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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions: * L \leq A_1 + A_2 + ... + A_N \leq R * When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal. Since the answer can be enormous, print it modulo 10^9+7. Constraints * All values in input are integers. * 1 \leq M < N \leq 3 \times 10^5 * 1 \leq L \leq R \leq 3 \times 10^5 Input Input is given from Standard Input in the following format: N M L R Output Print the number of sequences of N non-negative integers, modulo 10^9+7. Examples Input 4 2 3 7 Output 105 Input 2 1 4 8 Output 3 Input 141592 6535 89793 238462 Output 933832916 Submitted Solution: ``` # -*- coding: utf-8 -*- ############# # Libraries # ############# import sys input = sys.stdin.readline import math #from math import gcd import bisect import heapq from collections import defaultdict from collections import deque from collections import Counter from functools import lru_cache ############# # Constants # ############# MOD = 10**9+7 INF = float('inf') AZ = "abcdefghijklmnopqrstuvwxyz" ############# # Functions # ############# ######INPUT###### def I(): return int(input().strip()) def S(): return input().strip() def IL(): return list(map(int,input().split())) def SL(): return list(map(str,input().split())) def ILs(n): return list(int(input()) for _ in range(n)) def SLs(n): return list(input().strip() for _ in range(n)) def ILL(n): return [list(map(int, input().split())) for _ in range(n)] def SLL(n): return [list(map(str, input().split())) for _ in range(n)] ######OUTPUT###### def P(arg): print(arg); return def Y(): print("Yes"); return def N(): print("No"); return def E(): exit() def PE(arg): print(arg); exit() def YE(): print("Yes"); exit() def NE(): print("No"); exit() #####Shorten##### def DD(arg): return defaultdict(arg) #####Inverse##### def inv(n): return pow(n, MOD-2, MOD) ######Combination###### kaijo_memo = [] def kaijo(n): if(len(kaijo_memo) > n): return kaijo_memo[n] if(len(kaijo_memo) == 0): kaijo_memo.append(1) while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD) return kaijo_memo[n] gyaku_kaijo_memo = [] def gyaku_kaijo(n): if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n] if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1) while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD) return gyaku_kaijo_memo[n] def nCr(n,r): if n == r: return 1 if n < r or r < 0: return 0 ret = 1 ret = ret * kaijo(n) % MOD ret = ret * gyaku_kaijo(r) % MOD ret = ret * gyaku_kaijo(n-r) % MOD return ret ######Factorization###### def factorization(n): arr = [] temp = n for i in range(2, int(-(-n**0.5//1))+1): if temp%i==0: cnt=0 while temp%i==0: cnt+=1 temp //= i arr.append([i, cnt]) if temp!=1: arr.append([temp, 1]) if arr==[]: arr.append([n, 1]) return arr #####MakeDivisors###### def make_divisors(n): divisors = [] for i in range(1, int(n**0.5)+1): if n % i == 0: divisors.append(i) if i != n // i: divisors.append(n//i) return divisors #####MakePrimes###### def make_primes(N): max = int(math.sqrt(N)) seachList = [i for i in range(2,N+1)] primeNum = [] while seachList[0] <= max: primeNum.append(seachList[0]) tmp = seachList[0] seachList = [i for i in seachList if i % tmp != 0] primeNum.extend(seachList) return primeNum #####GCD##### def gcd(a, b): while b: a, b = b, a % b return a #####LCM##### def lcm(a, b): return a * b // gcd (a, b) #####BitCount##### def count_bit(n): count = 0 while n: n &= n-1 count += 1 return count #####ChangeBase##### def base_10_to_n(X, n): if X//n: return base_10_to_n(X//n, n)+[X%n] return [X%n] def base_n_to_10(X, n): return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X)))) def base_10_to_n_without_0(X, n): X -= 1 if X//n: return base_10_to_n_without_0(X//n, n)+[X%n] return [X%n] #####IntLog##### def int_log(n, a): count = 0 while n>=a: n //= a count += 1 return count ############# # Main Code # ############# N,M,L,R = IL() @lru_cache(maxsize=None) def P(t): ans = 0 for k in range(t//M+1): u = t-M*(k+1) if u>=0: if u%(k+1) == 0: ans += nCr(N-M,u//(k+1))*(-1)**((u//(k+1))%2) if k!