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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` N=int(input()) if len(input().replace("B",""))>N/2: print("Yes") else: print("No") ```
instruction
0
60,457
11
120,914
Yes
output
1
60,457
11
120,915
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` a = int(input()) b = input() if b.count("R") > a / 2 : print("Yes") else : print("No") ```
instruction
0
60,458
11
120,916
Yes
output
1
60,458
11
120,917
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` N = input() s = input() N1 = int(N)/2 import math math.ceil(N1) N1= int(N1) list_s = list(s) list_s.sort() X = list_s[N1] if X=='R': print('Yes') else: print('No') ```
instruction
0
60,459
11
120,918
No
output
1
60,459
11
120,919
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` n=int(input()) s = input().split() #print(s) r = 0 b = 0 for i in s: if i == "R": r += 1 if i == "B": b += 1 if r <= b: print("No") else: print("Yes") ```
instruction
0
60,460
11
120,920
No
output
1
60,460
11
120,921
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` N = int(input()) S = input() red = 0 blue = 0 for s in S: if s == 'R': red++ else: blue++ if red > blue: print('Yes') else print('No') ```
instruction
0
60,461
11
120,922
No
output
1
60,461
11
120,923
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. Constraints * 1 \leq N \leq 100 * |s| = N * s_i is `R` or `B`. Input Input is given from Standard Input in the following format: N s Output If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`. Examples Input 4 RRBR Output Yes Input 4 BRBR Output No Submitted Solution: ``` 4 RRBR ```
instruction
0
60,462
11
120,924
No
output
1
60,462
11
120,925
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,360
11
122,720
"Correct Solution: ``` X = int(input()) if X < 1200: print('ABC') else: print('ARC') ```
output
1
61,360
11
122,721
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,361
11
122,722
"Correct Solution: ``` N=int(input()) if N<1200: s="ABC" else: s="ARC" print(s) ```
output
1
61,361
11
122,723
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,362
11
122,724
"Correct Solution: ``` N = int(input()) if N < 1200: print('ABC') else: print('ARC') ```
output
1
61,362
11
122,725
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,363
11
122,726
"Correct Solution: ``` rate = int(input()) print("ABC") if rate < 1200 else print("ARC") ```
output
1
61,363
11
122,727
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,364
11
122,728
"Correct Solution: ``` m=int(input()) if m<1200:print("ABC") else:print("ARC") ```
output
1
61,364
11
122,729
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,365
11
122,730
"Correct Solution: ``` r = int(input()) if r >= 1200: print("ARC") else: print("ABC") ```
output
1
61,365
11
122,731
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,366
11
122,732
"Correct Solution: ``` print(('ARC','ABC')[int(input())<1200]) ```
output
1
61,366
11
122,733
Provide a correct Python 3 solution for this coding contest problem. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC
instruction
0
61,367
11
122,734
"Correct Solution: ``` print('ABC' if int(input().strip()) < 1200 else 'ARC') ```
output
1
61,367
11
122,735
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` #coding: utf-8 print(["ARC", "ABC"][int(input()) < 1200]) ```
instruction
0
61,368
11
122,736
Yes
output
1
61,368
11
122,737
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` x = int(input()) if x < 1200: r = 'ABC' else: r = 'ARC' print(r) ```
instruction
0
61,369
11
122,738
Yes
output
1
61,369
11
122,739
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` X = int(input()) print("ABC" if X<1200 else "ARC") ```
instruction
0
61,370
11
122,740
Yes
output
1
61,370
11
122,741
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` n = int(input()) print("ABC") if n < 1200 else print("ARC") ```
instruction
0
61,371
11
122,742
Yes
output
1
61,371
11
122,743
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` x = input() if int(x) >= 1200: print(ARC) else: print(ABC) ```
instruction
0
61,372
11
122,744
No
output
1
61,372
11
122,745
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` print(f"A{'RB'['12'>input()]}C") ```
instruction
0
61,373
11
122,746
No
output
1
61,373
11
122,747
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` print(“ABC” if int(input())<1200 else(“ARC”)) ```
instruction
0
61,374
11
122,748
No
output
1
61,374
11
122,749
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise. Constraints * 1 ≦ x ≦ 3{,}000 * x is an integer. Input The input is given from Standard Input in the following format: x Output Print the answer. Examples Input 1000 Output ABC Input 2000 Output ARC Submitted Solution: ``` n = int(input()) if x < 1200: print('ABC') else: print('ARC') ```
