
@param list: データ配列
@param num: 対象値
@return 対象値に最も近い値
"""
# リスト要素と対象値の差分を計算し最小値のインデックスを取得
idx = np.abs(np.asarray(list) - num).argmin()
return idx
def find_index(l, x, default=False):
if x in l:
return l.index(x)
else:
return default
"""
N, X = map(int, input().split())
x = [0]*N
x[:] = map(int, input().split())
P = [0]*N
Y = [0]*N
for n in range(N):
P[n], Y[n] = map(int, input().split())
all(nstr.count(c) for c in '753')
# 複数配列を並び替え
ABT = zip(A, B, totAB)
result = 0
# itemgetterには何番目の配列をキーにしたいか渡します
sorted(ABT,key=itemgetter(2))
A, B, totAB = zip(*ABT)
A.sort(reverse=True)
# 2進数のbit判定
(x >> i) & 1
# dp最小化問題
dp = [np.inf]*N
for n in range(N):
if n == 0:
dp[n] = 0
else:
for k in range(1,K+1):
if n-k >= 0:
dp[n] = min(dp[n], dp[n-k] + abs(h[n]-h[n-k]))
else:
break
"""
N, A, B = map(int, input().split())
p = [0]*N
p[:] = map(int, input().split())
p_end = find_index(p, 1)
cost = 0
for i in range(N-1):
n = i+2
i_idx = find_index(p,n)
tmp_p = p[i_idx]
if p_end == i_idx-1:
p_end += 1
elif p_end > i_idx:
cost += A
p[i_idx:p_end] = p[i_idx+1:p_end+1]
p[p_end] = tmp_p
else:
cost += B
p[p_end+2:i_idx+1] = p[p_end+1:i_idx]
p[p_end+1] = tmp_p
p_end += 1
print(n, p, p_end, i_idx)
print(cost)
```
No
| 8,247 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
n = input()
print('Hello World' if n == '1' else int(input()) + int(input()))
```
| 8,248 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
n,*l=map(int,open(0));print("Hello World" if n==1 else sum(l))
```
| 8,249 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
n=int(input())
if n==1:
print("Hello World")
else:
print(sum([int(input()), int(input())]))
```
| 8,250 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
import sys
if input()=="1": print("Hello World"),sys.exit()
print(int(input())+int(input()))
```
| 8,251 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
a = input()
if a == '1':
print('Hello World')
else:
print(int(input()) + int(input()))
```
| 8,252 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
n,*a=map(int,open(0).read().split()); print('Hello World' if n==1 else sum(a))
```
| 8,253 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
n = int(input())
print('Hello World' if n ==1 else int(input())+int(input()))
```
| 8,254 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
"Correct Solution:
```
N = int(input())
print('Hello World' if N == 1 else int(input()) + int(input()))
```
| 8,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
N = int(input())
print("Hello World") if N==1 else print(sum([int(input()) for _ in range(2)]))
```
Yes
| 8,256 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
N,*C=map(int,open(0).read().split())
if N == 2:
print(sum(C))
else:
print("Hello World")
```
Yes
| 8,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
if int(input()) == 1:
print("Hello World")
else:
print(int(input()) + int(input()))
```
Yes
| 8,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
x=int(input())
if (x==1):
print("Hello World")
else:
a=int(input())
b=int(input())
print(a+b)
```
Yes
| 8,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
N = int(input())
lst = [int(input()) for i in range(2)]
if (N == 1):
print('Hello World')
else:
print(sum(lst))
```
No
| 8,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
# | Its_Rashid | #
n = int(input())
if n == 1 :
print('Hello World')
else :
print(sum([int(input()) for i in range()]))
```
No
| 8,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
import random
def is_prime(q,k=50):
if q == 2: return True
if q < 2 or q&1 == 0: return False
d = (q-1)>>1
while d&1 == 0:
d >>= 1
for i in range(k):
a = random.randint(1,q-1)
t = d
y = pow(a,t,q)
while t != q-1 and y != 1 and y != q-1:
y = pow(y,2,q)
t <<= 1
if y != q-1 and t&1 == 0:
return False
return True
n,m=(int(i) for i in input().split(' '))
if m%n>100000:
for i in range(1,n):
if m%i==0 and is_prime(i) and is_prime(m//i) and (i==1 or m//i<n):
print('1')
exit()
for i in range(n,m+1):
if i==m:
print(1)
break
if m%i==0:
print(m//i)
break
if m-i-n>0 and m%(m-i-n)==0 and m//(m-i-n)>n:
print(m-i-n)
break
```
No
| 8,262 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
n = input()
if n == '1' :
print('Helli World')
else :
a,b = map(int,input().split())
print(a+b)
```
No
| 8,263 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
N = int(input())
A = list()
B = list()
m = float('inf')
for i in range(N):
a, b = list(map(int, input().split()))
A.append(a)
B.append(b)
if A == B: print(0)
else:
seriesSum = sum(A)
for i in range(N):
if A[i] > B[i]:
m = min(m, B[i])
print(seriesSum - m)
```
| 8,264 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
n = int(input())
INF = 10**9
csum = 0
min_b = INF
for _ in range(n):
a,b = map(int, input().split())
csum += a
if a > b:
min_b = min(min_b,b)
if min_b == INF:
print(0)
else:
ans = csum-min_b
print(ans)
# a_down_sum = 0
# top_diff = 0
# for _ in range(n):
# a,b = map(int, input().split())
# if a < b:
# a_down_sum += a
# else:
# top_diff = max(top_diff,a-b)
```
| 8,265 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
N = int(input())
AB = [[int(i) for i in input().split()] for _ in range(N)]
s = 0
m = float('inf')
for i, (A, B) in enumerate(AB) :
