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Provide tags and a correct Python 3 solution for this coding contest problem. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Tags: brute force, implementation Correct Solution: ``` import sys, os n, m = map(int, input().split()) mi = [] ma = [] if m == 0: if n == 1: print(1) else: print(-1) exit(0) sys.exit() os.abort() for i in range(m): a, b = map(int, input().split()) if b == 1: ma.append(1000000000) mi.append(a) if n <= a: print(1) exit(0) sys.exit() os.abort() else: mak = (a - 1) // (b - 1) ma.append(mak) if a % b == 0: mi.append(a // b) else: mi.append((a // b ) + 1) #print(mi, ma) mik = min(ma) mak = max(mi) if n % mik == 0: a = n // mik else: a = (n // mik) + 1 if n % mak == 0: b = n // mak else: b = (n // mak) + 1 if a == b: print(b) else: print(-1) ```
100,000
Provide tags and a correct Python 3 solution for this coding contest problem. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Tags: brute force, implementation Correct Solution: ``` def rec(i): global a return i import sys from collections import Counter sys.setrecursionlimit(10**6) #n=int(input()) n,m=list(map(int,input().split())) a=[[] for i in range(100)] b=[i for i in range(1,101)] for i in range(m): x,y=list(map(int,input().split())) z=b.copy() for i0 in z: if not(((x-1)>=(y-1)*i0)and((x-1)<y*i0)): b.remove(i0) a=set() for i0 in b: a.add((n-1)//i0) if len(a)==1: print(a.pop()+1) else: print(-1) ```
100,001
Provide tags and a correct Python 3 solution for this coding contest problem. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Tags: brute force, implementation Correct Solution: ``` n,m=map(int,input().split()) n-=1 k=[] f=[] p=0 for i in range(m): q=list(map(int,input().split())) k.append(q[0]) f.append(q[1]) for i in range(1,101): for j in range(m): if ((k[j]-1)//i)+1!=f[j]: break else: if not p: p=n//i+1 else: q=n//i+1 if p!=q: print(-1) exit() print(p) ```
100,002
Provide tags and a correct Python 3 solution for this coding contest problem. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Tags: brute force, implementation Correct Solution: ``` from math import ceil n, m = map(int, input().split()) k = [0 for i in range(m)] f = [0 for i in range(m)] x = -1 for i in range(m): k[i], f[i] = map(int, input().split()) i = 0 while i < m: j = i while j < m: if abs(k[i] - k[j]) == 1 and abs(f[i] - f[j]) == 1: if f[i] < f[j]: x = ceil(n / ceil(k[i] / f[i])) else: x = ceil(n / ceil(k[j] / f[j])) i = m - 1 j = i j += 1 i += 1 if x > 0: print(x) else: c = [] for i in range(1, 101): j = 0 b = True while j < m and b: if not (ceil(k[j] / i) == f[j]): b = False j += 1 if b: c.append(i) x = ceil(n / c[0]) for i in range(len(c)): if ceil(n / c[i]) != x: x = -1 break print(x) ```
100,003
Provide tags and a correct Python 3 solution for this coding contest problem. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Tags: brute force, implementation Correct Solution: ``` def xor(a,b,i): if (((a-1)//i)==(b-1)): return True elif(a<i) and (b==1): return True else: return False dannie = input() dannie = dannie.split(' ') level = int(dannie[0]) n = int(dannie[1]) levels = [] if level == 1: print(1) elif n == 0: print(-1) else: for i in range(0,n): info = input() info = info.split(' ') levels.append(info) fl = True i = 1 count = 0 while (i<=100) and (fl): for etaz in levels: fl2 = True flat = int(etaz[0]) floor = int(etaz[1]) if xor(flat, floor, i): continue else: fl2 = False if not(fl2): break if not(fl2): i = i + 1 elif (count==0) and fl2: count += 1 if level%i==0: answer = int(level/i) i = i + 1 else: answer = int((level//i) + 1) i = i + 1 elif fl2 and count!=0: if level%i==0: answer1 = level/i else: answer1 = level//i + 1 if answer1 == answer: i = i + 1 continue else: print(-1) fl = False else: i +=1 if fl : try: print(answer) except NameError: print(-1) ```
100,004
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` import math n,q = map(int,input().split()) lst = [ ] for i in range(q): x,y = map(int,input().split()) lst.append((x,y)) ans = set() for no_of_floors in range(1,101): flag = 1 for k,f in lst: temp_floor = math.ceil(k/no_of_floors) if(temp_floor!=f): flag = 0 if(flag): temp = math.ceil(n/no_of_floors) ans.add(temp) if(len(ans)!=1): print(-1) else: print(*ans) ``` Yes
100,005
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` import math def find_3(): etazh = [] for p in range(1, 101): if all((memory[0] + p - 1) // p == memory[1] for memory in mass): etazh.append(p) return etazh def find_2(): etazh = [] p = 1 while p < 100: for i in range (m): if math.ceil(mass[i][0]/p) != mass[i][1]: p += 1 break elif i==(m-1): etazh.append(p) p += 1 return etazh def find_1(): etazh = [] for p in range(100): suit = True for i in range(m): if math.ceil(mass[i][0]/p) != mass[i][1]: suit = False break if suit: etazh.append(p) return etazh n, m = map(int, input().split()) mass = [] for i in range (m): mass.append(list(map(int, input().split()))) etazh = find_3() if all((n+x-1)//x == (n+etazh[0]-1)//etazh[0] for x in etazh): print((n+etazh[0]-1)//etazh[0]) else: print(-1) ``` Yes
100,006
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` n, m = map(int, input().split()) k = [0] * 500 f = [0] * 500 for i in range(m): k[i], f[i] = map(int, input().split()) vars = set() for cnt in range(1, 201): flag = 1 for i in range(m): if (k[i] - 1) // cnt != f[i] - 1: flag = 0 break if flag: vars.add((n - 1) // cnt + 1) if n == 1: ans = 1 else: ans = -1 if len(vars) > 1 else max(vars) print(ans) ``` Yes
100,007
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` n, m = map(int, input().split()) ans = -1 k = [] f = [] for i in range(m): flat, level = map(int, input().split()) k.append(flat) f.append(level) mainflag = True for count in range(1, 101): flag = True for i in range(m): if (k[i] + count - 1) // count != f[i]: flag = False break if flag and ans != -1 and (n + count - 1) // count != (n + ans - 1) // ans: mainflag = False elif flag: ans = count if mainflag: print((n + ans - 1) // ans) else: print(-1) ``` Yes
100,008
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` n, m = map(int, input().split()) a = [] for i in range(m): k, f = map(int, input().split()) a.append((k, f)) cnt = 0 num = 0 for e in range(1, 101): can = True for k, f in a: if f != (k + e - 1) // e: can = False break if can: cnt += 1 num = e if cnt == 1: print((n + num - 1) // num) else: print(-1) ``` No
100,009
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` n, m = map(int, input().split()) k, f = map(int, input().split()) omin=(k-1)//f+1 omax = -1 if f > 1: omax=(k-1)//(f-1) for i in range(m-1): k, f = map(int, input().split()) localmin = (k-1) // f + 1 if f > 1: localmax = (k-1) // (f - 1) else: localmax = -1 omin = max(localmin, omin) if localmax != -1 and (localmax < omax or omax == -1): omax = localmax if omax == omin: print((n - 1) // omin + 1) elif omax == -1: print(1) else: print(-1) ``` No
100,010
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` from sys import stdout from sys import stdin def get(): return stdin.readline().strip() def getf(): return [int(i) for i in get().split()] def put(a, end = "\n"): stdout.write(str(a) + end) def putf(a, sep = " ", end = "\n"): stdout.write(sep.join(map(str, a)) + end) from math import ceil def main(): n, m = getf() fl = [[] for i in range(101)] for i in range(m): k, f = getf() fl[f] += [k] ans = -1 fl1 = -1 for i in range(1, 101): if(min(fl[i] or [-1]) <= n <= max(fl[i] or [-1])): ans = i for i in range(1, 100): f1, f2 = fl[i], fl[i + 1] if not(f1 and f2): continue k1, k2 = max(f1), min(f2) if(k2 - k1 == 1): fl1 = k1 // i if(ans == -1): if(fl1 != -1): put(int(ceil(n / fl1))) else: put(-1) else: put(ans) main() ``` No
100,011
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In a building where Polycarp lives there are equal number of flats on each floor. Unfortunately, Polycarp don't remember how many flats are on each floor, but he remembers that the flats are numbered from 1 from lower to upper floors. That is, the first several flats are on the first floor, the next several flats are on the second and so on. Polycarp don't remember the total number of flats in the building, so you can consider the building to be infinitely high (i.e. there are infinitely many floors). Note that the floors are numbered from 1. Polycarp remembers on which floors several flats are located. It is guaranteed that this information is not self-contradictory. It means that there exists a building with equal number of flats on each floor so that the flats from Polycarp's memory have the floors Polycarp remembers. Given this information, is it possible to restore the exact floor for flat n? Input The first line contains two integers n and m (1 ≀ n ≀ 100, 0 ≀ m ≀ 100), where n is the number of the flat you need to restore floor for, and m is the number of flats in Polycarp's memory. m lines follow, describing the Polycarp's memory: each of these lines contains a pair of integers ki, fi (1 ≀ ki ≀ 100, 1 ≀ fi ≀ 100), which means that the flat ki is on the fi-th floor. All values ki are distinct. It is guaranteed that the given information is not self-contradictory. Output Print the number of the floor in which the n-th flat is located, if it is possible to determine it in a unique way. Print -1 if it is not possible to uniquely restore this floor. Examples Input 10 3 6 2 2 1 7 3 Output 4 Input 8 4 3 1 6 2 5 2 2 1 Output -1 Note In the first example the 6-th flat is on the 2-nd floor, while the 7-th flat is on the 3-rd, so, the 6-th flat is the last on its floor and there are 3 flats on each floor. Thus, the 10-th flat is on the 4-th floor. In the second example there can be 3 or 4 flats on each floor, so we can't restore the floor for the 8-th flat. Submitted Solution: ``` from math import ceil def check(flats_on_level): for fl in flats: after = ceil(fl[0] / flats_on_level) if fl[1] != after: return False return True n, m = map(int, input().split()) flats = [] for _ in range(m): ki, fi = map(int, input().split()) flats.append((ki, fi)) answer = -1 cf = -1 if m == 0 and n == 1: print(1) exit(0) for i in range(1, 101): if check(i): if answer == -1: answer = i else: cf == i if ceil(n / answer) != ceil(n / cf): print(-1) else: print(ceil(n / answer)) ``` No