instruction
0
61,375
11
122,750
No
output
1
61,375
11
122,751
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,721
11
123,442
Tags: brute force, greedy Correct Solution: ``` x = int(input()) count = 0 for i in range(x): n1, n2, n3 = map(int, input().split()) if n1+n2+n3 > 1: count +=1 print(count) ```
output
1
61,721
11
123,443
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,722
11
123,444
Tags: brute force, greedy Correct Solution: ``` inp = input(str()) times = int(inp) list_one = list() while times>0: enter = input(str()) exp = enter.split(' ') list_one.append(exp) times-=1 count = 0 for element in list_one: if int(element[0]) + int(element[1]) + int(element[2]) >= 2: count += 1 else: continue print(count) ```
output
1
61,722
11
123,445
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,723
11
123,446
Tags: brute force, greedy Correct Solution: ``` number=int(input()) arr=[] for i in range(number): s=input().split(' ') arr.append(s) count=0 for i in arr: tmp=sum([int(x) for x in i]) if tmp>1: count+=1 print(count) ```
output
1
61,723
11
123,447
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,724
11
123,448
Tags: brute force, greedy Correct Solution: ``` n = int(input()) count = 0 if 1 <= n <= 1000: for i in range (1,n+1): a,b,c = map(int,input().split()) if a+b+c == 2 or a+b+c == 3: count+=1 print(count) ```
output
1
61,724
11
123,449
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,725
11
123,450
Tags: brute force, greedy Correct Solution: ``` n= int(input()) s = [] for _ in range(n): petya,vasya,tonya= map(int,input().split(" ")) s.append([petya,vasya,tonya]) l=0 for i in range(n): if((s[i][0]==1 and s[i][1]== 1) or (s[i][0]==1 and s[i][2]==1) or(s[i][1]==1 and s[i][2]==1)): l+=1 print(l) ```
output
1
61,725
11
123,451
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,726
11
123,452
Tags: brute force, greedy Correct Solution: ``` n=int(input()) count=0 while n: l=input() p=l.split(" ") count1=0 for x in p: if x=='1': count1+=1 if count1>=2: count+=1 n-=1 print(count) ```
output
1
61,726
11
123,453
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,727
11
123,454
Tags: brute force, greedy Correct Solution: ``` def moreThanTwo(arr): counter = 0 for num in arr: if num == 1: counter += 1 return counter >= 2 if __name__ == '__main__': n = int(input()) arr = [] for _ in range(n): str_arr = input().split(' ') a, b, c = int(str_arr[0]), int(str_arr[1]), int(str_arr[2]) arr.append([a, b, c]) ans = 0 for problem in arr: if moreThanTwo(problem): ans += 1 print(ans) ```
output
1
61,727
11
123,455
Provide tags and a correct Python 3 solution for this coding contest problem. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
instruction
0
61,728
11
123,456
Tags: brute force, greedy Correct Solution: ``` g=int(input()) c=0 for i in range(g): p,v,t=[int(x) for x in input().split()] if p+v+t>=2: c+=1 print(c) ```
output
1
61,728
11
123,457
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` a = int(input()) s = 0 p = 0 for i in range(a): a , b, c = map(int, input().split()) if a == 1: p+=1 if b == 1: p+=1 if c == 1: p+=1 if p >= 2: s+=1 p = 0 print(s) ```
instruction
0
61,729
11
123,458
Yes
output
1
61,729
11
123,459
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` n = int(input()) schetchik = 0 for i in range(n): r = map(int, input().split()) if sum(r) >= 2: schetchik += 1 print(schetchik) ```
instruction
0
61,730
11
123,460
Yes
output
1
61,730
11
123,461
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` n = int(input()) no = 0 for i in range(n): a, b, c = map(int, input().split()) if (a == 1 and b == 1) or (a == 1 and c == 1) or (b == 1 and c == 1): no += 1 print(no) ```
instruction
0
61,731
11
123,462
Yes
output
1
61,731
11
123,463
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` # cook your dish here n=int(input()) c=0 for i in range(0,n,1): a=[] a=list(map(int,input().split())) if(a.count(1)>=2): c=c+1 print(c) ```
instruction
0
61,732
11
123,464
Yes
output
1
61,732
11
123,465
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` for i in range(int(input())): s=list(map(int,input().rstrip().split())) print(s.count(1)) ```
instruction
0
61,733
11
123,466
No
output
1
61,733
11
123,467
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` import math num_of_lines = input("Enter The Number Of Problems To Filter It ") i = 0 count = 0 while i < int(num_of_lines): a = input("Enter Value Of a, b, and c Respectivley ") i = i + 1 if a.count('1') >= 2: count += 1 print(count) ```
instruction
0
61,734
11
123,468
No
output
1
61,734
11
123,469
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` t = int(input()) cnt = 0 temp = [] length = 0 while t: n = list(map(int, input().split())) length = len(n) for x in n: cnt += x #print(cnt) if cnt == 2 or cnt == 3: temp.append(1) cnt=0 t -= 1 print(sum(temp)//2) ```