s += A
if A > B :
m = min(m, B)
if m == float('inf') :
print(0)
else :
print(s - m)
```
| 8,266 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
N=int(input())
d=0
s=10**9+1
for i in range(N):
a,b=map(int, input().split())
d+=a
if a>b:
s=min(s,b)
if s==10**9+1:
print(0)
exit()
print(d-s)
```
| 8,267 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
#!/usr/bin/env python3
import sys
INF = float("inf")
def solve(N: int, A: "List[int]", B: "List[int]"):
ma = INF
mb = INF
for a, b in zip(A, B):
if a > b:
if mb > b:
ma = a
mb = b
if ma == INF:
print(0)
else:
print(sum(A)-mb)
return
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
A = [int()] * (N) # type: "List[int]"
B = [int()] * (N) # type: "List[int]"
for i in range(N):
A[i] = int(next(tokens))
B[i] = int(next(tokens))
solve(N, A, B)
if __name__ == '__main__':
main()
```
| 8,268 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
# ARC094E
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")
n = int(input())
a = [None] * n
b = [None] * n
ans = 0
same = 0
m = 0
diff = []
for i in range(n):
a[i],b[i] = map(int, input().split())
if a[i]<b[i]:
ans += (b[i] - a[i])
elif a[i]==b[i]:
same += a[i]
else:
diff.append(b[i])
m += b[i]
diff.sort()
c = 0
plus = 0
if ans==0:
print(0)
else:
plus = sum(diff[1:])
print(sum(a) - (m - plus))
```
| 8,269 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
N = int(input())
diff_a_b = []
for i in range(N):
a, b = map(int, input().split())
diff_a_b.append([a - b, a, b])
diff_a_b.sort()
if diff_a_b[0][0] == 0: # 最初から数列が等しい
print(0)
quit()
ans = 0
d = 0
temp = 10**10
for diff, a, b in diff_a_b:
if diff <= 0: # a <= b
ans += b
d -= diff
else:
ans += b
temp = min(temp, b)
print(ans - temp)
```
| 8,270 |
Provide a correct Python 3 solution for this coding contest problem.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
"Correct Solution:
```
n = int(input())
ab=[list(map(int,input().split())) for i in range(n)]
sm=0
bmin=10**10
for itm in ab:
a,b=itm
sm+=a
if a>b:
bmin=min(bmin,b)
if bmin==10**10:
print(0)
else:
print(sm-bmin)
```
| 8,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
def E():
N = int(input())
sum_ = 0
c = -1
for _ in range(N):
a, b = list(map(int, input().split(' ')))
if a > b:
if c == -1:
c = b
elif b < c:
c = b
sum_ += a
if c == -1:
print(0)
else:
print(sum_ - c)
if __name__ == '__main__':
E()
```
Yes
| 8,272 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
N = int(input())
c = -1
s = 0
for i in range(N):
ai,bi = map(int,input().split())
s += ai
if ai > bi:
if c == -1:
c = bi
if bi < c:
c = bi
if c == -1:
print(0)
else:
print(s-c)
```
Yes
| 8,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
n = int(input())
sum = 0
bmin = []
for _ in range(n):
a, b = map(int,input().split())
sum += a
if a > b:
bmin.append(b)
if len(bmin)==0:
print(0)
else:
print(sum - min(bmin))
```
Yes
| 8,274 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
n = int(input())
s = 0
m = 10**9+7
same = True
for _ in range(n):
a,b = map(int, input().split())
if a > b:
m = min(m, b)
s += a
same &= a==b
if same:
m = s
print(s-m)
```
Yes
| 8,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Contents : AtCoder Regular Contest094 e問題 未完成
Author : Kitaura Hiromi
LastUpdate : 20180420
Since : 20180414
"""
# 最終的に, 1組以外はすべて0になる
import numpy as np
N = int(input())
queA = []
queB = []
for i in range(N):
A, B = map(int, input().split(" "))
queA.append(A)
queB.append(B)
if np.allclose(queA, queB):
print(0)
else:
tmp = []
for i, B in enumerate(queB):
if queA[i] >= B:
tmp.append(B)
print(sum(queA) - min(tmp))
```
No
| 8,276 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
N = int(input())
ans = 0
for i in range(N):
a, b = map(int, input().split())
if a < b:
ans += b
print(ans)
```
No
| 8,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = []
b = []
res = 0
judge = True
for i in range(n):
aa,bb = map(int,input().split())
a.append(aa)
b.append(bb)
if aa != bb:
judge = False
if judge:
print(0)
else:
for i in range(n):
if a[i] <= b[i]:
res += b[i]
print(res)
```
No
| 8,278 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given sequences A and B consisting of non-negative integers. The lengths of both A and B are N, and the sums of the elements in A and B are equal. The i-th element in A is A_i, and the i-th element in B is B_i.
Tozan and Gezan repeats the following sequence of operations:
* If A and B are equal sequences, terminate the process.
* Otherwise, first Tozan chooses a positive element in A and decrease it by 1.
* Then, Gezan chooses a positive element in B and decrease it by 1.
* Then, give one candy to Takahashi, their pet.
Tozan wants the number of candies given to Takahashi until the process is terminated to be as large as possible, while Gezan wants it to be as small as possible. Find the number of candies given to Takahashi when both of them perform the operations optimally.
Constraints
* 1 \leq N \leq 2 × 10^5
* 0 \leq A_i,B_i \leq 10^9(1\leq i\leq N)
* The sums of the elements in A and B are equal.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the number of candies given to Takahashi when both Tozan and Gezan perform the operations optimally.