100,012
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` # -*- coding: utf-8 -*- import math import collections import bisect import heapq import time import random import itertools import sys """ created by shhuan at 2017/10/22 18:15 """ N = int(input()) S = input() def split(s, wc, pl, n): ans = [] if pl % 2 == 0: if any([x % 2 == 1 for x in wc.values()]): return [] for i in range(0, len(s), pl): t = s[i:i + pl] ans.append(t[::2] + t[::-1][::2]) return ans mid = [w for w, c in wc.items() if c % 2 == 1] if len(mid) > n: return [] midindex = 0 while len(ans) < n: t = '' u = mid[midindex] if midindex < len(mid) else '' midindex += 1 if not u: u = list([w for w, c in wc.items() if c > 0])[0] wc[u] -= 1 need = pl-1 while need > 0: pneed = need for w, c in [(w, c) for w, c in wc.items()]: if c > need: t += str(w) * (need//2) wc[w] -= need need = 0 break elif c > 1: c //= 2 t += str(w) * c wc[w] -= c * 2 need -= c*2 if need == pneed: return [] t = t + u + t[::-1] if len(t) < pl: return [] ans.append(t) return ans S = "".join(sorted(list(S))) WC = collections.Counter(S) for parts in range(1, N+1): if N % parts == 0: wc = {k: v for k, v in WC.items()} ans = split(S, wc, N//parts, parts) if ans: print(parts) print(" ".join(ans)) exit(0) ```
100,013
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` import sys import math from collections import defaultdict,deque def is_possible(odd,even): each = odd*2 if even % each == 0: return True return False n=int(sys.stdin.readline()) s=sys.stdin.readline()[:-1] dic=defaultdict(int) for i in range(n): dic[s[i]]+=1 odd,even = 0,0 for i in dic: if dic[i]%2 != 0: odd += 1 even += dic[i] - 1 else: even += dic[i] if odd == 0: #print(0) ans = [0 for x in range(even)] cnt = 0 for i in dic: while dic[i] > 0: ans[cnt] = i ans[even - cnt - 1] = i cnt += 1 dic[i] -= 2 #print(ans) print(1) print(''.join(x for x in ans)) sys.exit() while even >= 0: if is_possible(odd,even): break odd += 2 even -= 2 length = 1 + (even) // (odd) #print(length, 'length') ans = [deque() for _ in range(odd)] ansb = [[] for _ in range(odd)] ansa = [[] for _ in range(odd)] cnt = 0 #print(dic,'dic') for i in dic: if dic[i] % 2 !=0: ans[cnt].append(i) dic[i] -= 1 cnt += 1 #print() #print(ans,' after odd',cnt) for i in dic: while dic[i] > 0: if cnt >= odd: break ans[cnt].append(i) ans[cnt + 1].append(i) cnt += 2 dic[i] -= 2 if cnt >= odd: break cnt = 0 #print(ans,'ans') for i in dic: #print(cnt,'cnt',i,'i') #print(ans[cnt]) if dic[i] > 0: x = len(ans[cnt]) while dic[i] > 0: ans[cnt].appendleft(i) ans[cnt].append(i) #ans[cnt] = [i] + ans[cnt] + [i] dic[i] -= 2 x += 2 if x < length: pass else: cnt += 1 if cnt < odd: x = len(ans[cnt]) #print(ans,'ans') #print(odd,'odd') n = len(ans) #print(n) l =[] for i in range(n): l.append(''.join(x for x in ans[i])) print(n) print(*l) ```
100,014
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` n = int(input()) s = list(input()) d = {} for char in s: if char not in d.keys(): d[char] = 1 else: d[char] += 1 odd = [] even = [] for char in d.keys(): for i in range(d[char] // 2): even.append(char) even.append(char) for i in range(d[char] % 2): odd.append(char) # print(len(even), len(odd)) if len(odd) <= 1: x = ''.join([even[i] for i in range(1, len(even), 2)]) y = ''.join(odd) print(1) print(x + y + x[::-1]) elif len(even) % len(odd) == 0 and (len(even) // len(odd)) % 2 == 0 and (len(even) // len(odd)) > 0: print(len(odd)) x = [] ratio = len(even) // len(odd) indx = 0 for i in range(len(odd)): tail = '' while len(tail) < ratio // 2: tail += even[indx] indx += 2 x.append(tail + odd[i] + tail[::-1]) print(' '.join(x)) else: ratio = len(even)//len(odd) max_length = 1 for i in range(1,ratio+1,2): if (len(even)+len(odd)) % i == 0 and max_length < i: max_length = i parts = (len(even) + len(odd))//max_length print(parts) x = [] indx = 0 for i in range(len(odd)): tail = '' while len(tail) < max_length // 2: tail += even[indx] indx += 2 x.append(tail + odd[i] + tail[::-1]) middle = [] for i in range(parts - len(odd)): middle.append(even.pop()) for i in range(len(middle)): tail = '' while len(tail) < max_length // 2: tail += even[indx] indx += 2 x.append(tail + middle[i] + tail[::-1]) print(' '.join(x)) ```
100,015
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` mp,a,p={},[],[] n=int(input()) for i in input(): if i in mp:mp[i]+=1 else:mp[i]=1 odd=0 for i in mp: if mp[i]&1: a.append(i) mp[i]-=1 odd+=1 if mp[i]: p.append(i) m=max(1,odd) for i in range(m,n+1): if not n%i: d=n//i if odd and not d&1:continue print(i) for K in range(i-m): a.append(p[-1]) if mp[p[-1]]>1:mp[p[-1]]-=1 else:p.pop() for k in range(i): s='' for j in range(d//2): s+=p[-1] mp[p[-1]]-=2 if not mp[p[-1]]:p.pop() if odd:print(s+a.pop()+s[::-1],end=' ') else:print(s+s[::-1],end=' ') exit() ```
100,016
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` #!/usr/bin/env python3 def main(): import collections n = int(input()) s = input() alph = collections.Counter(s) odd = sum(x & 0x1 for x in alph.values()) dq = collections.deque() if odd == 0: print(1) for c, x in alph.items(): dq.append(c * (x >> 1)) dq.appendleft(c * (x >> 1)) print(*dq, sep="") else: for odd in range(odd, n): if (n - odd) % (odd << 1) == 0: print(odd) odds = [c for c, x in alph.items() if x & 0x1] items = list(alph.items()) while len(odds) != odd: for i, x in enumerate(items): if x[1] > 1: items[i] = (x[0], x[1] - 2) odds.append(x[0]) odds.append(x[0]) break req_length = (n - odd) // odd + 1 cur_length = 0 while odd > 0: if cur_length == 0: dq.append(odds[-1]) cur_length += 1 odds.pop() x = min(items[-1][1] >> 1, (req_length - cur_length) >> 1) dq.append(items[-1][0] * x) dq.appendleft(items[-1][0] * x) cur_length += x << 1 if items[-1][1] - (x << 1) <= 1: items.pop() else: items[-1] = (items[-1][0], items[-1][1] - (x << 1)) if cur_length == req_length: print(*dq, sep="", end=' ') odd -= 1 dq.clear() cur_length = 0 print() break else: print(n) print(*s) try: while True: main() except EOFError: pass ```
100,017
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` n=int(input()) s=list(input()) lets={} for i in s: try: lets[i]+=1 except: lets[i]=1 count=0 centre=[] rest=[] for i in lets.keys(): if(lets[i]%2): centre.append(i) count+=1 rest+=[i]*(lets[i]//2) if(count==0): t='' for i in lets.keys(): t=i*(lets[i]//2) + t + i*(lets[i]//2) print(1) print(t) else: extra=0 while(n%count!=0 or (n//count)%2==0): count+=2 extra+=1 for i in range(extra): centre+=[rest.pop()]*2 amt=len(rest)//(count) temp=rest.copy() rest='' for i in temp: rest+=i print(count) print(rest[0:amt] + centre[0] + rest[(1)*(amt)-1::-1],end=' ') for i in range(1,count): print(rest[i*(amt):(i+1)*amt] + centre[i] + rest[(i+1)*(amt)-1:(i)*amt-1:-1],end=' ') print() ```
100,018
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` n = int(input()) s = list(input()) d = {} for char in s: if char not in d.keys(): d[char] = 1 else: d[char] += 1 odd = [] even = [] for char in d.keys(): for i in range(d[char] // 2): even.append(char) even.append(char) for i in range(d[char] % 2): odd.append(char) if len(odd) <= 1: x = ''.join([even[i] for i in range(1, len(even), 2)]) y = ''.join(odd) print(1) print(x + y + x[::-1]) else: ratio = len(even)//len(odd) max_length = 1 for i in range(1,ratio+2,2): if n % i == 0 and max_length < i: max_length = i parts = n//max_length print(parts) x = [] indx = 0 for i in range(len(odd)): tail = '' while len(tail) < max_length // 2: tail += even[indx] indx += 2 x.append(tail + odd[i] + tail[::-1]) middle = [] for i in range(parts - len(odd)): middle.append(even.pop()) for i in range(len(middle)): tail = '' while len(tail) < max_length // 2: tail += even[indx] indx += 2 x.append(tail + middle[i] + tail[::-1]) print(' '.join(x)) ```
100,019
Provide tags and a correct Python 3 solution for this coding contest problem. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Tags: brute force, implementation, strings Correct Solution: ``` from collections import Counter n = int(input()) cnt = Counter(input()) odd = [] even = [] for s, c in cnt.items(): if c % 2 == 1: odd.append([s, c]) else: even.append([s, c]) k = max(1, len(odd)) for i in range(k, n + 1): h = n // i if i * h == n and (not odd and h % 2 == 0 or odd and h % 2 == 1): k = i break h = n // k res = [] while len(res) < k: if odd: s, c = odd.pop() res.append(s) c -= 1 if c: even.append([s, c]) elif h % 2: res.append(even[0][0]) res.append(even[0][0]) even[0][1] -= 2 if not even[0][1]: even.pop(0) else: res.append('') for i in range(k): s = res[i] while len(s) < h: t = (h - len(s)) // 2 a, c = even[0] if c <= 2 * t: c //= 2 s = a*c + s + a*c even.pop(0) else: s = a*t + s + a*t even[0][1] -= 2 * t res[i] = s print(len(res)) print(' '.join(res)) ```
100,020