instruction
0
61,735
11
123,470
No
output
1
61,735
11
123,471
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution. This contest offers n problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. Input The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. Output Print a single integer — the number of problems the friends will implement on the contest. Examples Input 3 1 1 0 1 1 1 1 0 0 Output 2 Input 2 1 0 0 0 1 1 Output 1 Note In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it. In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. Submitted Solution: ``` n = int(input()) n2 = [(input()) for i in range(n)] print(n2) solved = 0 for i in n2: if i.count("0") >= 2 : pass else: solved += 1 print(solved) ```
instruction
0
61,736
11
123,472
No
output
1
61,736
11
123,473
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are n people taking part in auction today. The rules of auction are classical. There were n bids made, though it's not guaranteed they were from different people. It might happen that some people made no bids at all. Each bid is define by two integers (ai, bi), where ai is the index of the person, who made this bid and bi is its size. Bids are given in chronological order, meaning bi < bi + 1 for all i < n. Moreover, participant never makes two bids in a row (no one updates his own bid), i.e. ai ≠ ai + 1 for all i < n. Now you are curious with the following question: who (and which bid) will win the auction if some participants were absent? Consider that if someone was absent, all his bids are just removed and no new bids are added. Note, that if during this imaginary exclusion of some participants it happens that some of the remaining participants makes a bid twice (or more times) in a row, only first of these bids is counted. For better understanding take a look at the samples. You have several questions in your mind, compute the answer for each of them. Input The first line of the input contains an integer n (1 ≤ n ≤ 200 000) — the number of participants and bids. Each of the following n lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109, bi < bi + 1) — the number of participant who made the i-th bid and the size of this bid. Next line contains an integer q (1 ≤ q ≤ 200 000) — the number of question you have in mind. Each of next q lines contains an integer k (1 ≤ k ≤ n), followed by k integers lj (1 ≤ lj ≤ n) — the number of people who are not coming in this question and their indices. It is guarenteed that lj values are different for a single question. It's guaranteed that the sum of k over all question won't exceed 200 000. Output For each question print two integer — the index of the winner and the size of the winning bid. If there is no winner (there are no remaining bids at all), print two zeroes. Examples Input 6 1 10 2 100 3 1000 1 10000 2 100000 3 1000000 3 1 3 2 2 3 2 1 2 Output 2 100000 1 10 3 1000 Input 3 1 10 2 100 1 1000 2 2 1 2 2 2 3 Output 0 0 1 10 Note Consider the first sample: * In the first question participant number 3 is absent so the sequence of bids looks as follows: 1. 1 10 2. 2 100 3. 1 10 000 4. 2 100 000 Participant number 2 wins with the bid 100 000. * In the second question participants 2 and 3 are absent, so the sequence of bids looks: 1. 1 10 2. 1 10 000 The winner is, of course, participant number 1 but the winning bid is 10 instead of 10 000 as no one will ever increase his own bid (in this problem). * In the third question participants 1 and 2 are absent and the sequence is: 1. 3 1 000 2. 3 1 000 000 The winner is participant 3 with the bid 1 000. Submitted Solution: ``` from collections import defaultdict import sys input = sys.stdin.readline import bisect n=int(input()) s=set() d=defaultdict(list) d[0].append(0) for i in range(n): x,y=map(int,input().split()) s.add(x) d[x].append(y) q=int(input()) for i in range(q): l=list(map(int,input().split())) s1=s.difference(set(l[1:])) if len(s1)==0: print(0,0) else: m1,m2=0,0 for i in s1: if d[i][-1]>d[m1][-1]: m2=m1 m1=i j=bisect.bisect_right(d[m1],d[m2][-1]) print(m1,d[m1][j]) ```
instruction
0
61,944
11
123,888
No
output
1
61,944
11
123,889