Examples
Input
2
1 2
3 2
Output
2
Input
3
8 3
0 1
4 8
Output
9
Input
1
1 1
Output
0
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
def main():
n=int(input())
ab=[list(map(int, input().split())) for _ in range(n)]
ab.sort(key=lambda x:x[0]-x[1])
#print(ab)
re=0
ans=0
for a,b in ab:
if a<b:
ans+=b
re+=b-a
if a>=b:
if re>=a:
ans+=b
re+=b-a
else:
ans+=b-a+re
#print(a,b,re,ans)
print(ans)
main()
```
No
| 8,279 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
"Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
ope=[[] for i in range(n)]
Q=int(input())
for i in range(Q):
l,r=map(int,input().split())
ope[r-1].append(l-1)
res=b.count(0)
Data=[(-1)**((b[i]==1)+1) for i in range(n)]
for i in range(1,n):
Data[i]+=Data[i-1]
Data=[0]+Data
for i in range(n):
ope[i].sort(reverse=True)
# N: 処理する区間の長さ
N=n+1
N0 = 2**(N-1).bit_length()
data = [None]*(2*N0)
INF = (-2**31, -2**31)
# 区間[l, r+1)の値をvに書き換える
# vは(t, value)という値にする (新しい値ほどtは大きくなる)
def update(l, r, v):
L = l + N0; R = r + N0
while L < R:
if R & 1:
R -= 1
if data[R-1]:
data[R-1] = max(v,data[R-1])
else:
data[R-1]=v
if L & 1:
if data[L-1]:
data[L-1] = max(v,data[L-1])
else:
data[L-1]=v
L += 1
L >>= 1; R >>= 1
# a_iの現在の値を取得
def _query(k):
k += N0-1
s = INF
while k >= 0:
if data[k]:
s = max(s, data[k])
k = (k - 1) // 2
return s
# これを呼び出す
def query(k):
return _query(k)[1]
for i in range(n+1):
update(i,i+1,(-Data[i],-Data[i]))
if ope[0]:
update(1,2,(0,0))
for i in range(1,n):
val=query(i)
update(i+1,i+2,(val+Data[i]-Data[i+1],val+Data[i]-Data[i+1]))
for l in ope[i]:
val=query(l)
update(l+1,i+2,(val,val))
print(n-(res+query(n)+Data[n]))
```
| 8,280 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
B = list(map(int,readline().split()))
Q = int(readline())
m = map(int,read().split())
LR = sorted(zip(m,m))
class MaxSegTree():
def __init__(self,N):
self.Nelem = N
self.size = 1<<(N.bit_length()) # 葉の要素数
self.data = [0] * (2*self.size)
def build(self,raw_data):
# raw_data は 0-indexed
for i,x in enumerate(raw_data):
self.data[self.size+i] = x
for i in range(self.size-1,0,-1):
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
def update(self,i,x):
i += self.size
self.data[i] = x
i >>= 1
while i:
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
i >>= 1
def get_data(self,i):
return self.data[i+self.size]
def get_max(self,L,R):
# [L,R] に対する値を返す
L += self.size
R += self.size + 1
# [L,R) に変更
x = 0
while L < R:
if L&1:
y = self.data[L]
if x < y: x = y
L += 1
if R&1:
R -= 1
y = self.data[R]
if x < y: x = y
L >>= 1; R >>= 1
return x
"""
・ある場所まで確定して、そこから先を全て1で埋めた場合
・ある場所まで確定して、そこから先を全て0で埋めた場合
の加算スコア
"""
add0 = [0] * (N+1)
add1 = [0] * (N+1)
x = sum(B)
add1[0] = x
add0[0] = N-x
for i,x in enumerate(B,1):
if x == 0:
add0[i]=add0[i-1]-1; add1[i] = add1[i-1]
else:
add1[i]=add1[i-1]-1; add0[i] = add0[i-1]
"""
ある場所を右端としてとった時点での
・残りをすべて0で埋めたときのスコア
・残りをすべて1で埋めたときのスコア
dp0, dp1 をseg木で管理
"""
dp0 = MaxSegTree(N+1)
dp0.build([add0[0]] + [0] * N)
dp1 = MaxSegTree(N+1)
dp1.build([add1[0]] + [0] * N)
for L,R in LR:
# dp1[R] を計算したい。[L,inf) が1で埋まったときのスコア
x = dp1.get_data(R)
y = dp0.get_max(0,L-1) + add1[L-1] - add0[L-1] # 0埋め部分を1埋めに修正してdp1[R]に遷移
z = dp1.get_max(L,R-1) # そのままdp1[R]に遷移
if y < z: y = z
if x < y:
dp1.update(R,y)
dp0.update(R,y - add1[R] + add0[R])
# 一致を数えていたので、距離に修正
answer = N - dp0.data[1]
print(answer)
```
| 8,281 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
B = list(map(int,readline().split()))
Q = int(readline())
m = map(int,read().split())
LR = sorted(zip(m,m))
class MaxSegTree():
def __init__(self,raw_data):
N = len(raw_data)
self.size = 1<<(N.bit_length()) # 葉の要素数
self.data = [0] * (2*self.size)
self.build(raw_data)
def build(self,raw_data):
for i,x in enumerate(raw_data):
self.data[self.size+i] = x
for i in range(self.size-1,0,-1):
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
def update(self,i,x):
i += self.size
self.data[i] = x
i >>= 1
while i:
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
i >>= 1
def get_data(self,i):
return self.data[i+self.size]
def get_max(self,L,R):
# [L,R] に対する値を返す
L += self.size
R += self.size + 1
# [L,R) に変更
x = 0