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` from collections import Counter n = int(input()) inp = input() c = Counter(inp) t = len(list(filter(lambda x: x % 2 == 1, list(c.values())))) if t == 0: temp = sum([[list(c.items())[j][0]] * (list(c.items())[j][1] // 2) for j in range(len(c.items()))], []) print(1) print(''.join(temp) + ''.join(temp[::-1])) else: ones = list() for j in range(len(c.items())): i = list(c.items())[j] if i[1] % 2 == 1: c[list(c.keys())[j]] -= 1 ones.append(i[0]) others = sum([[list(c.items())[j][0]] * (list(c.items())[j][1] // 2) for j in range(len(c.items()))], []) while n % t != 0 or n // t % 2 == 0: t += 2 o = others.pop() ones.append(o) ones.append(o) L = n // t ans = list() ot = 0 for i in range(t): mid = str() if L % 2 == 1: mid = ones[i] ln = L // 2 ans.append(''.join(others[ot: ot + ln]) + mid + ''.join(others[ot: ot + ln][::-1])) ot += ln print(t) print(' '.join(ans)) ``` Yes
100,021
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` n = int(input()) s = input() d = {} for c in s: if c not in d: d[c] = 0 d[c] += 1 odd = [] sum = 0 for key in d: if d[key] % 2 == 1: odd.append(key) d[key] -= 1 sum += d[key] if sum // 2 < len(odd): print(len(s)) for c in s: print(c+' ', end='') exit(0) while True: if len(odd) == 0 or (sum // 2) % len(odd) == 0: break sum -= 2 if sum // 2 < len(odd): print(len(s)) for c in s: print(c + ' ', end='') exit(0) for key in d: odd.append(key) odd.append(key) d[key] -= 2 if d[key] == 0: d.pop(key, None) break if len(odd) == 0: odd.append('') even = [] for key in d: even += [key] * (d[key] // 2) l = (sum // 2) // len(odd) print(len(odd)) for i in range(len(odd)): s = odd[i] for j in range(i * l, (i + 1) * l): s = even[j] + s + even[j] print(s + ' ', end='') ``` Yes
100,022
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` n = int(input()) string = input() char = [] charPair = [] charImpair = [] for car in string: if car not in char: z = string.count(car) while z>1: charPair.append(car) z-=2 if(z==1): charImpair.append(car) char.append(car) if len(charImpair) ==0 : String1 = '' for x in charPair: String1+= x if len(charImpair) ==1 : String1 += charImpair[0] for x in charPair[::-1]: String1 += x print('1') print(String1) else : nbPalin = len(charImpair) while len(charPair)%nbPalin != 0 : nbPalin += 2 charImpair.append(charPair[-1]) charImpair.append(charPair[-1]) del(charPair[-1]) lenPalindrome = n/nbPalin Palin = [] for i in range(nbPalin): String2 = '' indice = i * int(((lenPalindrome-1)/2)) for x in charPair[indice : int(indice + ((lenPalindrome-1)/2))]: String2 += x String2 += charImpair[i] for x in range(indice + int(((lenPalindrome-1)/2))-1, indice-1 ,-1): # charPair[i + int(((lenPalindrome-1)/2))-1: i-1 :-1]: String2 += charPair[x] Palin.append(String2) print(nbPalin) String3 = ' '.join(Palin) print(String3) ``` Yes
100,023
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` import sys from collections import Counter def gen(counter, skip): for c, n in counter: has = n // 2 if skip > has: skip -= has has = 0 elif skip > 0: has -= skip skip = 0 for i in range(has): yield c def gen_o(counter, take): for c, n in counter: for i in range(n % 2): yield c for c, n in counter: has = (n // 2) * 2 if take > has: take -= has elif take > 0: has = take take = 0 for i in range(has): yield c N = int(next(sys.stdin)) s = next(sys.stdin).strip() c = Counter(s) even = 0 odd = 0 items = list(c.items()) for _, n in items: even += n // 2 odd += n % 2 generator = gen(items, 0) generator_o = gen_o(items, 0) odd_c = odd if odd == 0: print(1) front = [next(generator) for i in range(len(s) // 2)] fs = "".join(front) print(fs + fs[::-1]) else: skip = 0 while even != 0 and even % odd != 0: skip += 1 odd += 2 even -= 1 generator = gen(items, skip) generator_o = gen_o(items, skip * 2) print(odd) per = even // odd strs = [] for i in range(odd): front = [next(generator) for i in range(per)] fs = "".join(front) strs.append(fs + next(generator_o) + fs[::-1]) print(" ".join(strs)) ``` Yes
100,024
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` nlet = int(input()) let = [x for x in input()] uno = [x for x in let if let.count(x)==1] due = sorted([x for x in let if let.count(x)>1])[::2] palin = len(uno) totale = "" if palin == 0: totale += "".join(due+due[::-1]) else: grup = len(due)//palin j = 0 for i in range(1,palin+1): f = i*grup ini = due[j:f] fin = ini[::-1] totale += "".join(ini+[uno[i-1]]+fin+[" "]) j = f print (totale) ``` No
100,025
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` n = int(input()) s = list(input())[:n] d = {} for char in s: if char not in d.keys(): d[char] = 1 else: d[char] += 1 odd = [] even = [] for char in d.keys(): for i in range(d[char] // 2): even.append(char) even.append(char) for i in range(d[char] % 2): odd.append(char) if len(odd) <= 1: x = ''.join([even[i] for i in range(1,len(even),2)]) y = ''.join(odd) print(1) print(x+y+x[::-1]) elif (len(even)//len(odd)) % 2 == 0 and (len(even)//len(odd)) > 0: print(len(odd)) x = [] ratio = len(even)//len(odd) indx = 0 for i in range(len(odd)): tail = '' while len(tail) < ratio//2: tail += even[indx] indx += 2 x.append(tail+odd[i]+tail[::-1]) print(' '.join(x)) else: print(len(even)//2 + len(odd)) print((' '.join(even)+' '+ ' '.join(odd)).strip()) ``` No
100,026
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` from collections import Counter import sys n = int(input()) s = input() c = Counter(s) a = [] odd = 0 for i in c: if c[i] % 2: odd += 1 a.append(i) c[i] -= 1 even = n - odd oddawal = odd if odd != 0: while even > 0: if (even//2) % odd: even -= 2 odd += 2 else: break else: s = '' for i in c: s += i * (c[i]//2) s += s[::-1] print(1) print(s) sys.exit() ubah = odd-oddawal d = {} for i in c: if ubah > 0: if c[i] % 2 == 0: if ubah >= c[i]: for _ in range(c[i]): a.append(i) ubah -= c[i] d[i] = 0 else: for _ in range(c[i]-ubah): a.append(i) ubah = 0 d[i] = c[i]-ubah else: break for i in d: c[i] = d[i] print(len(a)) s = '' for i in c: s += i * (c[i]//2) su = (len(s)//2)//len(a) for i in range(len(a)): a[i] = s[:su+1] + a[i] + s[:su+1][::-1] s = s[su+1:] print(*a) ``` No
100,027
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Kolya has a string s of length n consisting of lowercase and uppercase Latin letters and digits. He wants to rearrange the symbols in s and cut it into the minimum number of parts so that each part is a palindrome and all parts have the same lengths. A palindrome is a string which reads the same backward as forward, such as madam or racecar. Your task is to help Kolya and determine the minimum number of palindromes of equal lengths to cut s into, if it is allowed to rearrange letters in s before cuttings. Input The first line contains an integer n (1 ≀ n ≀ 4Β·105) β€” the length of string s. The second line contains a string s of length n consisting of lowercase and uppercase Latin letters and digits. Output Print to the first line an integer k β€” minimum number of palindromes into which you can cut a given string. Print to the second line k strings β€” the palindromes themselves. Separate them by a space. You are allowed to print palindromes in arbitrary order. All of them should have the same length. Examples Input 6 aabaac Output 2 aba aca Input 8 0rTrT022 Output 1 02TrrT20 Input 2 aA Output 2 a A Submitted Solution: ``` n = int(input()) s = input() let = list(set(list(s))) m = len(let) cnt = [] for c in let: cn = [0] for sc in s: cn.append(cn[-1] + (sc == c)) cnt.append(cn) def check(i, j): c = 0 for p in range(m): c += (cnt[p][j + 1] - cnt[p][i]) % 2 == 1 return c < 2 def create(i, j): c = [] mid = None for p in range(m): h = cnt[p][j + 1] - cnt[p][i] if h and h % 2 == 0: c.append((let[p], h)) if h % 2: mid = let[p], h res = [] for l, h in c: res.extend([l]*(h//2)) if mid: res.extend([mid[0]]*mid[1]) for l, h in reversed(c): res.extend([l]*(h//2)) return ''.join(res) for c in range(1, n + 1): if n % c == 0: ch = True l = n // c for p in range(c): ch = ch and check(p * l, (p + 1) * l - 1) if ch: res = [] for p in range(c): res.append(create(p * l, (p + 1) * l - 1)) print(' '.join(res)) break ``` No
100,028
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` n = int(input()) alpha = [0]*(26) c = 0 flag = 0 for i in range(n): a,b = input().split() if a=='.' : b = list(set(b)) for i in b: alpha[ord(i)-97] = -1 elif a=='!': if alpha.count(max(alpha))==1: if flag: c+=1 else: flag = 1 elif not flag: b = list(set(b)) for i in b: if alpha[ord(i)-97] !=-1: alpha[ord(i)-97] += 1 else: if flag : if (ord(b)-97)!=alpha.index(max(alpha)) : c+=1 else: alpha[ord(b)-97] = -1 if alpha.count(max(alpha))==1 and not flag: flag = 1 print(c) ```
100,029
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` from collections import * import os, sys from io import BytesIO, IOBase 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") BUFSIZE = 8192 sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") rstr = lambda: input().strip() rstrs = lambda: [str(x) for x in input().split()] rstr_2d = lambda n: [rstr() for _ in range(n)] rint = lambda: int(input()) rints = lambda: [int(x) for x in input().split()] rint_2d = lambda n: [rint() for _ in range(n)] rints_2d = lambda n: [rints() for _ in range(n)] ceil1 = lambda a, b: (a + b - 1) // b mem, ans = set('abcdefghijklmnopqrstuvwxyz'), 0 for _ in range(rint() - 1): ch, word = rstrs() if ch == '.': mem -= set(word) elif ch == '?': if len(mem) > 1: mem.discard(word) elif mem: ans += 1 else: if len(mem) == 1: ans += 1 else: mem &= set(word) print(ans) ```
100,030