=0: if u%k == 0: ans -= nCr(N-M,u//k)*(-1)**((u//k)%2) ans %= MOD return ans @lru_cache(maxsize=None) def Q(t): ans = nCr(t+N,N) return ans def f(MAX): ans = 0 for t in range(MAX+1): ans += P(t)*Q(MAX-t) ans %= MOD return ans def F(x): return (Q(x)-nCr(N,M)*f(x))%MOD print((F(R)-F(L-1))%MOD) ```
instruction
0
93,304
5
186,608
No
output
1
93,304
5
186,609
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions: * L \leq A_1 + A_2 + ... + A_N \leq R * When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal. Since the answer can be enormous, print it modulo 10^9+7. Constraints * All values in input are integers. * 1 \leq M < N \leq 3 \times 10^5 * 1 \leq L \leq R \leq 3 \times 10^5 Input Input is given from Standard Input in the following format: N M L R Output Print the number of sequences of N non-negative integers, modulo 10^9+7. Examples Input 4 2 3 7 Output 105 Input 2 1 4 8 Output 3 Input 141592 6535 89793 238462 Output 933832916 Submitted Solution: ``` import numpy as np N,M,L,R = map(int,input().split()) MOD = 10 ** 9 + 7 U = 6*10**5 + 100 fact = [1] * (U+1) for n in range(1,U+1): fact[n] = fact[n-1]*n%MOD fact_inv = [1] * (U+1) fact_inv[U] = pow(fact[U],MOD-2,MOD) for n in range(U,0,-1): fact_inv[n-1] = fact_inv[n] * n % MOD fact = np.array(fact,dtype=np.int64) fact_inv = np.array(fact_inv,dtype=np.int64) U = N - M comb = fact[U] * fact_inv[:U+1] % MOD * fact_inv[U::-1] % MOD comb[1::2] *= (-1) F = np.zeros(N+M+1,dtype=np.int64) F[M:N+1] = comb P = np.zeros(R+1,dtype=np.int64) for i,x in enumerate(F[1:],1): P[::i] += x P[:M-1:-1] -= P[-M-1::-1] P %= MOD # (1-x)^{N+1}で割る Q = fact[N:N+R+1] * fact_inv[:R+1] % MOD * fact_inv[N] % MOD def f(R): total = fact[R+N] * fact_inv[R] % MOD * fact_inv[N] % MOD x = (Q[R::-1] * P[:R+1] % MOD).sum() % MOD coef = fact[N] * fact_inv[M] % MOD * fact_inv[N-M] % MOD return (total - x*coef)%MOD answer = (f(R) - f(L-1))%MOD print(answer) ```
instruction
0
93,305
5
186,610
No
output
1
93,305
5
186,611
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions: * L \leq A_1 + A_2 + ... + A_N \leq R * When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal. Since the answer can be enormous, print it modulo 10^9+7. Constraints * All values in input are integers. * 1 \leq M < N \leq 3 \times 10^5 * 1 \leq L \leq R \leq 3 \times 10^5 Input Input is given from Standard Input in the following format: N M L R Output Print the number of sequences of N non-negative integers, modulo 10^9+7. Examples Input 4 2 3 7 Output 105 Input 2 1 4 8 Output 3 Input 141592 6535 89793 238462 Output 933832916 Submitted Solution: ``` # -*- coding: utf-8 -*- ############# # Libraries # ############# import sys input = sys.stdin.readline import math #from math import gcd import bisect import heapq from collections import defaultdict from collections import deque from collections import Counter from functools import lru_cache ############# # Constants # ############# MOD = 10**9+7 INF = float('inf') AZ = "abcdefghijklmnopqrstuvwxyz" ############# # Functions # ############# ######INPUT###### def I(): return int(input().strip()) def S(): return input().strip() def IL(): return list(map(int,input().split())) def SL(): return list(map(str,input().split())) def ILs(n): return list(int(input()) for _ in range(n)) def SLs(n): return list(input().strip() for _ in range(n)) def ILL(n): return [list(map(int, input().split())) for _ in range(n)] def SLL(n): return [list(map(str, input().split())) for _ in range(n)] ######OUTPUT###### def P(arg): print(arg); return def Y(): print("Yes"); return def N(): print("No"); return def E(): exit() def PE(arg): print(arg); exit() def YE(): print("Yes"); exit() def NE(): print("No"); exit() #####Shorten##### def DD(arg): return defaultdict(arg) #####Inverse##### def inv(n): return pow(n, MOD-2, MOD) ######Combination###### kaijo_memo = [] def kaijo(n): if(len(kaijo_memo) > n): return kaijo_memo[n] if(len(kaijo_memo) == 0): kaijo_memo.append(1) while(len(kaijo_memo) <= n): kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD) return kaijo_memo[n] gyaku_kaijo_memo = [] def gyaku_kaijo(n): if(len(gyaku_kaijo_memo) > n): return gyaku_kaijo_memo[n] if(len(gyaku_kaijo_memo) == 0): gyaku_kaijo_memo.append(1) while(len(gyaku_kaijo_memo) <= n): gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD) return gyaku_kaijo_memo[n] def nCr(n,r): if n == r: return 1 if n < r or r < 0: return 0 ret = 1 ret = ret * kaijo(n) % MOD ret = ret * gyaku_kaijo(r) % MOD ret = ret * gyaku_kaijo(n-r) % MOD return ret ######Factorization###### def factorization(n): arr = [] temp = n for i in range(2, int(-(-n**0.5//1))+1): if temp%i==0: cnt=0 while temp%i==0: cnt+=1 temp //= i arr.append([i, cnt]) if temp!=1: arr.append([temp, 1]) if arr==[]: arr.append([n, 1]) return arr #####MakeDivisors###### def make_divisors(n): divisors = [] for i in range(1, int(n**0.5)+1): if n % i == 0: divisors.append(i) if i != n // i: divisors.append(n//i) return divisors #####MakePrimes###### def make_primes(N): max = int(math.sqrt(N)) seachList = [i for i in range(2,N+1)] primeNum = [] while seachList[0] <= max: primeNum.append(seachList[0]) tmp = seachList[0] seachList = [i for i in seachList if i % tmp != 0] primeNum.extend(seachList) return primeNum #####GCD##### def gcd(a, b): while b: a, b = b, a % b return a #####LCM##### def lcm(a, b): return a * b // gcd (a, b) #####BitCount##### def count_bit(n): count = 0 while n: n &= n-1 count += 1 return count #####ChangeBase##### def base_10_to_n(X, n): if X//n: return base_10_to_n(X//n, n)+[X%n] return [X%n] def base_n_to_10(X, n): return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X)))) def base_10_to_n_without_0(X, n): X -= 1 if X//n: return base_10_to_n_without_0(X//n, n)+[X%n] return [X%n] #####IntLog##### def int_log(n, a): count = 0 while n>=a: n //= a count += 1 return count ############# # Main Code # ############# N,M,L,R = IL() M = max(M,N-M) @lru_cache(maxsize=None) def P(t): ans = 0 for k in range(t//M+1): u = t-M*(k+1) if u>=0: if u%(k+1) == 0: ans += nCr(N-M,u//(k+1))*(-1)**((u//(k+1))%2) if k!=0: if u%k == 0: ans -= nCr(N-M,u//k)*(-1)**((u//k)%2) ans %= MOD return ans @lru_cache(maxsize=None) def Q(t): ans = nCr(t+N,N) return ans def f(MAX): ans = 0 for t in range(MAX+1): ans += P(t)*Q(MAX-t) ans %= MOD return ans def F(x): return (Q(x)-nCr(N,M)*f(x))%MOD print((F(R)-F(L-1))%MOD) ```
instruction
0
93,306
5
186,612
No
output
1
93,306
5
186,613
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,339
5
186,678
"Correct Solution: ``` l = list(map(int,input().split())) k = int(input()) print(sum(l)+max(l)*(2**k-1)) ```
output
1
93,339
5
186,679
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,340
5
186,680
"Correct Solution: ``` l=list(map(int,input().split())) n=int(input()) print(sum(l)+max(l)*((2**n)-1)) ```
output
1
93,340
5
186,681
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,341
5
186,682
"Correct Solution: ``` A = [int(x) for x in input().split()] K = int(input()) print(sum(A)+max(A)*(2**K-1)) ```
output
1
93,341
5
186,683
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,342
5
186,684
"Correct Solution: ``` num=list(map(int,input().split())) print(sum(num)-max(num)+max(num)*2**int(input())) ```