while L < R:
if L&1:
y = self.data[L]
if x < y: x = y
L += 1
if R&1:
R -= 1
y = self.data[R]
if x < y: x = y
L >>= 1; R >>= 1
return x
"""
・あるから先を全て0で埋めた場合
・あるから先を全て1で埋めた場合
"""
add1 = [0] * (N+1)
add1[0] = sum(B)
for i,x in enumerate(B,1):
if x:
add1[i] = add1[i-1] - 1
else:
add1[i] = add1[i-1]
add0 = [N - i - x for i,x in enumerate(add1)]
add0,add1
# ある場所を最後の右端まで使ったとする。残りを全て 0 / 1 で埋めたときのスコア
dp0 = MaxSegTree([0] * (N+1))
dp1 = MaxSegTree([0] * (N+1))
dp0.update(0,add0[0])
dp1.update(0,add1[0])
for L,R in LR:
a = dp1.get_data(R)
b = dp0.get_max(0,L-1) + add1[L-1] - add0[L-1]
c = dp1.get_max(L,R-1)
if b < c:
b = c
if a < b:
dp1.update(R, b)
dp0.update(R, b - add1[R] + add0[R])
answer = N - dp0.get_max(0,N)
print(answer)
```
| 8,282 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
"Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
b=list(map(int,input().split()))
ope=[[] for i in range(n)]
Q=int(input())
for i in range(Q):
l,r=map(int,input().split())
ope[r-1].append(l-1)
res=b.count(0)
Data=[(-1)**((b[i]==1)+1) for i in range(n)]
for i in range(1,n):
Data[i]+=Data[i-1]
Data=[0]+Data
for i in range(n):
ope[i].sort(reverse=True)
# N: 処理する区間の長さ
N=n+1
N0 = 2**(N-1).bit_length()
data = [None]*(2*N0)
INF = (-2**31, -2**31)
# 区間[l, r+1)の値をvに書き換える
# vは(t, value)という値にする (新しい値ほどtは大きくなる)
def update(l, r, v):
L = l + N0; R = r + N0
while L < R:
if R & 1:
R -= 1
if data[R-1]:
data[R-1] = max(v,data[R-1])
else:
data[R-1]=v
if L & 1:
if data[L-1]:
data[L-1] = max(v,data[L-1])
else:
data[L-1]=v
L += 1
L >>= 1; R >>= 1
# a_iの現在の値を取得
def _query(k):
k += N0-1
s = INF
while k >= 0:
if data[k]:
s = max(s, data[k])
k = (k - 1) // 2
return s
# これを呼び出す
def query(k):
return _query(k)[1]
for i in range(n+1):
update(i,i+1,(-Data[i],-Data[i]))
if ope[0]:
update(1,2,(0,0))
for i in range(1,n):
val=query(i)
update(i+1,i+2,(val+Data[i]-Data[i+1],val+Data[i]-Data[i+1]))
for l in ope[i]:
val=query(l)
update(l+1,i+2,(val,val))
print(n-(res+query(n)+Data[n]))
```
| 8,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
class SegtreeMin():
def __init__(self, n):
self.inf = 10 ** 16
tree_width = 2
while tree_width < n:
tree_width *= 2
self.tree_width = tree_width
self.tree = [self.inf] * (tree_width * 2 - 1)
def update(self, i, a):
seg_i = self.tree_width - 1 + i
self.tree[seg_i] = a
while seg_i != 0:
seg_i = (seg_i - 1) // 2
self.tree[seg_i] = min(self.tree[seg_i * 2 + 1], self.tree[seg_i * 2 + 2])
def element(self, i):
return self.tree[self.tree_width - 1 + i]
# [l,r)の最小値
def min(self, l, r, seg_i=0, segL=0, segR=-1):
if segR == -1: segR = self.tree_width
if r <= segL or segR <= l: return self.inf
if l <= segL and segR <= r: return self.tree[seg_i]
segM = (segL + segR) // 2
ret0 = self.min(l, r, seg_i * 2 + 1, segL, segM)
ret1 = self.min(l, r, seg_i * 2 + 2, segM, segR)
return min(ret0, ret1)
def main():
n = int(input())
bb = LI()
q = int(input())
lr = defaultdict(list)
for _ in range(q):
l, r = map(int1, input().split())
lr[l].append(r)
# print(lr)
dp = SegtreeMin(n + 1)
dp.update(0, 0)
for i in range(n):
for r in lr[i]:
dp.update(r + 1, dp.min(i, r + 2))
dp.update(i + 1, min(dp.element(i + 1), dp.element(i) + bb[i] * 2 - 1))
print(bb.count(0) + dp.element(n))
main()
```
No
| 8,284 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
"""
https://atcoder.jp/contests/arc085/tasks/arc085_d
dp[i][x] = indexがxになるまでは1が連続する際の最小スコア
何もない場合
i <= x では1を選ばなくてはいけない bが1の時0 bが0の時1
i > x では0を選ばなくてはいけない bが0の時0 1の時1
i == lがある場合
dp[i][r] = min(dp[i-1][t] + l番目で1を選んだ際のスコア)
t<r
つまり、区間加算と全体最小値がわかればいい
→遅延評価セグメント木…
"""
#遅延評価セグメント木(tree&lazy)を作る。初期値lisを渡す
#defaultで初期化する
def make_LST(lis,default = float("inf")):
n = 1
while n < len(lis):
n *= 2
tree = [default] * (2*n-1)
lazy = [0] * (2*n-1)
for i in range(len(lis)):
tree[i+n-1] = lis[i]
for i in range(n-2,-1,-1):
tree[i] = min(tree[i*2+1] , tree[i*2+2])
return tree,lazy
def eval_LST(k,l,r,tree,lazy):
if lazy[k] != 0:
tree[k] += lazy[k]
if r-l > 1:
lazy[2*k+1] += lazy[k]
lazy[2*k+2] += lazy[k]
lazy[k] = 0
#add x to [a,b)
def add_LST(a,b,x,tree,lazy,k=0,l=0,r=-1):
if r < 0:
n = (len(tree)+1)//2
r = n