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` n = int(input()) d = set() for i in range(26): d.add(chr(ord('a')+i)) chocks = 0 done = -1 for i in range(n): s = input().split() toR = [] if s[0] == '!': chocks += 1 for x in d: if not x in s[1]: toR.append(x) for x in toR: d.remove(x) if s[0] == '.': for x in d: if x in s[1]: toR.append(x) for x in toR: d.remove(x) if s[0] == '?': if i!=n-1: chocks += 1 if s[1] in d: d.remove(s[1]) if len(d) == 1 and done == -1: done = chocks if done == -1: print(0) else: print(chocks-done) ```
100,031
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` '''input 5 ! abc . ad . b ! cd ? c ''' t = [0] * 26 e = 0 n = int(input()) if n == 1: print(0) quit() for i in range(n-1): x, y = input().split() if x == '!': if 1 in t: c = [ord(p) - 97 for p in set(y)] for x in range(26): if not(t[x] == 1 and x in c): t[x] = -1 else: for l in set(y): if t[ord(l) - 97] == 0: t[ord(l) - 97] = 1 elif x == '.': for l in set(y): t[ord(l) - 97] = -1 else: t[ord(y) - 97] = -1 if t.count(1) == 1 or (t.count(0) == 1 and max(t) == 0): break g = [False] * 26 for _ in range(i+1, n-1): x, y = input().split() if x == '!' or x == '?': e += 1 print(e) ```
100,032
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` n = int(input()) flag = False ans = 0 curr = set([chr(x) for x in range(ord('a'), ord('z')+1)]) for i in range(n): c,s = input().split(' ') if c[0] == '.': letters = set(s) curr = curr - letters if len(curr) == 1: flag = True continue if c[0] == '!': if flag: ans += 1 else: letters = set(s) curr = curr & letters if len(curr) == 1: flag = True continue if c[0] == '?': guess = s[0] if flag: ans += 1 else: curr = curr - set(guess) if len(curr) == 1: flag = True if i == n-1 and len(curr) == 1 and guess in curr: ans -= 1 continue print(ans) ```
100,033
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` # reading input from stdin numActions = int(input()) actions = [] for i in range(numActions): tempStr = input() if tempStr[0] == '!': action = 'shock' elif tempStr[0] == '.': action = 'none' else: action = 'guess' string = tempStr[2:len(tempStr)] actions.append([action, string]) # algorithm shockList = set() cleanList = set() numExtraShocks = 0 firstShock = True for i in range(len(actions)): action = actions[i][0] string = actions[i][1] if action == 'shock': if len(shockList) == 1 or len(cleanList) == 25: numExtraShocks += 1 for letter in string: if firstShock == True: if letter not in cleanList: shockList.add(letter) elif letter not in cleanList and letter not in shockList: cleanList.add(letter) if firstShock == True: firstShock = False newShockList = shockList.copy() for i in range(len(list(shockList))): string1 = list(shockList)[i] for letter1 in string1: if letter1 not in string: newShockList.remove(letter1) cleanList.add(letter1) shockList = newShockList.copy() elif action == 'none': for letter in string: if letter in shockList: shockList.remove(letter) cleanList.add(letter) elif action == 'guess': if (len(shockList) == 1 or len(cleanList) == 25) and i != len(actions) - 1: numExtraShocks += 1 if i != len(actions) - 1: if string[0] in shockList: shockList.remove(string[0]) cleanList.add(string[0]) print(numExtraShocks) ```
100,034
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` n = int(input()) q = [] for _ in range(n): q.append(input().split()) s = set('abcdefghijklmnopqrstuvwxyz') letter = q[-1][-1] for i in range(n-1): if q[i][0] == '!': s &= set(q[i][1]) else: s -= set(q[i][1]) if len(s) == 1: cnt = 0 for j in range(i+1, n-1): if q[j][0] != '.': cnt += 1 print(cnt) exit(0) print(0) ```
100,035
Provide tags and a correct Python 3 solution for this coding contest problem. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Tags: implementation, strings Correct Solution: ``` n = int(input()) s = set ('abcdefghijklmnopqrstuvwxyz') count = 0 for i in range (n): strg = input().split() if strg[0] == "?": if len(s)==1 and i != n - 1: count += 1 s -= set(strg[1]) if strg[0] == "!": if len(s)==1: count += 1 s = s & set(strg[1]) if strg[0] == ".": s -= set(strg[1]) print (count) ```
100,036
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` n = int(input()) candidates = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'} count = 0 for i in range(n - 1): action, word = input().split() if len(candidates) == 1: if action != '.': count += 1 else: if action == '?': candidates.discard(word) elif action == '.': for letter in word: candidates.discard(letter) else: toremove = set() for letter in candidates: if letter not in word: toremove.add(letter) candidates = candidates.difference(toremove) print(count) ``` Yes
100,037
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` n = int(input()) sym = [] word = [] dict = {} for z in range(26): dict[chr(97 + z)] = 1 for x in range(n): a, b = input().split() sym.append(a) word.append(b) q = -1 flag = 0 for x in range(n): if sym[x] == '!': flag = 1 sety = set() for i in range(26): if (dict[chr(97 + i)] == 0): sety.add(chr(97 + i)) dict[chr(97 + i)] = 0 for s in word[x]: dict[s] = 1 for s in sety: dict[s] = 0 if sym[x] == '.': for s in word[x]: dict[s] = 0 if sym[x] == '?': for s in word[x]: dict[s] = 0 j = 0 for i in range(26): j += dict[chr(97 + i)] if j == 1: q = x break count = 0 if q==-1: print(0) else: for p in range(q+1,n-1): if sym[p] == '!' or sym[p]=='?': count+=1 print(count) ``` Yes
100,038
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` from collections import * import os, sys from io import BytesIO, IOBase 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") BUFSIZE = 8192 sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") rstr = lambda: input().strip() rstrs = lambda: [str(x) for x in input().split()] rstr_2d = lambda n: [rstr() for _ in range(n)] rint = lambda: int(input()) rints = lambda: [int(x) for x in input().split()] rint_2d = lambda n: [rint() for _ in range(n)] rints_2d = lambda n: [rints() for _ in range(n)] ceil1 = lambda a, b: (a + b - 1) // b mem, ans = set('abcdefghijklmnopqrstuvwxyz'), 0 for _ in range(rint() - 1): ch, word = rstrs() if ch == '.': mem.difference_update(list(word)) elif ch == '?': if len(mem) > 1: mem.discard(word) elif mem: ans += 1 else: if len(mem) == 1: ans += 1 else: mem.intersection_update(list(word)) print(ans) ``` Yes
100,039
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` import math,sys,bisect,heapq from collections import defaultdict,Counter,deque from itertools import groupby,accumulate #sys.setrecursionlimit(200000000) int1 = lambda x: int(x) - 1 input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__ ilele = lambda: map(int,input().split()) alele = lambda: list(map(int, input().split())) ilelec = lambda: map(int1,input().split()) alelec = lambda: list(map(int1, input().split())) def list2d(a, b, c): return [[c] * b for i in range(a)] def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)] #MOD = 1000000000 + 7 def Y(c): print(["NO","YES"][c]) def y(c): print(["no","yes"][c]) def Yy(c): print(["No","Yes"][c]) extra = -1;s = set();mark = {} for _ in range(int(input())): a,b = input().split(" ") if len(s) == 1 or len(mark) == 25: if a!='.': extra += 1 continue if a == '!': if not(s): s = set(b) else: c = set(b) l = [] for i in c: if i in s: l.append(i) new_s = set() #print(l) for i in s: if i in l: new_s.add(i) else: mark[i] = 1 s = new_s.copy() elif a == '.': c = set(b) l = [] for i in c: mark[i] = 1 if i in s: l.append(i) new_s = set() #print(l) for i in s: if i not in l: new_s.add(i) s = new_s.copy() else: mark[b] = 1 t = set() for i in s: if mark.get(i,-1) == -1: t.add(i) s = t.copy() #print(s,mark,len(mark)) print(max(0,extra)) ``` Yes
100,040
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` # reading input from stdin numActions = int(input()) actions = [] for i in range(numActions): tempStr = input() if tempStr[0] == '!': action = 'shock' elif tempStr[0] == '.': action = 'none' else: action = 'guess' string = tempStr[2:len(tempStr)] actions.append([action, string]) # algorithm shockList = set() cleanList = set() numExtraShocks = 0 for i in range(len(actions)): action = actions[i][0] string = actions[i][1] if action == 'shock': if len(shockList) == 1: numExtraShocks += 1 for letter in string: if i == 0: shockList.add(letter) elif letter not in cleanList and letter not in shockList: cleanList.add(letter) newShockList = shockList.copy() for i in range(len(list(shockList))): string1 = list(shockList)[i] for letter1 in string1: if letter1 not in string: newShockList.remove(letter1) cleanList.add(letter1) shockList = newShockList.copy() elif action == 'none': for letter in string: if letter in shockList: shockList.remove(letter) cleanList.add(letter) elif action == 'guess': if i != len(actions) - 1: if string[0] in shockList: shockList.remove(string[0]) cleanList.add(string[0]) if len(shockList) == 1 and i != len(actions) - 1: numExtraShocks += 1 print(numExtraShocks) ``` No
100,041
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` def main(): n = int(input()) yes = set(list(map(chr, range(97, 123)))) no = set() cnt = 0 for i in range(n): mark, word = [x for x in input().split()] tmp_yes = set() tmp_no = set() if (mark == '!'): if (len(yes) == 1): cnt += 1 for c in word: tmp_yes.add(c) yes = yes & tmp_yes elif (mark == '.'): for c in word: tmp_no.add(c) no = no | tmp_no elif (mark == '?' and i != n-1): if (len(yes) == 1): cnt += 1 tmp_no.add(word) no = no | tmp_no for c in (yes & no): yes.remove(c) no.add(c) print(yes) print(no) print(cnt) main() ``` No
100,042