output
1
93,342
5
186,685
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,343
5
186,686
"Correct Solution: ``` a,b,c = sorted(map(int,input().split())) k = int(input()) print(a + b + c*(2**k)) ```
output
1
93,343
5
186,687
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,344
5
186,688
"Correct Solution: ``` a,b,c = map(int,input().split()) k = int(input()) x = max(a,b,c) print(a+b+c+x*2**k-x) ```
output
1
93,344
5
186,689
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,345
5
186,690
"Correct Solution: ``` x=[int(i) for i in input().split()] x.sort() n=int(input()) print(x[2]*2**n+x[0]+x[1]) ```
output
1
93,345
5
186,691
Provide a correct Python 3 solution for this coding contest problem. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22
instruction
0
93,346
5
186,692
"Correct Solution: ``` A,B,C=map(int, input().split()) K=int(input()) a=max(A,B,C) b=A+B+C-a print(b+a*(2**K)) ```
output
1
93,346
5
186,693
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` L=sorted(list(map(int,input().split()))) k=int(input()) L[-1]=L[-1]*(2**k) print(sum(L)) ```
instruction
0
93,347
5
186,694
Yes
output
1
93,347
5
186,695
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` L=list(map(int,input().split())) K=int(input()) N=sorted(L) print(N[0]+N[1]+N[2]*(2**K)) ```
instruction
0
93,348
5
186,696
Yes
output
1
93,348
5
186,697
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` a, b, c, K = map(int, open(0).read().split()) m = max(a, b, c) print(m*2**K+a+b+c-m) ```
instruction
0
93,349
5
186,698
Yes
output
1
93,349
5
186,699
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` a,b,c = map(int,input().split()) k = int(input()) print(max(a,b,c)*(2**k-1)+a+b+c) ```
instruction
0
93,350
5
186,700
Yes
output
1
93,350
5
186,701
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` a,b,c = map(int,input().split()) k = int(input()) m = a count = 0 if b>m: m=b count=1 if c>m: m=c count=2 if count==0: print(a**k+b+c) if count==1: print(a+b**k+c) if count==2: print(a+b+c**k) ```
instruction
0
93,351
5
186,702
No
output
1
93,351
5
186,703
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` n = list(map(int,input().split())) k = int(input()) n.sort(reverse=True) print(n[0]**k+n[1]+n[2]) ```
instruction
0
93,352
5
186,704
No
output
1
93,352
5
186,705
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` A, B, C = map(int, input().split()) K = int(input()) print(max(A, B, C)*(2**K)+B+C) ```
instruction
0
93,353
5
186,706
No
output
1
93,353
5
186,707
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? Constraints * A, B and C are integers between 1 and 50 (inclusive). * K is an integer between 1 and 10 (inclusive). Input Input is given from Standard Input in the following format: A B C K Output Print the largest possible sum of the integers written on the blackboard after K operations by E869220. Examples Input 5 3 11 1 Output 30 Input 3 3 4 2 Output 22 Submitted Solution: ``` a, b, c = map(int, input().split()) k = int(input()) m = max(a, b, c) print(sum([a, b, c])-m+m*2*k) ```
instruction
0
93,354
5
186,708
No
output
1
93,354
5
186,709
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,512
5
187,024
"Correct Solution: ``` # -*- coding: utf-8 -*- """ Dictionary - Map: Range Search http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP2_8_C&lang=jp """ from bisect import insort, bisect_right, bisect_left class Range_map: def __init__(self): self.rm = dict() self.lr = [] def insert(self, x, y): if x not in self.rm: insort(self.lr, x) self.rm[x] = y def get(self, x): print(self.rm.get(x, 0)) def delete(self, x): if x in self.rm: self.rm[x] = 0 def dump(self, l, r): lb = bisect_left(self.lr, l) ub = bisect_right(self.lr, r) for i in range(lb, ub): k = self.lr[i] if k in self.rm and self.rm[k] != 0: print(f'{k} {self.rm[k]}') rm = Range_map() for _ in range(int(input())): op, x, y = (input() + ' 1').split()[:3] if op == '0': rm.insert(x, int(y)) elif op == '1': rm.get(x) elif op == '2': rm.delete(x) else: rm.dump(x, y) ```