eval_LST(k,l,r,tree,lazy)
if b <= l or r <= a:
return
if a <= l and r <= b:
lazy[k] += x
eval_LST(k,l,r,tree,lazy)
else:
add_LST(a,b,x,tree,lazy,2*k+1,l,(l+r)//2)
add_LST(a,b,x,tree,lazy,2*k+2,(l+r)//2,r)
tree[k] = min(tree[2*k+1] , tree[2*k+2])
#range_minimum_query(def)
def rmq_LST(a,b,tree,lazy,k=0,l=0,r=-1):
if r < 0:
n = (len(tree)+1)//2
r = n
if b <= l or r <= a:
return float("inf")
eval_LST(k,l,r,tree,lazy)
if a <= l and r <= b:
return tree[k]
vl = rmq_LST(a,b,tree,lazy,2*k+1,l,(l+r)//2)
vr = rmq_LST(a,b,tree,lazy,2*k+2,(l+r)//2,r)
return min(vl,vr)
def getfrom_index_LST(index,tree,lazy):
return rmq_LST(index,index+1,tree,lazy)
def upd_point_LST(index,newnum,tree,lazy):
oldnum = getfrom_index_LST(index,tree,lazy)
difference = newnum - oldnum
add_LST(index,index+1,difference,tree,lazy)
from sys import stdin
N = int(stdin.readline())
b = list(map(int,stdin.readline().split()))
Q = int(stdin.readline())
rlis = [ [] for i in range(N+1) ]
for i in range(Q):
l,r = map(int,stdin.readline().split())
rlis[l].append(r)
tmp = [10**9]*(N+1)
tmp[0] = 0
tree,lazy = make_LST(tmp,10**9)
for i in range(1,N+1):
#print (tree,lazy)
#i <= x では1を選ばなくてはいけない bが1の時0 bが0の時1
#i > x では0を選ばなくてはいけない bが0の時0 1の時1
for r in rlis[i]:
newnum = rmq_LST(0,r,tree,lazy)
oldnum = getfrom_index_LST(r,tree,lazy)
if newnum < oldnum:
upd_point_LST(r,newnum,tree,lazy)
if b[i-1] == 0:
add_LST(i,N+1,1,tree,lazy)
else:
add_LST(0,i,1,tree,lazy)
#print (tree,lazy)
print (rmq_LST(0,N+1,tree,lazy))
"""
#verify
n,q = map(int,stdin.readline().split())
tree,lazy = make_LST([0]*n,0)
for i in range(q):
SS = stdin.readline()
if SS[0] == "0":
tmp,s,t,x = map(int,SS.split())
add_LST(s-1,t,x,tree,lazy)
else:
tmp,s,t = map(int,SS.split())
print ( rsq_LST(s-1,t,tree,lazy) )
"""
```
No
| 8,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
class SegtreeMin():
def __init__(self, n):
self.inf = 10 ** 16
tree_width = 2
while tree_width < n:
tree_width *= 2
self.tree_width = tree_width
self.tree = [self.inf] * (tree_width * 2 - 1)
def update(self, i, a):
seg_i = self.tree_width - 1 + i
self.tree[seg_i] = a
while seg_i != 0:
seg_i = (seg_i - 1) // 2
self.tree[seg_i] = min(self.tree[seg_i * 2 + 1], self.tree[seg_i * 2 + 2])
def element(self, i):
return self.tree[self.tree_width - 1 + i]
# [l,r)の最小値
def min(self, l, r, seg_i=0, segL=0, segR=-1):
if segR == -1: segR = self.tree_width
if r <= segL or segR <= l: return self.inf
if l <= segL and segR <= r: return self.tree[seg_i]
segM = (segL + segR) // 2
ret0 = self.min(l, r, seg_i * 2 + 1, segL, segM)
ret1 = self.min(l, r, seg_i * 2 + 2, segM, segR)
return min(ret0, ret1)
def main():
n = int(input())
bb = LI()
q = int(input())
lr = defaultdict(list)
for _ in range(q):
l, r = MI()
lr[l-1].append(r-1)
# print(lr)
dp = SegtreeMin(n + 1)
dp.update(0, 0)
for i in range(n):
for r in lr[i]:
dp.update(r + 1, dp.min(i, r + 2))
case0 = dp.element(i) + bb[i] * 2 - 1
if case0 < dp.element(i + 1): dp.update(i + 1, case0)
print(bb.count(0) + dp.element(n))
main()
```
No
| 8,286 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
INF = 10**6
class Rmin():
def __init__(self, size):
#the number of nodes is 2n-1
self.n = 1
while self.n < size:
self.n *= 2
self.node = [INF] * (2*self.n-1)
def Access(self, x):
return self.node[x+self.n-1]
def Update(self, x, val):
x += self.n-1
self.node[x] = val
while x > 0:
x = (x-1)//2
self.node[x] = min(self.node[2*x+1], self.node[2*x+2])
return
#[l, r)
def Get(self, l, r):
L, R = l+self.n, r+self.n
s = INF
while L<R:
if R & 1:
R -= 1
s = min(s, self.node[R-1])
if L & 1:
s = min(s, self.node[L-1])
L += 1
L >>= 1
R >>= 1
return s
n = int(input())
b = list(map(int, input().split()))
cum = [0]
for x in b:
if x == 1:
cum.append(cum[-1] + 1)
else:
cum.append(cum[-1])
Q = int(input())
q = sorted([list(map(int, input().split())) for _ in range(Q)], key=lambda x: x[1])
dp = Rmin(n+1)
dp.Update(0, cum[-1])
ans = cum[-1]
for l, r in q:
mi = dp.Get(0, l)
num = cum[r] - cum[l-1]
res = r-l+1 - num*2 + mi
dp.Update(r, res)
ans = min(ans, res)
print(ans)
```
No
| 8,287 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.