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` from sys import stdin n = int(stdin.readline().rstrip('\n')) not_in_ans = set() in_ans = set() count = 0 f = False for i in range(n): line = stdin.readline().rstrip('\n').split() if line[0] == chr(46): # . if len(in_ans) == 0: not_in_ans |= set(line[1]) else: in_ans -= set(line[1]) in_ans -= not_in_ans not_in_ans = set() if len(in_ans) == 1: f = True elif line[0] == chr(33): # ! if f: count += 1 else: if len(in_ans) == 0: in_ans = set(line[1]) if len(in_ans) == 1: f = True in_ans &= set(line[1]) elif line[0] == chr(63): # ? if f and i + 1 != n: count += 1 else: if i + 1 != n: in_ans -= set(line[1]) if len(in_ans) == 1: f = True print(count) ``` No
100,043
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly. Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks. Input The first line contains a single integer n (1 ≀ n ≀ 105) β€” the number of actions Valentin did. The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types: 1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and w is the word that Valentin said. 2. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and w is the word that Valentin said. 3. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and s is the guess β€” a lowercase English letter. All words consist only of lowercase English letters. The total length of all words does not exceed 105. It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter. Output Output a single integer β€” the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. Examples Input 5 ! abc . ad . b ! cd ? c Output 1 Input 8 ! hello ! codeforces ? c . o ? d ? h . l ? e Output 2 Input 7 ! ababahalamaha ? a ? b ? a ? b ? a ? h Output 0 Note In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces the word cd and gets an excessive electric shock. In the second test case after the first two electric shocks we understand that the selected letter is e or o. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is e, but Valentin makes 3 more actions before he makes a correct hypothesis. In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks. Submitted Solution: ``` '''input 5 . abcdefghijklmnopqrstuvwxy ? b ? c ? d ? z ''' t = [0] * 26 n = int(input()) if n == 1: print(0) quit() for i in range(n-1): x, y = input().split() if x == '!': if 1 in t: c = [ord(p) - 97 for p in set(y)] for i in range(26): if t[i] != 1 or i not in c: t[i] = -1 else: for l in set(y): if t[ord(l) - 97] == 0: t[ord(l) - 97] = 1 elif x == '.': for l in set(y): t[ord(l) - 97] = -1 else: t[ord(y) - 97] = -1 # print(t) if t.count(1) == 1 or (t.count(0) == 1 and max(t) == 0): break e = 0 for _ in range(i+1, n-1): x, y = input().split() if x == '!' or x == '?': e += 1 print(e) ``` No
100,044
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` import sys n,m = [int(x) for x in input().split(' ')] a=[] for i in range(n): b=[] s = input().strip() for j in range(m): if s[j]=='#': b.append(j) if b not in a: a.append(b) c=[0]*m ans=True #print(a) for i in a: for j in i: c[j]+=1 if c[j]>1: ans=False break if ans: print("Yes") else: print("No") ```
100,045
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` par = input() par = list(map(int, par.split())) n, m = par[0], par[1] mat = [] for i in range(n): row = list(input()) mat.append(row) def posColor(n, m, mt): for i in range(n): for j in range(m): if mt[i][j] == '#': for x in range(i + 1, n): if mt[x][j] == '#': for y in range(m): if mt[i][y] != mt[x][y]: return 'No' return 'Yes' print(posColor(n, m, mat)) ```
100,046
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` from collections import defaultdict r,c=map(int,input().split()) h=defaultdict(list) flg=[] for i in range(r): st=input() st=str(st) y=[] for j in range(c): if st[j]=='#': y.append(j) if y in list(h.values()): flg.append(1) # print(1) else: for k in y: for item in list(h.values()): if k in item: flg.append(0) # print(2) break h[i]=y # print(list(h.values()),y) if 0 in flg: print('No') else: print('Yes') ```
100,047
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` N, M = map(int, input().split()) grid = [] for _ in range(N): grid.append(input()) S = [set() for _ in range(N)] for i in range(N): for j in range(M): if grid[i][j] == "#": S[i].add(j) solved = [False for _ in range(N)] used = [False for _ in range(M)] for i in range(N): if not solved[i]: for c in S[i]: if used[c]: print("No") quit() used[c] = True for j in range(N): if S[i] == S[j]: solved[j] = True print("Yes") ```
100,048
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` used =[] a = [] def dfs(v, c): vert = v + 100*c if used[vert] == 0: used[vert] = 1 a[c].append(v) for i in range(0, len(g[vert])): if used[g[vert][i]] == 0: dfs(g[vert][i] - (1-c)*100, 1-c) n,m = map(int, input().split()) r = [] for i in range(0,n): s = input() r.append(s) g = [] for i in range (0,200): g.append([]) used.append(0) for i in range (0,n): for j in range (0,m): if(r[i][j] == "#"): g[i].append(100 + j) g[100+j].append(i) #print(g) for i in range(0,n): a = [[],[]] dfs(i,0) #print(a) for j in range (0, len(a[0])): for k in range (0, len(a[1])): if(r[a[0][j]][a[1][k]] != "#"): print ("NO") raise SystemExit print("YES") ```
100,049
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` n, m = map(int, input().split()) x = [] for i in range(n): q = [] y = input() for j in y: q.append(j) x.append(q) ind = True for i in range(n): for j in range(m): if x[i][j] == "#": for k in range(i + 1, n): if x[k][j] == "#" and x[k] != x[i]: ind = False break if ind: print("Yes") else: print("No") ```
100,050
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` def partial_match(r1, r2): n = len(r1) match_found = False mismatch_found = False for i in range(n): if r1[i] == '#' and r2[i] == '#': match_found = True if r1[i] == '#' and r2[i] == '.': mismatch_found = True if r2[i] == '#' and r1[i] == '.': mismatch_found = True return match_found and mismatch_found n , m = tuple(map(int, input().split())) l = [] for i in range(n): l.append(input().strip()) invalid = False for i in range(n): for j in range(n): if partial_match(l[i], l[j]): invalid = True break if invalid: print("No") else: print("Yes") ```
100,051
Provide tags and a correct Python 3 solution for this coding contest problem. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Tags: greedy, implementation Correct Solution: ``` def func(ss): occured = set() for pr, s in zip(['']+ss[:-1], ss): if pr == s: continue for i, el in enumerate(s): if el == "#": if i in occured: return False occured.add(i) return True n, m = [int(x) for x in input().split()] ss = [input() for _ in range(n)] ss.sort() print("Yes" if func(ss) else "No") ```
100,052
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n, m = map(int, input().split()) st = [] for i in range(n): s = input() x = set() for j in range(m): if s[j] == '#': x.add(j) for i in st: if i & x and i != x: print('No') exit() st.append(x) print('Yes') ``` Yes
100,053
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n, m = map( int, input().split() ) M = [] for i in range(n): M.append(input()) Adj = [[] for i in range(n+m)] for i in range(n): for j in range(m): if M[i][j] == '#': Adj[i].append(n+j) Adj[n+j].append(i) p = [-1]*(n+m) comp = [] answer = 'Yes' for i in range(n+m): if p[i] != -1: continue L = [i] comp.append([]) numEdges = 0 numA = 0 numB = 0 while L: v = L.pop() comp[-1].append((v, 'A') if v < n else (v-n, 'B')) if v < n: numA += 1 else: numB += 1 for u in Adj[v]: numEdges += 1 if p[u] == -1 and u != i: p[u] = v L.append(u) #print(comp[-1], numA, numB, numEdges/2) if numEdges//2 != numA*numB: answer = 'No' print(answer) ``` Yes
100,054
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n,m = map(int,input().split()) a = [] f = 0 for i in range(n):a.append(input()) for i in range(n): for j in range(i+1,n): r1 = a[i] r2 = a[j] f1 = f2 = 0 for k in range(m): if r1[k] != r2[k]:f1 = 1 if (r1[k] == '#') and (r2[k]=='#'):f2 = 1 if f1 == 1: if f2 == 1: f = 1 if f:print('No') else:print('Yes') ``` Yes
100,055
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n,m=map(int,input().split()) a=[] b=[] for i in range(n): a.append(input()) s='' for j in a[i]: if j=='#': s+='1' else: s+='0' b.append(s) for i in range(n): for j in range(i,n): if a[i]!=a[j]: if '2' in str(int(b[i]) + int(b[j])): print('No') exit() print('Yes') ``` Yes
100,056
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` def table(n, m, lst): for i in range(n): for j in range(n): if lst[i] != lst[j]: for k in range(m): if lst[i][z] == lst[j][z] == "#": return "No" return "Yes" n, m = [int(i) for i in input().split()] b = list() for i in range(n): z = list(input()) b.append(z) ``` No
100,057
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` from collections import defaultdict as dd, deque r,c = map(int,input().split()) A = [list(input()) for i in range(r)] At = [[A[i][j] for i in range(r)] for j in range(c)] R=r C=c def find(): for i in range(R): for j in range(C): if A[i][j] == "#": return i,j return () while True: pos = find() if pos: i,j = pos rows = [i] cols = [j] for x in range(c): if A[i][x] == '#': cols.append(x) for x in range(r): if A[x][j] == '#': rows.append(x) fail = False row = A[rows[0]] for x in range(1,len(rows)): if row != A[rows[x]]: fail = True col = At[cols[0]] for x in range(1,len(cols)): if col != At[cols[x]]: fail = True if fail: print('No') break #print('fill') for r in rows: for c in cols: A[r][c] = '.' At[c][r] = '.' #print(cols) #print(rows) #for a in A: # print(''.join(a)) #for a in At: # print(''.join(a)) else: print('Yes') break ``` No
100,058