output
1
93,512
5
187,025
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,513
5
187,026
"Correct Solution: ``` import bisect keys=[] q=int(input()) d={} def find(x): index=bisect.bisect_left(keys,x) if index==len(keys): return -1 if keys[index]==x: return index else: return -1 for _ in range(q): query=input().split(" ") if query[0]=="0": if find(query[1])!=-1: pass else: bisect.insort_left(keys, query[1]) d[query[1]]=query[2] elif query[0]=="1": if find(query[1])!=-1: print(d[query[1]]) else: print(0) elif query[0]=="2": while find(query[1])>=0: keys.pop(find(query[1])) else: L,R=query[1],query[2] indL=bisect.bisect_left(keys,L) indR=bisect.bisect_right(keys,R) for s in keys[indL:indR]: print(s,d[s]) ```
output
1
93,513
5
187,027
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,514
5
187,028
"Correct Solution: ``` # AOJ ITP2_8_C: Map: Range Search # Python3 2018.6.24 bal4u from bisect import bisect_left, bisect_right, insort_left dict = {} keytbl = [] q = int(input()) for i in range(q): a = list(input().split()) ki = a[1] if a[0] == '0': if ki not in dict: insort_left(keytbl, ki) dict[ki] = int(a[2]) elif a[0] == '1': print(dict[ki] if ki in dict else 0) elif a[0] == '2': if ki in dict: dict[ki] = 0 else: L = bisect_left (keytbl, a[1]) R = bisect_right(keytbl, a[2], L) for j in range(L, R): if dict[keytbl[j]] > 0: print(keytbl[j], dict[keytbl[j]]) ```
output
1
93,514
5
187,029
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,515
5
187,030
"Correct Solution: ``` from bisect import * import sys n = int(input()) dic = {} dic2 = {} box = [] for i in range(n): a, b, *c = sys.stdin.readline().split() if a == '0': insort_left(box, b) dic[b] = int(c[0]) elif a == '1': print(dic[b] if b in dic else 0) elif a == '2': if b in dic: del dic[b] else: L = bisect_left(box,b) R = bisect_right(box, c[0]) while L<R: if box[L] in dic: print(box[L], dic[box[L]]) L = bisect_right(box, box[L]) ```
output
1
93,515
5
187,031
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,516
5
187,032
"Correct Solution: ``` def resolve(): import sys import bisect input = sys.stdin.readline n = int(input()) ans = dict() need_sort = True sorted_keys = None for _ in range(n): q, key, *x = input().split() if q == "0": if key in ans: ans[key] = x[0] else: ans[key] = x[0] need_sort = True elif q == "1": if key in ans: print(ans[key]) else: print(0) elif q == "2": ans[key] = "0" else: if sorted_keys is None: sorted_keys = sorted(ans.keys()) if need_sort: sorted_keys = sorted(ans.keys()) need_sort = False l, r = key, x[0] fr = bisect.bisect_left(sorted_keys, l) to = bisect.bisect_right(sorted_keys, r) for i in range(fr, to): if ans[sorted_keys[i]] != "0": print(sorted_keys[i], ans[sorted_keys[i]]) resolve() ```
output
1
93,516
5
187,033
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,517
5
187,034
"Correct Solution: ``` import sys import bisect n = int(input()) arr = [] lines = sys.stdin.readlines() ans = [None] * n d = {} for i in range(n): q, *arg = lines[i].split() key = arg[0] idx = bisect.bisect_left(arr, arg[0]) if q == '0': # insert if idx == len(arr) or arr[idx] != key: arr.insert(idx, key) d[key] = arg[1] elif q == '1': # get ans[i] = d[key] if idx != len(arr) and arr[idx] == key else '0' elif q == '2': # delete if idx < len(arr) and arr[idx] == key: del arr[idx] del d[key] elif q == '3': # dump L R r_idx = bisect.bisect_right(arr, arg[1]) ans[i] = '\n'.join([f'{key_} {d[key_]}' for key_ in arr[idx:r_idx]]) if (r_idx - idx) != 0 else None [print(x) for x in ans if x is not None] ```