On a number line, there are N copies of Takahashi, numbered 1 through N. The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.
Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.
Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja. The walking velocity of a copy does not change when it transforms. From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.
Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time? Find the count modulo 10^9+7.
Constraints
* 1 ≤ N ≤ 200000
* 1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)
* X_i and V_i are integers.
* All X_i are distinct.
* All V_i are distinct.
Input
The input is given from Standard Input in the following format:
N
X_1 V_1
:
X_N V_N
Output
Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.
Examples
Input
3
2 5
6 1
3 7
Output
6
Input
4
3 7
2 9
8 16
10 8
Output
9
"Correct Solution:
```
class BIT():
def __init__(self,n,mod):
self.BIT=[0]*(n+1)
self.num=n
self.mod=mod
def query(self,idx):
res_sum = 0
mod=self.mod
while idx > 0:
res_sum += self.BIT[idx]
res_sum%=mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod=self.mod
while idx <= self.num:
self.BIT[idx] += x
self.BIT[idx]%=mod
idx += idx&(-idx)
return
import sys,bisect
input=sys.stdin.readline
mod=10**9+7
inv=pow(2,mod-2,mod)
N=int(input())
taka=[tuple(map(int,input().split())) for i in range(N)]
taka.sort()
V=[0]+[taka[i][1] for i in range(N)]+[10**15]
cummin=[V[i] for i in range(N+2)]
for i in range(N,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
cummax=[V[i] for i in range(N+2)]
for i in range(1,N+2):
cummax[i]=max(cummax[i],cummax[i-1])
const=[]
for i in range(1,N+1):
R=bisect.bisect_right(cummin,V[i])-1
L=bisect.bisect_left(cummax,V[i])
const.append((L,R))
Lconst=[(i,10**15) for i in range(N+1)]
for L,R in const:
pL,pR=Lconst[L]
Lconst[L]=(L,min(R,pR))
_const=[]
for i in range(1,N+1):
L,R=Lconst[i]
if R!=10**15:
_const.append((L,R))
const=[]
for L,R in _const:
while const and const[-1][1]>=R:
const.pop()
const.append((L,R))
M=len(const)
const=[(-10**15,0)]+const
Rconst=[const[i][1] for i in range(M+1)]
B=BIT(M,mod)
dp=[1]*(M+1)
for i in range(1,M+1):
L,R=const[i]
id=bisect.bisect_left(Rconst,L)
res=(B.query(i-1)-B.query(id-1))%mod
l,r=const[id]
l=max(l,Rconst[id-1]+1)
res+=dp[id-1]*(pow(2,r-L+1,mod)-1)*pow(2,L-l,mod)%mod
res%=mod
dp[i]=res
if i!=M:
nR=Rconst[i+1]
add=dp[i]*(pow(2,nR-R,mod)-1)%mod
add%=mod
B.update(i,add)
cnt=0
data=[0]*(N+2)
for i in range(1,M+1):
L,R=const[i]
data[L]+=1
data[R+1]+=-1
for i in range(1,N+2):
data[i]+=data[i-1]
for i in range(1,N+1):
if data[i]==0:
cnt+=1
ans=pow(2,cnt,mod)*dp[M]
ans%=mod
print(ans)
```
| 8,288 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.
On a number line, there are N copies of Takahashi, numbered 1 through N. The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.
Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.
Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja. The walking velocity of a copy does not change when it transforms. From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.
Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time? Find the count modulo 10^9+7.
Constraints
* 1 ≤ N ≤ 200000
* 1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)
* X_i and V_i are integers.
* All X_i are distinct.
* All V_i are distinct.
Input
The input is given from Standard Input in the following format:
N
X_1 V_1
:
X_N V_N
Output
Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.
Examples
Input
3
2 5
6 1
3 7
Output
6
Input
4
3 7
2 9
8 16
10 8
Output
9
Submitted Solution:
```
import sys
input=sys.stdin.readline
mod=10*9+7
N=int(input())
taka=[tuple(map(int,input().split())) for i in range(N)]
taka.sort()
V=[0]+[taka[i][1] for i in range(N)]+[10**15]
cummin=[V[i] for i in range(N+2)]
for i in range(N,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
cummax=[V[i] for i in range(N+2)]
for i in range(1,N+2):
cummax[i]=max(cummax[i],cummax[i-1])
newV=[]
for i in range(1,N+1):
if not (V[i]>cummax[i-1] and cummin[i+1]>V[i]):
newV.append(V[i])
V=newV
N=len(V)
all=pow(2,N,mod)
highid,lowid=-1,-1
Vmax=cummax[-2]
Vmin=cummin[1]
for i in range(0,N):
if V[i]==Vmax:
highid=i+1
elif V[i]==Vmin:
lowid=i+1
high=pow(2,highid-1,mod)
low=pow(2,N-lowid,mod)
if highid>lowid:
highlow=pow(2,N+1-highid-lowid,mod)
else:
highlow=1
ans=all+highlow-high-low
print(ans%mod)
```
No
| 8,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.
On a number line, there are N copies of Takahashi, numbered 1 through N. The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.
Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.
Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja. The walking velocity of a copy does not change when it transforms. From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.
Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time? Find the count modulo 10^9+7.
Constraints
* 1 ≤ N ≤ 200000
* 1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)
* X_i and V_i are integers.
* All X_i are distinct.
* All V_i are distinct.
Input
The input is given from Standard Input in the following format:
N
X_1 V_1
:
X_N V_N
Output
Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.