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are n people taking part in auction today. The rules of auction are classical. There were n bids made, though it's not guaranteed they were from different people. It might happen that some people made no bids at all. Each bid is define by two integers (ai, bi), where ai is the index of the person, who made this bid and bi is its size. Bids are given in chronological order, meaning bi < bi + 1 for all i < n. Moreover, participant never makes two bids in a row (no one updates his own bid), i.e. ai ≠ ai + 1 for all i < n. Now you are curious with the following question: who (and which bid) will win the auction if some participants were absent? Consider that if someone was absent, all his bids are just removed and no new bids are added. Note, that if during this imaginary exclusion of some participants it happens that some of the remaining participants makes a bid twice (or more times) in a row, only first of these bids is counted. For better understanding take a look at the samples. You have several questions in your mind, compute the answer for each of them. Input The first line of the input contains an integer n (1 ≤ n ≤ 200 000) — the number of participants and bids. Each of the following n lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109, bi < bi + 1) — the number of participant who made the i-th bid and the size of this bid. Next line contains an integer q (1 ≤ q ≤ 200 000) — the number of question you have in mind. Each of next q lines contains an integer k (1 ≤ k ≤ n), followed by k integers lj (1 ≤ lj ≤ n) — the number of people who are not coming in this question and their indices. It is guarenteed that lj values are different for a single question. It's guaranteed that the sum of k over all question won't exceed 200 000. Output For each question print two integer — the index of the winner and the size of the winning bid. If there is no winner (there are no remaining bids at all), print two zeroes. Examples Input 6 1 10 2 100 3 1000 1 10000 2 100000 3 1000000 3 1 3 2 2 3 2 1 2 Output 2 100000 1 10 3 1000 Input 3 1 10 2 100 1 1000 2 2 1 2 2 2 3 Output 0 0 1 10 Note Consider the first sample: * In the first question participant number 3 is absent so the sequence of bids looks as follows: 1. 1 10 2. 2 100 3. 1 10 000 4. 2 100 000 Participant number 2 wins with the bid 100 000. * In the second question participants 2 and 3 are absent, so the sequence of bids looks: 1. 1 10 2. 1 10 000 The winner is, of course, participant number 1 but the winning bid is 10 instead of 10 000 as no one will ever increase his own bid (in this problem). * In the third question participants 1 and 2 are absent and the sequence is: 1. 3 1 000 2. 3 1 000 000 The winner is participant 3 with the bid 1 000. Submitted Solution: ``` def upper_bound(l, v): lo = 0 hi = len(l) while lo <= hi: mi = (lo + hi) // 2 if v <= l[mi]: hi = mi - 1 else: lo = mi + 1 return l[lo] q = int(input()) price = {} state = {} for i in range(q): test = list(str(input()).split(' ')) index = int(test[0]) pri = int(test[1]) if index in price: price[index].append(pri) else: price[index] = [pri] state[index] = True quer = int(input()) for i in range(quer): k = list(str(input()).split(' ')) ss = [] for x in k: if x == 0: continue t = int(x) ss.append(t) state[t] = False index1 = 0 index2 = 0 price[0] = [0] for x in price: if x == 0: continue if state[x]: if price[x][-1] > price[index1][-1]: index2 = index1 index1 = x elif price[x][-1] > price[index2][-1]: index2 = x ans = upper_bound(price[index1], price[index2][-1]) print(index1, ans) for x in ss: state[x] = True ```
instruction
0
61,945
11
123,890
No
output
1
61,945
11
123,891
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are n people taking part in auction today. The rules of auction are classical. There were n bids made, though it's not guaranteed they were from different people. It might happen that some people made no bids at all. Each bid is define by two integers (ai, bi), where ai is the index of the person, who made this bid and bi is its size. Bids are given in chronological order, meaning bi < bi + 1 for all i < n. Moreover, participant never makes two bids in a row (no one updates his own bid), i.e. ai ≠ ai + 1 for all i < n. Now you are curious with the following question: who (and which bid) will win the auction if some participants were absent? Consider that if someone was absent, all his bids are just removed and no new bids are added. Note, that if during this imaginary exclusion of some participants it happens that some of the remaining participants makes a bid twice (or more times) in a row, only first of these bids is counted. For better understanding take a look at the samples. You have several questions in your mind, compute the answer for each of them. Input The first line of the input contains an integer n (1 ≤ n ≤ 200 000) — the number of participants and bids. Each of the following n lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109, bi < bi + 1) — the number of participant who made the i-th bid and the size of this bid. Next line contains an integer q (1 ≤ q ≤ 200 000) — the number of question you have in mind. Each of next q lines contains an integer k (1 ≤ k ≤ n), followed by k integers lj (1 ≤ lj ≤ n) — the number of people who are not coming in this question and their indices. It is guarenteed that lj values are different for a single question. It's guaranteed that the sum of k over all question won't exceed 200 000. Output For each question print two integer — the index of the winner and the size of the winning bid. If there is no winner (there are no remaining bids at all), print two zeroes. Examples Input 6 1 10 2 100 3 1000 1 10000 2 100000 3 1000000 3 1 3 2 2 3 2 1 2 Output 2 100000 1 10 3 1000 Input 3 1 10 2 100 1 1000 2 2 1 2 2 2 3 Output 0 0 1 10 Note Consider the first sample: * In the first question participant number 3 is absent so the sequence of bids looks as follows: 1. 1 10 2. 2 100 3. 1 10 000 4. 2 100 000 Participant number 2 wins with the bid 100 000. * In the second question participants 2 and 3 are absent, so the sequence of bids looks: 1. 1 10 2. 1 10 000 The winner is, of course, participant number 1 but the winning bid is 10 instead of 10 000 as no one will ever increase his own bid (in this problem). * In the third question participants 1 and 2 are absent and the sequence is: 1. 3 1 000 2. 3 1 000 000 The winner is participant 3 with the bid 1 000. Submitted Solution: ``` # ---------------------------iye ha aam zindegi--------------------------------------------- import math import random import heapq, bisect import sys from collections import deque, defaultdict from fractions import Fraction import sys import threading from collections import defaultdict threading.stack_size(10**8) mod = 10 ** 9 + 7 mod1 = 998244353 # ------------------------------warmup---------------------------- import os import sys from io import BytesIO, IOBase sys.setrecursionlimit(300000) BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") # -------------------game starts now----------------------------------------------------import math class TreeNode: def __init__(self, k, v): self.key = k self.value = v self.left = None self.right = None self.parent = None self.height = 1 self.num_left = 1 self.num_total = 1 class AvlTree: def __init__(self): self._tree = None def add(self, k, v): if not self._tree: self._tree = TreeNode(k, v) return node = self._add(k, v) if node: self._rebalance(node) def _add(self, k, v): node = self._tree while node: if k < node.key: if node.left: node = node.left else: node.left = TreeNode(k, v) node.left.parent = node return node.left elif node.key < k: if node.right: node = node.right else: node.right = TreeNode(k, v) node.right.parent = node return node.right else: node.value = v return @staticmethod def get_height(x): return x.height if x else 0 @staticmethod def get_num_total(x): return x.num_total if x else 0 def _rebalance(self, node): n = node while n: lh = self.get_height(n.left) rh = self.get_height(n.right) n.height = max(lh, rh) + 1 balance_factor = lh - rh n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right) n.num_left = 1 + self.get_num_total(n.left) if balance_factor > 1: if self.get_height(n.left.left) < self.get_height(n.left.right): self._rotate_left(n.left) self._rotate_right(n) elif balance_factor < -1: if self.get_height(n.right.right) < self.get_height(n.right.left): self._rotate_right(n.right) self._rotate_left(n) else: n = n.parent def _remove_one(self, node): """ Side effect!!! Changes node. Node should have exactly one child """ replacement = node.left or node.right if node.parent: if AvlTree._is_left(node): node.parent.left = replacement else: node.parent.right = replacement replacement.parent = node.parent node.parent = None else: self._tree = replacement replacement.parent = None node.left = None node.right = None node.parent = None self._rebalance(replacement) def _remove_leaf(self, node): if node.parent: if AvlTree._is_left(node): node.parent.left = None else: node.parent.right = None self._rebalance(node.parent) else: self._tree = None node.parent = None node.left = None node.right = None def remove(self, k): node = self._get_node(k) if not node: return if AvlTree._is_leaf(node): self._remove_leaf(node) return if node.left and node.right: nxt = AvlTree._get_next(node) node.key = nxt.key node.value = nxt.value if self._is_leaf(nxt): self._remove_leaf(nxt) else: self._remove_one(nxt) self._rebalance(node) else: self._remove_one(node) def get(self, k): node = self._get_node(k) return node.value if node else -1 def _get_node(self, k): if not self._tree: return None node = self._tree while node: if k < node.key: node = node.left elif node.key < k: node = node.right else: return node return None def get_at(self, pos): x = pos + 1 node = self._tree while node: if x < node.num_left: node = node.left elif node.num_left < x: x -= node.num_left node = node.right