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n, m = map(int, input().split()) array = [] state = True for i in range(n): string = input() if string.find('##')!=-1: state = False if state==True: print('Yes') else: print('No') ``` No
100,059
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a rectangular grid of n rows of m initially-white cells each. Arkady performed a certain number (possibly zero) of operations on it. In the i-th operation, a non-empty subset of rows Ri and a non-empty subset of columns Ci are chosen. For each row r in Ri and each column c in Ci, the intersection of row r and column c is coloured black. There's another constraint: a row or a column can only be chosen at most once among all operations. In other words, it means that no pair of (i, j) (i < j) exists such that <image> or <image>, where <image> denotes intersection of sets, and <image> denotes the empty set. You are to determine whether a valid sequence of operations exists that produces a given final grid. Input The first line contains two space-separated integers n and m (1 ≀ n, m ≀ 50) β€” the number of rows and columns of the grid, respectively. Each of the following n lines contains a string of m characters, each being either '.' (denoting a white cell) or '#' (denoting a black cell), representing the desired setup. Output If the given grid can be achieved by any valid sequence of operations, output "Yes"; otherwise output "No" (both without quotes). You can print each character in any case (upper or lower). Examples Input 5 8 .#.#..#. .....#.. .#.#..#. #.#....# .....#.. Output Yes Input 5 5 ..#.. ..#.. ##### ..#.. ..#.. Output No Input 5 9 ........# #........ ..##.#... .......#. ....#.#.# Output No Note For the first example, the desired setup can be produced by 3 operations, as is shown below. <image> For the second example, the desired setup cannot be produced, since in order to colour the center row, the third row and all columns must be selected in one operation, but after that no column can be selected again, hence it won't be possible to colour the other cells in the center column. Submitted Solution: ``` n, m = map(int, input().split()) l = [input() for i in range(n)] flag = True for i in range(n): if not flag: break for j in range(i + 1, n): if l[i] != l[j]: for q in range(m): if l[j][q] == l[i][q] == '#': print('NO') flag = False break else: print('YES') ``` No
100,060
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` n = int(input()) seats = list(map(int, input().split())) sarr = list(zip(seats, range(1, n+1))) sarr.sort() sarr = [i[1] for i in sarr] ans = [] k = 0 for i in input(): if i == '0': print(sarr[k], end = " ") ans.append(sarr[k]) k+=1 if i == '1': print(ans[-1], end = " ") ans.pop() ```
100,061
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` # -*- coding: utf-8 -*- # Baqir Khan # Software Engineer (Backend) from sys import stdin inp = stdin.readline n = int(inp()) a = [(int(ele), i + 1) for i, ele in enumerate(inp().split())] s = inp().strip() ans = [0] * (2* n) partially_filled = [] a.sort() start = 0 for j in range(2*n): if s[j] == "0": partially_filled.append(a[start]) ans[j] = a[start][1] start += 1 else: ans[j] = partially_filled[-1][1] partially_filled.pop() print(*ans) ```
100,062
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` n = int(input()) w = [[-int(x), i + 1] for (i, x) in enumerate(input().split(' '))] w.sort(key = lambda x : x[0]) m = input() e = []; res = []; for x in m: if x == '0': res.append(w[len(w) - 1][1]) e.append(w.pop()[1]) else: res.append(e.pop()); print(' '.join(map(str, res))) ```
100,063
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` n=int(input()) l=list(map(int,input().split())) a={} for t in range(n): a[l[t]]=t+1 c=a.items() b=sorted(c,reverse=True) f=[] p=[] k=input() for i in k: if i=="0": x=b.pop() f.append(x) p.append(x[1]) else: y=f.pop() p.append(y[1]) print(*p) ```
100,064
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` n = int(input()) widths = [(int(x), c) for c, x in enumerate(input().split())] widths.sort() passengers = input() seating = "" stack = [] i = 0 for pi in passengers: if pi == '1': seating += str(stack[-1][1]+1) + " " stack.pop() else: stack.append(widths[i]) seating += str(stack[-1][1]+1) + " " i += 1 print(seating[:-1]) ```
100,065
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` def main(): n = int(input()) w = input().split() w = sorted([[int(w[i]), i + 1] for i in range(n)]) introindex = 0 result = [] introPlaces = [] for passenger in input(): if passenger == '0': result.append(str(w[introindex][1])) introPlaces.append(w[introindex][1]) introindex += 1 else: result.append(str(introPlaces.pop())) print(' '.join(result)) if __name__ == "__main__": main() ```
100,066
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` N = int(input()) Width = [int(x) for x in input().split()] Input_Order = [int(x) for x in list(input())] Sorted_Width = sorted(Width) Data = {Width[x]:x for x in range(0,N)} A = [] B = [] position = 0 for x in range(0,2*N): if Input_Order[x]==0: A.append(Data[Sorted_Width[position]]+1) B.append(Data[Sorted_Width[position]]+1) position+=1 else : A.append(B[-1]) B.pop() for x in A: print(x,end =' ') ```
100,067
Provide tags and a correct Python 3 solution for this coding contest problem. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Tags: data structures, greedy, implementation Correct Solution: ``` """Problem B - Bus of Characters. http://codeforces.com/contest/982/problem/B In the Bus of Characters there are `n` rows of seat, each having `2` seats. The width of both seats in the `i`-th row is `w_i` centimeters. All integers `w_i` are distinct. Initially the bus is empty. On each of `2n` stops one passenger enters the bus. There are two types of passengers: - an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; - an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input: The first line contains a single integer `n` (`1 <= n <= 200000`) β€” the number of rows in the bus. The second line contains the sequence of integers `w_1, w_2, ..., w_n` (`1 <= w_i <= 10^{9}`), where `w_i` is the width of each of the seats in the `i`-th row. It is guaranteed that all `w_i` are distinct. The third line contains a string of length `2n`, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the `j`-th character is '0', then the passenger that enters the bus on the `j`-th stop is an introvert. If the `j`-th character is '1', the the passenger that enters the bus on the `j`-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal `n`), and for each extrovert there always is a suitable row. Output: Print `2n` integers β€” the rows the passengers will take. The order of passengers should be the same as in input. """ def solve(w, s): empty = sorted(enumerate(w), key=lambda x: x[1], reverse=True) non_empty = [] result = [] for p in s: if p == '0': row = empty.pop() non_empty.append(row) else: row = non_empty.pop() result.append(row[0] + 1) return result def main(): _ = input() w = [int(x) for x in input().strip().split()] s = input().strip() result = solve(w, s) print(' '.join(map(str, result))) if __name__ == '__main__': main() ```
100,068
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` n = int(input()) width = [(int(w), i) for i, w in enumerate(input().split(" "))] order = input() width = sorted(width, key=lambda x: x[0]) no_passengers = [w[1] for w in width] one_passenger = [] i = 0 for person in order: if person == "0": print(no_passengers[i] + 1, end=" ") one_passenger.append(no_passengers[i]) i += 1 else: print(one_passenger.pop() + 1, end=" ") ``` Yes
100,069
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` def main(): n = int(input()) w = input().split() w = sorted([(int(w[i]), i + 1) for i in range(n)]) introindex = 0 result = [] introPlaces = [] for passenger in input(): if passenger == '0': result.append(str(w[introindex][1])) introPlaces.append(w[introindex][1]) introindex += 1 else: result.append(str(introPlaces.pop())) print(' '.join(result)) if __name__ == "__main__": main() ``` Yes
100,070
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` class heap: def __init__(self, maxn): self.a = [0] * maxn self.size = 0 def shift_down(self, i): while 2 * i + 1 < self.size: l = 2 * i + 1 r = 2 * i + 2 j = l if r < self.size and self.a[r] < self.a[l]: j = r if self.a[i] <= self.a[j]: break self.a[i], self.a[j] = self.a[j], self.a[i] i = j def shift_up(self, i): while i and self.a[i] < self.a[(i - 1) // 2]: self.a[i], self.a[(i - 1) // 2] = self.a[(i - 1) // 2], self.a[i] i = (i - 1) // 2 def erase_min(self): mn = self.a[0] self.a[0] = self.a[self.size - 1] self.size -= 1 self.shift_down(0) return mn def insert(self, val): self.size += 1 self.a[self.size - 1] = val self.shift_up(self.size - 1) n = int(input()) ar = [int(j) for j in input().split()] s0 = heap(400000 + 100) s1 = heap(400000 + 100) for i in range(n): s0.insert((ar[i], i + 1)) for c in input(): if c == '0': v = s0.erase_min() print(v[1], end=' ') s1.insert((-v[0], v[1])) else: print(s1.erase_min()[1], end=' ') ``` Yes
100,071
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` import heapq row = int(input()) width = [int(x) for x in input().split()] Introvert = [] Extrovert = [] for i in range(row): heapq.heappush(Introvert,(width[i],i+1)) sequence = input() Ans = [] for i in sequence: if(i=="0"): #Introvert x,y = heapq.heappop(Introvert) heapq.heappush(Extrovert,(-x,y)) Ans.append(y) else: x,y = heapq.heappop(Extrovert) Ans.append(y) print(*Ans) ``` Yes
100,072