Examples
Input
3
2 5
6 1
3 7
Output
6
Input
4
3 7
2 9
8 16
10 8
Output
9
Submitted Solution:
```
class BIT():
def __init__(self,n,mod):
self.BIT=[0]*(n+1)
self.num=n
self.mod=mod
def query(self,idx):
res_sum = 0
mod=self.mod
while idx > 0:
res_sum += self.BIT[idx]
res_sum%=mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod=self.mod
while idx <= self.num:
self.BIT[idx] += x
self.BIT[idx]%=mod
idx += idx&(-idx)
return
import sys,bisect
input=sys.stdin.readline
mod=10*9+7
N=int(input())
taka=[tuple(map(int,input().split())) for i in range(N)]
taka.sort()
V=[0]+[taka[i][1] for i in range(N)]+[10**15]
cummin=[V[i] for i in range(N+2)]
for i in range(N,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
cummax=[V[i] for i in range(N+2)]
for i in range(1,N+2):
cummax[i]=max(cummax[i],cummax[i-1])
const=[]
for i in range(1,N+1):
R=bisect.bisect_right(cummin,V[i])-1
L=bisect.bisect_left(cummax,V[i])
const.append((L,R))
Lconst=[(i,10**15) for i in range(N+1)]
for L,R in const:
pL,pR=Lconst[L]
Lconst[L]=(L,min(R,pR))
_const=[]
for i in range(1,N+1):
L,R=Lconst[i]
if R!=10**15:
_const.append((L,R))
const=[]
for L,R in _const:
while const and const[-1][1]>=R:
const.pop()
const.append((L,R))
M=len(const)
const=[(-10**15,-10**15)]+const
Rconst=[const[i][1] for i in range(M+1)]
B=BIT(M,mod)
dp=[1]*(M+1)
for i in range(1,M+1):
L,R=const[i]
id=bisect.bisect_left(Rconst,L)
res=(B.query(i-1)-B.query(id-1))%mod
R=Rconst[id]
res+=dp[id-1]*(pow(2,R-L+1,mod)-1)%mod
res%=mod
dp[i]=res
if i!=M:
nR=Rconst[i+1]
add=dp[i]*(pow(2,nR-R,mod)-1)*pow(2,nR-L+1,mod)%mod
add%=mod
B.update(i,add)
cnt=0
data=[0]*(N+2)
for i in range(1,M+1):
L,R=const[i]
data[L]+=1
data[R+1]+=-1
for i in range(1,N+2):
data[i]+=data[i-1]
for i in range(1,N+1):
if data[i]==0:
cnt+=1
ans=pow(2,cnt,mod)*dp[M]
ans%=mod
print(ans)
```
No
| 8,290 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.
On a number line, there are N copies of Takahashi, numbered 1 through N. The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.
Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.
Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja. The walking velocity of a copy does not change when it transforms. From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.
Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time? Find the count modulo 10^9+7.
Constraints
* 1 ≤ N ≤ 200000
* 1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)
* X_i and V_i are integers.
* All X_i are distinct.
* All V_i are distinct.
Input
The input is given from Standard Input in the following format:
N
X_1 V_1
:
X_N V_N
Output
Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.
Examples
Input
3
2 5
6 1
3 7
Output
6
Input
4
3 7
2 9
8 16
10 8
Output
9
Submitted Solution:
```
class BIT():
def __init__(self,n,mod):
self.BIT=[0]*(n+1)
self.num=n
self.mod=mod
def query(self,idx):
res_sum = 0
mod=self.mod
while idx > 0:
res_sum += self.BIT[idx]
res_sum%=mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod=self.mod
while idx <= self.num:
self.BIT[idx] += x
self.BIT[idx]%=mod
idx += idx&(-idx)
return
import sys,bisect
input=sys.stdin.readline
mod=10*9+7
inv=pow(2,mod-2,mod)
N=int(input())
taka=[tuple(map(int,input().split())) for i in range(N)]
taka.sort()
V=[0]+[taka[i][1] for i in range(N)]+[10**15]
cummin=[V[i] for i in range(N+2)]
for i in range(N,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
cummax=[V[i] for i in range(N+2)]
for i in range(1,N+2):
cummax[i]=max(cummax[i],cummax[i-1])
const=[]
for i in range(1,N+1):
R=bisect.bisect_right(cummin,V[i])-1
L=bisect.bisect_left(cummax,V[i])
const.append((L,R))
Lconst=[(i,10**15) for i in range(N+1)]
for L,R in const:
pL,pR=Lconst[L]
Lconst[L]=(L,min(R,pR))
_const=[]
for i in range(1,N+1):
L,R=Lconst[i]
if R!=10**15:
_const.append((L,R))
const=[]
for L,R in _const:
while const and const[-1][1]>=R:
const.pop()
const.append((L,R))
M=len(const)
const=[(-10**15,0)]+const
Rconst=[const[i][1] for i in range(M+1)]
B=BIT(M,mod)
dp=[1]*(M+1)
for i in range(1,M+1):
L,R=const[i]
id=bisect.bisect_left(Rconst,L)
res=(B.query(i-1)-B.query(id-1))%mod
r=Rconst[id]
res+=dp[id-1]*(pow(2,r-L+1,mod)-1)*pow(2,L-Rconst[id-1]-1,mod)%mod
res%=mod
dp[i]=res
if i!=M:
nR=Rconst[i+1]
add=dp[i]*(pow(2,nR-R,mod)-1)%mod
add%=mod
B.update(i,add)
cnt=0
data=[0]*(N+2)
for i in range(1,M+1):
L,R=const[i]
data[L]+=1
data[R+1]+=-1
for i in range(1,N+2):
data[i]+=data[i-1]
for i in range(1,N+1):
if data[i]==0:
cnt+=1
ans=pow(2,cnt,mod)*dp[M]
ans%=mod
print(ans)
```
No
| 8,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is an expert of Clone Jutsu, a secret art that creates copies of his body.
On a number line, there are N copies of Takahashi, numbered 1 through N. The i-th copy is located at position X_i and starts walking with velocity V_i in the positive direction at time 0.