else: return (node.key, node.value) raise IndexError("Out of ranges") @staticmethod def _is_left(node): return node.parent.left and node.parent.left == node @staticmethod def _is_leaf(node): return node.left is None and node.right is None def _rotate_right(self, node): if not node.parent: self._tree = node.left node.left.parent = None elif AvlTree._is_left(node): node.parent.left = node.left node.left.parent = node.parent else: node.parent.right = node.left node.left.parent = node.parent bk = node.left.right node.left.right = node node.parent = node.left node.left = bk if bk: bk.parent = node node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1 node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right) node.num_left = 1 + self.get_num_total(node.left) def _rotate_left(self, node): if not node.parent: self._tree = node.right node.right.parent = None elif AvlTree._is_left(node): node.parent.left = node.right node.right.parent = node.parent else: node.parent.right = node.right node.right.parent = node.parent bk = node.right.left node.right.left = node node.parent = node.right node.right = bk if bk: bk.parent = node node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1 node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right) node.num_left = 1 + self.get_num_total(node.left) @staticmethod def _get_next(node): if not node.right: return node.parent n = node.right while n.left: n = n.left return n # -----------------------------------------------binary seacrh tree--------------------------------------- class SegmentTree1: def __init__(self, data, default=2**51, func=lambda a, b: a & b): """initialize the segment tree with data""" self._default = default self._func = func self._len = len(data) self._size = _size = 1 << (self._len - 1).bit_length() self.data = [default] * (2 * _size) self.data[_size:_size + self._len] = data for i in reversed(range(_size)): self.data[i] = func(self.data[i + i], self.data[i + i + 1]) def __delitem__(self, idx): self[idx] = self._default def __getitem__(self, idx): return self.data[idx + self._size] def __setitem__(self, idx, value): idx += self._size self.data[idx] = value idx >>= 1 while idx: self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) idx >>= 1 def __len__(self): return self._len def query(self, start, stop): if start == stop: return self.__getitem__(start) stop += 1 start += self._size stop += self._size res = self._default while start < stop: if start & 1: res = self._func(res, self.data[start]) start += 1 if stop & 1: stop -= 1 res = self._func(res, self.data[stop]) start >>= 1 stop >>= 1 return res def __repr__(self): return "SegmentTree({0})".format(self.data) # -------------------game starts now----------------------------------------------------import math class SegmentTree: def __init__(self, data, default=-1, func=lambda a, b: max(a , b)): """initialize the segment tree with data""" self._default = default self._func = func self._len = len(data) self._size = _size = 1 << (self._len - 1).bit_length() self.data = [default] * (2 * _size) self.data[_size:_size + self._len] = data for i in reversed(range(_size)): self.data[i] = func(self.data[i + i], self.data[i + i + 1]) def __delitem__(self, idx): self[idx] = self._default def __getitem__(self, idx): return self.data[idx + self._size] def __setitem__(self, idx, value): idx += self._size self.data[idx] = value idx >>= 1 while idx: self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1]) idx >>= 1 def __len__(self): return self._len def query(self, start, stop): if start == stop: return self.__getitem__(start) stop += 1 start += self._size stop += self._size res = self._default while start < stop: if start & 1: res = self._func(res, self.data[start]) start += 1 if stop & 1: stop -= 1 res = self._func(res, self.data[stop]) start >>= 1 stop >>= 1 return res def __repr__(self): return "SegmentTree({0})".format(self.data) # -------------------------------iye ha chutiya zindegi------------------------------------- class Factorial: def __init__(self, MOD): self.MOD = MOD self.factorials = [1, 1] self.invModulos = [0, 1] self.invFactorial_ = [1, 1] def calc(self, n): if n <= -1: print("Invalid argument to calculate n!") print("n must be non-negative value. But the argument was " + str(n)) exit() if n < len(self.factorials): return self.factorials[n] nextArr = [0] * (n + 1 - len(self.factorials)) initialI = len(self.factorials) prev = self.factorials[-1] m = self.MOD for i in range(initialI, n + 1): prev = nextArr[i - initialI] = prev * i % m self.factorials += nextArr return self.factorials[n] def inv(self, n): if n <= -1: print("Invalid argument to calculate n^(-1)") print("n must be non-negative value. But the argument was " + str(n)) exit() p = self.MOD pi = n % p if pi < len(self.invModulos): return self.invModulos[pi] nextArr = [0] * (n + 1 - len(self.invModulos)) initialI = len(self.invModulos) for i in range(initialI, min(p, n + 1)): next = -self.invModulos[p % i] * (p // i) % p self.invModulos.append(next) return self.invModulos[pi] def invFactorial(self, n): if n <= -1: print("Invalid argument to calculate (n^(-1))!") print("n must be non-negative value. But the argument was " + str(n)) exit() if n < len(self.invFactorial_): return self.invFactorial_[n] self.inv(n) # To make sure already calculated n^-1 nextArr = [0] * (n + 1 - len(self.invFactorial_)) initialI = len(self.invFactorial_) prev = self.invFactorial_[-1] p = self.MOD for i in range(initialI, n + 1): prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p self.invFactorial_ += nextArr return self.invFactorial_[n] class Combination: def __init__(self, MOD): self.MOD = MOD self.factorial = Factorial(MOD) def ncr(self, n, k): if k < 0 or n < k: return 0 k = min(k, n - k) f = self.factorial return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD # --------------------------------------iye ha combinations ka zindegi--------------------------------- def powm(a, n, m): if a == 1 or n == 0: return 1 if n % 2 == 0: s = powm(a, n // 2, m) return s * s % m else: return a * powm(a, n - 1, m) % m # --------------------------------------iye ha power ka zindegi--------------------------------- def sort_list(list1, list2): zipped_pairs = zip(list2, list1) z = [x for _, x in sorted(zipped_pairs)] return z # --------------------------------------------------product---------------------------------------- def product(l): por = 1 for i in range(len(l)): por *= l[i] return por # --------------------------------------------------binary---------------------------------------- def binarySearchCount(arr, n, key): left = 0 right = n - 1 count = 0 while (left <= right): mid = int((right + left) / 2) # Check if middle element is # less than or equal to key if (arr[mid] < key): count = mid + 1 left = mid + 1 # If key is smaller, ignore right half else: right = mid - 1 return count # --------------------------------------------------binary---------------------------------------- def countdig(n): c = 0 while (n > 0): n //= 10 c += 1 return c def binary(x, length): y = bin(x)[2:] return y if len(y) >= length else "0" * (length - len(y)) + y def countGreater(arr, n, k): l = 0 r = n - 1 # Stores the index of the left most element # from the array which is greater than k leftGreater = n # Finds number of elements greater than k while (l <= r): m = int(l + (r - l) / 2) if (arr[m] >= k): leftGreater = m r = m - 1 # If mid element is less than # or equal to k update l else: l = m + 1 # Return the count of elements # greater than k return (n - leftGreater) # --------------------------------------------------binary------------------------------------ n=int(input()) w=defaultdict(int) d=defaultdict(list) cost=[0]*n ma=[-1]*(n+1) for i in range(n): a,b=map(int,input().split()) d[a].append(i) w[i]=a cost[i]=b for i in d: ma[i]=d[i][-1] s=SegmentTree(ma) q=int(input()) for i in range(q): l=list(map(int,input().split())) k=l[0] l=l[1:] l.append(n) l.sort() m=-1 last=0 for j in range(k+1): m=max(m,s.query(last,l[j]-1)) last=l[j]+1 if m==-1: print(0,0) continue ind=w[m] l.append(ind) l.sort() m1=-1 last=0 for j in range(k+2): m1 = max(m1, s.query(last, l[j] - 1)) last = l[j] + 1 x=binarySearchCount(d[ind],len(d[ind]),m1) print(ind,cost[d[ind][x]]) ```
instruction
0
61,946
11
123,892
No
output
1
61,946
11
123,893
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are n people taking part in auction today. The rules of auction are classical. There were n bids made, though it's not guaranteed they were from different people. It might happen that some people made no bids at all. Each bid is define by two integers (ai, bi), where ai is the index of the person, who made this bid and bi is its size. Bids are given in chronological order, meaning bi < bi + 1 for all i < n. Moreover, participant never makes two bids in a row (no one updates his own bid), i.e. ai ≠ ai + 1 for all i < n. Now you are curious with the following question: who (and which bid) will win the auction if some participants were absent? Consider that if someone was absent, all his bids are just removed and no new bids are added. Note, that if during this imaginary exclusion of some participants it happens that some of the remaining participants makes a bid twice (or more times) in a row, only first of these bids is counted. For better understanding take a look at the samples. You have several questions in your mind, compute the answer for each of them. Input The first line of the input contains an integer n (1 ≤ n ≤ 200 000) — the number of participants and bids. Each of the following n lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109, bi < bi + 1) — the number of participant who made the i-th bid and the size of this bid. Next line contains an integer q (1 ≤ q ≤ 200 000) — the number of question you have in mind. Each of next q lines contains an integer k (1 ≤ k ≤ n), followed by k integers lj (1 ≤ lj ≤ n) — the number of people who are not coming in this question and their indices. It is guarenteed that lj values are different for a single question. It's guaranteed that the sum of k over all question won't exceed 200 000. Output For each question print two integer — the index of the winner and the size of the winning bid. If there is no winner (there are no remaining bids at all), print two zeroes. Examples Input 6 1 10 2 100 3 1000 1 10000 2 100000 3 1000000 3 1 3 2 2 3 2 1 2 Output 2 100000 1 10 3 1000 Input 3 1 10 2 100 1 1000 2 2 1 2 2 2 3 Output 0 0 1 10 Note Consider the first sample: * In the first question participant number 3 is absent so the sequence of bids looks as follows: 1. 1 10 2. 2 100 3. 1 10 000 4. 2 100 000 Participant number 2 wins with the bid 100 000. * In the second question participants 2 and 3 are absent, so the sequence of bids looks: 1. 1 10 2. 1 10 000 The winner is, of course, participant number 1 but the winning bid is 10 instead of 10 000 as no one will ever increase his own bid (in this problem). * In the third question participants 1 and 2 are absent and the sequence is: 1. 3 1 000 2. 3 1 000 000 The winner is participant 3 with the bid 1 000. Submitted Solution: ``` n = int(input()) data = list(reversed([list(map(int, input().split())) for i in range(n)])) k = int(input()) for i in range(k): q = list(map(int, input().split())) q.pop(0) q = set(q) xx = 0 yy = 0 for x, y in filter(lambda v: v[0] not in q, data): if not xx: xx = x yy = y elif x == xx: yy = y else: break print(xx, yy) ```
instruction
0
61,947
11
123,894
No
output
1
61,947
11
123,895
Provide a correct Python 3 solution for this coding contest problem. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0
instruction
0
62,061
11
124,122
"Correct Solution: ``` x,y = map(int,input().split()) k = [0,3,2,1] + [0]*1000 ans = k[x]+k[y] if x == y == 1: ans += 4 print(ans*100000) ```
output
1
62,061
11
124,123
Provide a correct Python 3 solution for this coding contest problem. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0
instruction
0
62,063
11
124,126
"Correct Solution: ``` x, y = map(int, input().split()) ans = max(-x + 4, 0) + max(-y + 4, 0) print(ans * 10**5 if ans != 6 else 10**6) ```
output
1
62,063
11
124,127
Provide a correct Python 3 solution for this coding contest problem. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0
instruction
0
62,064
11
124,128
"Correct Solution: ``` x,y=map(int,input().split()) if(x==y==1): print(1000000) else: ans=max(4-x,0)*100000+max(4-y,0)*100000 print(ans) ```
output
1
62,064
11
124,129
Provide a correct Python 3 solution for this coding contest problem. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0
instruction
0
62,065
11
124,130
"Correct Solution: ``` x,y=map(int,input().split()) ans=max(0,4-x)+max(0,4-y) if ans==6: ans+=4 print(ans*100000) ```
output
1
62,065
11
124,131
Provide a correct Python 3 solution for this coding contest problem. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0
instruction
0
62,066
11
124,132
"Correct Solution: ``` X, Y = map(int, input().split()) ans = 0 if X <= 3: ans += 4 - X if Y <= 3: ans += 4 - Y if X == 1 and Y == 1: ans += 4 ans *= 100000 print(ans) ```
output
1
62,066
11
124,133
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0 Submitted Solution: ``` X, Y = map(int, input().split()) ans = max(4 - X, 0) + max(4 - Y, 0) + max(12 - (X + Y) * 4, 0) print(ans * 100000) ```
instruction
0
62,069
11
124,138
Yes
output
1
62,069
11
124,139
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0 Submitted Solution: ``` x, y = map(int, input().split()) ans = 0 ans += max((4-x)*100000, 0) ans += max((4-y)*100000, 0) print(ans+400000 if x == 1 and y == 1 else ans) ```
instruction
0
62,070
11
124,140
Yes
output
1
62,070
11
124,141
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned. Constraints * 1 \leq X \leq 205 * 1 \leq Y \leq 205 * X and Y are integers. Input Input is given from Standard Input in the following format: X Y Output Print the amount of money DISCO-Kun earned, as an integer. Examples Input 1 1 Output 1000000 Input 3 101 Output 100000 Input 4 4 Output 0 Submitted Solution: ``` x,y=map(int,input().split()) ans=0 ans+=max(4-x,0)*100000 ans+=max(4-y,0)*100000 if x==1 and y==1: ans+=400000 print(ans) ```
instruction
0
62,071
11
124,142
Yes
output
1
62,071
11
124,143