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` n = int(input().strip()) temp = input().strip().split() w = [] for j in range(len(temp)): w.append([int(temp[j]),j]) w.sort() #print(w) passenger = input().strip() biggest_intro = [] smallest_index = 0 for i in range(2*n): if passenger[i] == '0': #introvert #smallest_empty = w[smallest_index][1] #actual index print(w[smallest_index][1]+1, end='') biggest_intro.append(smallest_index) #non-actual index, bigger width at the end smallest_index+=1 elif passenger[i] == '1': b = biggest_intro.pop() print(w[b][1]+1,end='') #print(biggest_intro) ``` No
100,073
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` import queue n=int(input()) l=list(map(int,input().split())) a={} for t in range(n): a[l[t]]=t+1 c=a.items() b=sorted(c) L=queue.Queue(maxsize=n) for u in b: L.put(u) f=[] p=[] k=input() for i in k: if i=="0": x=L.get() f.append(x) p.append(x[1]) else: y=f.pop() p.append(x[1]) print(*p) ``` No
100,074
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` n = int(input()) temp = list(zip([int(x) for x in input().split()], range(1, n+1))) temp.sort(key=lambda x: x[0]) type1 = [int(x) for x in list(input())] a = [] ans = [] for i in type1: if i == 0: ans.append(temp[0][1]) a.append(temp[0]) del temp[0] else: ans.append(a[-1][1]) # temp = [a[-1]] + temp del a[-1] print(ans) ``` No
100,075
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. In the Bus of Characters there are n rows of seat, each having 2 seats. The width of both seats in the i-th row is w_i centimeters. All integers w_i are distinct. Initially the bus is empty. On each of 2n stops one passenger enters the bus. There are two types of passengers: * an introvert always chooses a row where both seats are empty. Among these rows he chooses the one with the smallest seats width and takes one of the seats in it; * an extrovert always chooses a row where exactly one seat is occupied (by an introvert). Among these rows he chooses the one with the largest seats width and takes the vacant place in it. You are given the seats width in each row and the order the passengers enter the bus. Determine which row each passenger will take. Input The first line contains a single integer n (1 ≀ n ≀ 200 000) β€” the number of rows in the bus. The second line contains the sequence of integers w_1, w_2, ..., w_n (1 ≀ w_i ≀ 10^{9}), where w_i is the width of each of the seats in the i-th row. It is guaranteed that all w_i are distinct. The third line contains a string of length 2n, consisting of digits '0' and '1' β€” the description of the order the passengers enter the bus. If the j-th character is '0', then the passenger that enters the bus on the j-th stop is an introvert. If the j-th character is '1', the the passenger that enters the bus on the j-th stop is an extrovert. It is guaranteed that the number of extroverts equals the number of introverts (i. e. both numbers equal n), and for each extrovert there always is a suitable row. Output Print 2n integers β€” the rows the passengers will take. The order of passengers should be the same as in input. Examples Input 2 3 1 0011 Output 2 1 1 2 Input 6 10 8 9 11 13 5 010010011101 Output 6 6 2 3 3 1 4 4 1 2 5 5 Note In the first example the first passenger (introvert) chooses the row 2, because it has the seats with smallest width. The second passenger (introvert) chooses the row 1, because it is the only empty row now. The third passenger (extrovert) chooses the row 1, because it has exactly one occupied seat and the seat width is the largest among such rows. The fourth passenger (extrovert) chooses the row 2, because it is the only row with an empty place. Submitted Solution: ``` import heapq n=int(input()) get=lambda:map(int,input().split()) w=list(get()) s=input() intro=[(i,e) for e,i in enumerate(w)] exto=[] heapq.heapify(w) for i in s: if i=='0': e,i=heapq.heappop(intro) heapq.heappush(exto,(-e,i)) print(i+1,end=' ') else: e,i=heapq.heappop(exto) print(i+1,end=' ') ``` No
100,076
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` class BIT(): __slots__ = ["n", "data"] def __init__(self, length_or_list): if isinstance(length_or_list, int): self.n = length_or_list + 1 self.data = [0] * self.n else: self.n = len(length_or_list) + 1 self.data = [0] + length_or_list for i in range(1, self.n): if i + (i & -i) < self.n: self.data[i + (i & -i)] += self.data[i] def point_append(self, index, delta): index += 1 while index < self.n: self.data[index] += delta index += index & -index def prefix_folded(self, end): res = 0 while end > 0: res += self.data[end] end -= end & -end return res def folded(self, begin, end): ret = 0 while begin < end: ret += self.data[end] end -= end & -end while end < begin: ret -= self.data[begin] begin -= begin & -begin return ret def main(): import sys input = sys.stdin.buffer.readline read = sys.stdin.buffer.read N, Q = map(int, input().split()) seg = BIT(list(map(int, input().split()))) for _ in range(Q): a, b, c = map(int, input().split()) if a: print(seg.folded(b, c)) else: seg.point_append(b, c) if __name__ == "__main__": main() ```
100,077
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` class BIT: def __init__(self, n): self.n = n self.bit = [0]*(self.n+1) # 1-indexed def init(self, init_val): for i, v in enumerate(init_val): self.add(i, v) def add(self, i, x): # i: 0-indexed i += 1 # to 1-indexed while i <= self.n: self.bit[i] += x i += (i & -i) def sum(self, i, j): # return sum of [i, j) # i, j: 0-indexed return self._sum(j) - self._sum(i) def _sum(self, i): # return sum of [0, i) # i: 0-indexed res = 0 while i > 0: res += self.bit[i] i -= i & (-i) return res def lower_bound(self, x): s = 0 pos = 0 depth = self.n.bit_length() v = 1 << depth for i in range(depth, -1, -1): k = pos + v if k <= self.n and s + self.bit[k] < x: s += self.bit[k] pos += v v >>= 1 return pos n, q = map(int, input().split()) A = list(map(int, input().split())) bit = BIT(n) bit.init(A) for _ in range(q): i, j, k = map(int, input().split()) if i == 0: bit.add(j, k) else: print(bit.sum(j, k)) ```
100,078
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` import sys class Bit: def __init__(self, n, arr): self.size = n self.tree = [0] + arr for i in range(1, n+1): x = i + (i & -i) if x < n + 1: self.tree[x] += self.tree[i] 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 def range_sum(self, l, r): return self.sum(r) - self.sum(l) def main(): n, q = map(int, sys.stdin.buffer.readline().split()) bit = Bit(n, list(map(int, sys.stdin.buffer.readline().split()))) for x in sys.stdin.buffer.readlines(): q, p, x = map(int, x.split()) if q: print(bit.range_sum(p, x)) else: bit.add(p+1, x) if __name__ == "__main__": main() ```
100,079
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` import sys input = lambda: sys.stdin.readline() class BIT: def __init__(self, n): self.n = n self.data = [0] * (n + 1) def sum(self, i): s = 0 while i > 0: s += self.data[i] i -= i & -i return s def add(self, i, x): while i <= self.n: self.data[i] += x i += i & -i n, m = map(int, input().split()) c = list(map(int, input().split())) b = BIT(n) for i in range(n): b.add(i + 1, c[i]) res = [] for i in range(m): t, x, y = map(int, input().split()) if t == 1: res.append(b.sum(y) - b.sum(x)) else: b.add(x + 1, y) print("\n".join(map(str, res))) ```
100,080
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` import sys sys.setrecursionlimit(10 ** 8) input = sys.stdin.readline def main(): def segfunc(x, y): return x + y def init(init_val): for i in range(n): seg[i + num - 1] = init_val[i] for i in range(num - 2, -1, -1): seg[i] = segfunc(seg[2 * i + 1], seg[2 * i + 2]) def update(k, x): k += num - 1 seg[k] = x while k: k = (k - 1) // 2 seg[k] = segfunc(seg[k * 2 + 1], seg[k * 2 + 2]) def query(p, q): if q <= p: return ide_ele p += num - 1 q += num - 2 res = ide_ele while q - p > 1: if p & 1 == 0: res = segfunc(res, seg[p]) if q & 1 == 1: res = segfunc(res, seg[q]) q -= 1 p = p // 2 q = (q - 1) // 2 if p == q: res = segfunc(res, seg[p]) else: res = segfunc(segfunc(res, seg[p]), seg[q]) return res N, Q = [int(x) for x in input().split()] A = [int(x) for x in input().split()] # seg treeεˆζœŸε€€ (ε˜δ½ε…ƒ) n = N ide_ele = 0 num = 2 ** (n - 1).bit_length() seg = [ide_ele] * 2 * num init(A) for _ in range(Q): q, p, x = [int(x) for x in input().split()] if q == 0: A[p] += x update(p, A[p]) else: print(query(p, x)) if __name__ == '__main__': main() ```
100,081
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` def main(): def f(i,x): while i<=n: b[i]+=x i+=i&-i def g(i): s=0 while i: s+=b[i] i-=i&-i return s (n,q,*w),a,*t=[map(int,t.split())for t in open(0)] b=[0]*-~n *map(f,range(1,n+1),a), for q,l,r in t: if q:w+=g(r)-g(l), else:f(l+1,r) print(' '.join(map(str,w))) main() ```
100,082
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` class fenwick_tree(): def __init__(self, n:int): self.__n = n self.__data = [0] * self.__n def add(self, p:int, x:int): assert (0 <= p) & (p < self.__n) p+=1 while( p<= self.__n): self.__data[p-1] += x p += p & -p def sum(self, l:int, r:int): assert (0 <= l) & (l <= r) & (r <= self.__n) return self.__sum_mod0(r) - self.__sum_mod0(l) def __sum_mod0(self, r:int): s = 0 while(r > 0): s += self.__data[r-1] r -= r & -r return s import sys read = sys.stdin.buffer.read readline = sys.stdin.buffer.readline readlines = sys.stdin.buffer.readlines n,q = map(int,readline().split()) a = list(map(int,readline().split())) query = list(map(int,read().split())) ft = fenwick_tree(n) ans = [] for i,ai in enumerate(a): ft.add(i,ai) i = 0 for _ in range(q): if(query[i]==0): p,x = query[i+1:i+3] ft.add(p,x) else: l,r = query[i+1:i+3] ans.append(ft.sum(l,r)) i += 3 print('\n'.join(map(str,ans))) ```