Kenus is a master of Transformation Jutsu, and not only can he change into a person other than himself, but he can also transform another person into someone else.
Kenus can select some of the copies of Takahashi at time 0, and transform them into copies of Aoki, another Ninja. The walking velocity of a copy does not change when it transforms. From then on, whenever a copy of Takahashi and a copy of Aoki are at the same coordinate, that copy of Takahashi transforms into a copy of Aoki.
Among the 2^N ways to transform some copies of Takahashi into copies of Aoki at time 0, in how many ways will all the copies of Takahashi become copies of Aoki after a sufficiently long time? Find the count modulo 10^9+7.
Constraints
* 1 ≤ N ≤ 200000
* 1 ≤ X_i,V_i ≤ 10^9(1 ≤ i ≤ N)
* X_i and V_i are integers.
* All X_i are distinct.
* All V_i are distinct.
Input
The input is given from Standard Input in the following format:
N
X_1 V_1
:
X_N V_N
Output
Print the number of the ways that cause all the copies of Takahashi to turn into copies of Aoki after a sufficiently long time, modulo 10^9+7.
Examples
Input
3
2 5
6 1
3 7
Output
6
Input
4
3 7
2 9
8 16
10 8
Output
9
Submitted Solution:
```
class BIT():
def __init__(self,n,mod):
self.BIT=[0]*(n+1)
self.num=n
self.mod=mod
def query(self,idx):
res_sum = 0
mod=self.mod
while idx > 0:
res_sum += self.BIT[idx]
res_sum%=mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod=self.mod
while idx <= self.num:
self.BIT[idx] += x
self.BIT[idx]%=mod
idx += idx&(-idx)
return
import sys,bisect
input=sys.stdin.readline
mod=10*9+7
N=int(input())
taka=[tuple(map(int,input().split())) for i in range(N)]
taka.sort()
V=[0]+[taka[i][1] for i in range(N)]+[10**15]
cummin=[V[i] for i in range(N+2)]
for i in range(N,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
cummax=[V[i] for i in range(N+2)]
for i in range(1,N+2):
cummax[i]=max(cummax[i],cummax[i-1])
const=[]
for i in range(1,N+1):
R=bisect.bisect_right(cummin,V[i])-1
L=bisect.bisect_left(cummax,V[i])
const.append((L,R))
Lconst=[(i,10**15) for i in range(N+1)]
for L,R in const:
pL,pR=Lconst[L]
Lconst[L]=(L,min(R,pR))
Rconst=[(-1,i) for i in range(N+1)]
for i in range(1,N+1):
L,R=Lconst[i]
if R!=10**15:
pL,pR=Rconst[R]
Rconst[R]=(max(pL,L),R)
_const=[]
for i in range(1,N+1):
L,R=Rconst[i]
if L!=-1:
_const.append((L,R))
_const.sort()
const=[]
for L,R in _const:
while const and const[-1][1]>=R:
cosnt.pop()
const.append((L,R))
M=len(const)
const=[(-10**15,-10**15)]+const
Rconst=[const[i][1] for i in range(M+1)]
B=BIT(M,mod)
dp=[1]*(M+1)
for i in range(1,M+1):
L,R=const[i]
id=bisect.bisect_left(Rconst,L)
res=(B.query(i-1)-B.query(id-1))%mod
R=Rconst[id]
res+=dp[id-1]*(pow(2,R-L+1,mod)-1)%mod
res%=mod
dp[i]=res
if i!=M:
nR=Rconst[i+1]
add=dp[i]*(pow(2,nR-R,mod)-1)%mod
add%=mod
B.update(i,add)
cnt=0
data=[0]*(N+2)
for i in range(1,M+1):
L,R=const[i]
data[L]+=1
data[R+1]+=-1
for i in range(1,N+2):
data[i]+=data[i-1]
for i in range(1,N+1):
if data[i]==0:
cnt+=1
ans=pow(2,cnt,mod)*dp[M]
ans%=mod
print(ans)
```
No
| 8,292 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
s = input()
print('First' if (len(s)%2) ^ (s[0] == s[-1]) else 'Second')
```
| 8,293 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
S = input()
print('First' if (S[0] == S[-1]) ^ (len(S) % 2) else 'Second')
```
| 8,294 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
s=input()
n=len(s)
if s[0]==s[n-1] and n%2==0 or s[0]!=s[n-1] and n%2==1:
print("First")
else:
print("Second")
```
| 8,295 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
S = input()
if (len(S)%2)^(S[0]==S[-1]):
print("First")
else:
print("Second")
```
| 8,296 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
s = input()
if (s[0] == s[-1] and len(s) % 2 == 0) or (s[0] != s[-1] and len(s) % 2 == 1):
print("First")
else:
print("Second")
```
| 8,297 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
s = input()
ans=0
if s[0] == s[-1]:
ans +=1
ans = (len(s)-ans) %2
if ans == 0:
print("Second")
else:
print("First")
```
| 8,298 |
Provide a correct Python 3 solution for this coding contest problem.
There is a string s of length 3 or greater. No two neighboring characters in s are equal.
Takahashi and Aoki will play a game against each other. The two players alternately performs the following operation, Takahashi going first:
* Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.
The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.
Constraints
* 3 ≤ |s| ≤ 10^5
* s consists of lowercase English letters.
* No two neighboring characters in s are equal.
Input
The input is given from Standard Input in the following format:
s
Output
If Takahashi will win, print `First`. If Aoki will win, print `Second`.
Examples
Input
aba
Output
Second
Input
abc
Output
First
Input
abcab
Output
First
"Correct Solution:
```
s=input()
if s[0]==s[-1]:
print("Second" if len(s)%2 else "First")
else:
print("Second" if not len(s)%2 else "First")
```
| 8,299 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.