100,083
Provide a correct Python 3 solution for this coding contest problem. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 "Correct Solution: ``` n, q = map(int, input().split()) BIT = [0] * (n + 1) def add(i, x): while i <= n: BIT[i] += x i += i & -i def query(i): s = 0 while i > 0: s += BIT[i] i -= i & -i return s a = list(map(int, input().split())) for i in range(n): add(i + 1, a[i]) for i in range(q): x, y, z = map(int, input().split()) if x == 0: add(y + 1, z) else: print(query(z) - query(y)) ```
100,084
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` import sys input = sys.stdin.buffer.readline class FenwickTree: """FenwickTree (Binary Indexed Tree, 0-index) Queries: 1. add(i, val): add val to i-th value 2. sum(n): sum(bit[0] + ... + bit[n-1]) complexity: O(log n) See: http://hos.ac/slides/20140319_bit.pdf """ def __init__(self, a_list): self.N = len(a_list) self.bit = a_list[:] for _ in range(self.N, 1 << (self.N - 1).bit_length()): self.bit.append(0) for i in range(self.N-1): self.bit[i | (i+1)] += self.bit[i] def add(self, i, val): while i < self.N: self.bit[i] += val i |= i + 1 def sum(self, n): ret = 0 while n >= 0: ret += self.bit[n] n = (n & (n + 1)) - 1 return ret def query(self, low, high): return self.sum(high) - self.sum(low) def yosupo(): # https://judge.yosupo.jp/problem/point_add_range_sum _, Q = map(int, input().split()) fwt = FenwickTree([int(x) for x in input().split()]) ans = [] for _ in range(Q): type_, l, r = map(int, input().split()) if type_ == 0: fwt.add(l, r) else: ans.append(fwt.query(l-1, r-1)) print(*ans, sep="\n") def aoj(): # https://onlinejudge.u-aizu.ac.jp/courses/library/3/DSL/2/DSL_2_B N, Q = map(int, input().split()) fwt = FenwickTree([0] * N) for _ in range(Q): type_, l, r = map(int, input().split()) if type_ == 0: fwt.add(l-1, r) else: print(fwt.query(l-2, r-1)) if __name__ == "__main__": yosupo() # aoj() ``` Yes
100,085
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` import sys class Fenwick_Tree: def __init__(self, n): self._n = n self.data = [0] * n def add(self, p, x): assert 0 <= p < self._n p += 1 while p <= self._n: self.data[p - 1] += x p += p & -p def sum(self, l, r): assert 0 <= l <= r <= self._n return self._sum(r) - self._sum(l) def _sum(self, r): s = 0 while r > 0: s += self.data[r - 1] r -= r & -r return s def main(): input = sys.stdin.readline n, q = map(int, input().split()) fw = Fenwick_Tree(n) for i, a in enumerate(map(int, input().split())): fw.add(i, a) for _ in range(q): t, a, b = map(int, input().split()) if t == 0: fw.add(a, b) else: print(fw.sum(a, b)) if __name__ == "__main__": main() ``` Yes
100,086
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` class SegmentTree: def __init__(self, a): self.padding = 0 self.n = len(a) self.N = 2 ** (self.n-1).bit_length() self.seg_data = [self.padding]*(self.N-1) + a + [self.padding]*(self.N-self.n) for i in range(2*self.N-2, 0, -2): self.seg_data[(i-1)//2] = self.seg_data[i] + self.seg_data[i-1] def __len__(self): return self.n def update(self, i, x): idx = self.N - 1 + i self.seg_data[idx] += x while idx: idx = (idx-1) // 2 self.seg_data[idx] = self.seg_data[2*idx+1] + self.seg_data[2*idx+2] def query(self, i, j): # [i, j) if i == j: return self.seg_data[self.N - 1 + i] else: idx1 = self.N - 1 + i idx2 = self.N - 2 + j # ι–‰εŒΊι–“γ«γ™γ‚‹ result = self.padding while idx1 < idx2 + 1: if idx1&1 == 0: # idx1γŒεΆζ•° result = result + self.seg_data[idx1] if idx2&1 == 1: # idx2がε₯‡ζ•° result = result + self.seg_data[idx2] idx2 -= 1 idx1 //= 2 idx2 = (idx2 - 1)//2 return result @property def data(self): return self.seg_data[self.N-1:self.N-1+self.n] N, Q = map(int, input().split()) A = list(map(int, input().split())) st = SegmentTree(A) ans = [] for _ in range(Q): t, i, j = map(int, input().split()) if t: ans.append(st.query(i, j)) else: st.update(i, j) print(*ans, sep='\n') ``` Yes
100,087
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` class BinaryIndexedTree: # a[i] = [0] * n def __init__(self, n): self.size = n self.data = [0] * (n+1) # return sum(a[0:i]) def cumulative_sum(self, i): ans = 0 while i > 0: ans += self.data[i] i -= i & -i return ans # a[i] += x def add(self, i, x): i += 1 while i <= self.size: self.data[i] += x i += i & -i def main(): from sys import stdin input = stdin.buffer.readline n, q = map(int, input().split()) a = list(map(int, input().split())) bit = BinaryIndexedTree(n) for i, ai in enumerate(a): bit.add(i, ai) ans = [] for _ in range(q): typ, i, j = map(int, input().split()) if typ == 0: bit.add(i, j) else: ans.append(bit.cumulative_sum(j) - bit.cumulative_sum(i)) for i in ans: print(i) main() ``` Yes
100,088
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` n, q = list(map(lambda x: int(x), input().split())) list_a = list(map(lambda x: int(x), input().split())) # element of ith index is the sum of subarray: list_a[:i+1] current_sum = 0 sum_list_a = [] for ele in list_a: # print(ele, current_sum) current_sum += ele sum_list_a.append(current_sum) for _ in range(q): i, x, y = list(map(lambda x: int(x), input().split())) if i == 0: for i in range(x, len(x)): sum_list_a[i] += y continue if (x == 0): print(sum_list_a[y-1]) continue print(sum_list_a[y-1] - sum_list_a[x-1]) ``` No
100,089
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` import sys 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 def main(): n, q = map(int, sys.stdin.buffer.readline().split()) bit = Bit(n+1) a = list(map(int, sys.stdin.buffer.readline().split())) for i in range(n): bit.add(i+1, a[i]) for x in sys.stdin.buffer.readlines(): q, p, x = map(int, x.split()) if q: print(bit.sum(x) - bit.sum(p)) else: bit.add(p, x) if __name__ == "__main__": main() ``` No
100,090
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` class binaryindexedtree(): def __init__(self,n): self.n=n self.tree=[0]*n def add(self,a,w): x=a while x<=self.n: self.tree[x-1]+=w x+=x&(-x) def sums(self,a): x=a S=0 while x!=0: S+=self.tree[x-1] x-=x&(-x) return S n,q=map(int,input().split()) a=list(map(int,input().split())) bit=binaryindexedtree(n) for i,x in enumerate(a,1): bit.add(i,x) ans=[] for _ in range(q): x,y,z=map(int,input().split()) if x:ans+=[bit.sums(z)-bit.sums(y)] else:bit.add(y,z) print(*ans,sep='\n') ``` No
100,091
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given an array a_0, a_1, ..., a_{N-1} of length N. Process Q queries of the following types. * `0 p x`: a_p \gets a_p + x * `1 l r`: Print \sum_{i = l}^{r - 1}{a_i}. Constraints * 1 \leq N, Q \leq 500,000 * 0 \leq a_i, x \leq 10^9 * 0 \leq p < N * 0 \leq l_i < r_i \leq N * All values in Input are integer. Input Input is given from Standard Input in the following format: N Q a_0 a_1 ... a_{N - 1} \textrm{Query}_0 \textrm{Query}_1 : \textrm{Query}_{Q - 1} Output For each query of the latter type, print the answer. Example Input 5 5 1 2 3 4 5 1 0 5 1 2 4 0 3 10 1 0 5 1 0 3 Output 15 7 25 6 Submitted Solution: ``` class FenwickTree(): def __init__(self, n): self.n = n self.data = [0] * n def build(self, arr): #assert len(arr) <= n for i, a in enumerate(arr): self.data[i] = a for i in range(1, self.n + 1): if i + (i & -i) < self.n: self.data[i + (i & -i) - 1] += self.data[i - 1] def add(self, p, x): #assert 0 <= p < self.n p += 1 while p <= self.n: self.data[p - 1] += x p += p & -p def sum(self, r): s = 0 while r: s += self.data[r - 1] r -= r & -r return s def range_sum(self, l, r): #assert 0 <= l <= r <= self.n return self.sum(r) - self.sum(l) import sys input = sys.stdin.buffer.readline N, Q = map(int, input().split()) A = tuple(map(int, input().split())) ft = FenwickTree(N) ft.build(A) res = [] for _ in range(Q): q, x, y = map(int, input().split()) if q: res.append(str(ft.range_sum(x, y))) else: ft.add(x, y) print('\n'.join(res)) ``` No
100,092
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x=int(input()) for i in range(120): for j in range(-62,i): if i**5-j**5==x:print(i,j);exit() ```
100,093
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x=int(input()) for a in range(201): for b in range(-200,201): if a**5-b**5==x: print(a,b) exit() ```
100,094
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` A= int(input()) for i in range(-119,119): for j in range(-119,119): if i**5-j**5 == A: print(i,j) exit() ```
100,095
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x=int(input()) for a in range(120): for b in range(-120,a): if a**5 - b**5 == x: print(a,b) exit() ```
100,096
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x=int(input()) for i in range(-400,400): for j in range(-400,400): if i**5-j**5==x: print(i,j) exit() ```
100,097
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x=int(input()) for i in range(120): for j in range(-54,i): if i**5-j**5==x:print(i,j);exit() ```
100,098
Provide a correct Python 3 solution for this coding contest problem. Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. Constraints * 1 \leq X \leq 10^9 * X is an integer. * There exists a pair of integers (A, B) satisfying the condition in Problem Statement. Input Input is given from Standard Input in the following format: X Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Output Print A and B, with space in between. If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them. A B Examples Input 33 Output 2 -1 Input 1 Output 0 -1 "Correct Solution: ``` x = int(input()) for i in range(238): a = i-118 for j in range(238): b = j-119 if a**5-b**5==x: print(a, b) exit() ```
100,099