output
1
93,517
5
187,035
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,518
5
187,036
"Correct Solution: ``` from bisect import bisect_left, bisect_right, insort_left dict = {} keytbl = [] q = int(input()) for i in range(q): a = list(input().split()) ki = a[1] if a[0] == '0': if ki not in dict: insort_left(keytbl, ki) dict[ki] = int(a[2]) elif a[0] == '1': print(dict[ki] if ki in dict else 0) elif a[0] == '2': if ki in dict: dict[ki] = 0 else: L = bisect_left (keytbl, a[1]) R = bisect_right(keytbl, a[2], L) for j in range(L, R): if dict[keytbl[j]] > 0: print(keytbl[j], dict[keytbl[j]]) ```
output
1
93,518
5
187,037
Provide a correct Python 3 solution for this coding contest problem. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5
instruction
0
93,519
5
187,038
"Correct Solution: ``` from bisect import bisect,bisect_left,insort dic = {} l = [] for i in range(int(input())): order = list(input().split()) if order[0] == "0": if order[1] in dic: dic[order[1]] = int(order[2]) else: dic[order[1]] = int(order[2]) insort(l,order[1]) elif order[0] == "1": if order[1] in dic: print(dic[order[1]]) else: print(0) elif order[0] == "2": if order[1] in dic: dic[order[1]] = 0 elif order[0] == "3": L = bisect_left(l,order[1]) R = bisect(l,order[2]) for i in range(L,R): if dic[l[i]]: print(l[i],dic[l[i]]) ```
output
1
93,519
5
187,039
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` import bisect q = int(input()) M = {} sortedList = [] for i in range(q): query, *inp = input().split() key = inp[0] # insert if query == "0": x = int(inp[1]) if key not in M: bisect.insort_left(sortedList, key) M[key] = x # get elif query == "1": if key in M: if M[key] != None: print(M[key]) else: print(0) else: print(0) # delete elif query == "2": if key in M: M[key] = None # dump else: L = inp[0] R = inp[1] index_left = bisect.bisect_left(sortedList, L) index_right = bisect.bisect_right(sortedList, R) for i in range(index_left, index_right): key_ans = sortedList[i] if M[key_ans] != None: print(key_ans, M[key_ans]) ```
instruction
0
93,520
5
187,040
Yes
output
1
93,520
5
187,041
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` import bisect q = int(input()) M = {} list_key_sorted = [] for _ in range(q): command, *list_num = input().split() if command == "0": # insert(key, x) key = list_num[0] x = int(list_num[1]) if key not in M: bisect.insort_left(list_key_sorted, key) M[key] = x elif command == "1": # get(key) key = list_num[0] if key in M: if M[key] == None: print(0) else: print(M[key]) else: print(0) elif command == "2": # delete(key) key = list_num[0] if key in M: M[key] = None elif command == "3": # dump(L, R) L = list_num[0] R = list_num[1] index_left = bisect.bisect_left(list_key_sorted, L) index_right = bisect.bisect_right(list_key_sorted, R) for index in range(index_left, index_right): key_ans = list_key_sorted[index] if M[key_ans] != None: print(key_ans, M[key_ans]) else: raise ```
instruction
0
93,521
5
187,042
Yes
output
1
93,521
5
187,043
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from enum import Enum class Color(Enum): BLACK = 0 RED = 1 @staticmethod def flip(c): return [Color.RED, Color.BLACK][c.value] class Node: __slots__ = ('key', 'left', 'right', 'size', 'color', 'value') def __init__(self, key, value): self.key = key self.value = value self.left = Leaf self.right = Leaf self.size = 1 self.color = Color.RED def is_red(self): return self.color == Color.RED def is_black(self): return self.color == Color.BLACK def __str__(self): if self.color == Color.RED: key = '*{}'.format(self.key) else: key = '{}'.format(self.key) return "{}[{}] ({}, {})".format(key, self.size, self.left, self.right) class LeafNode(Node): def __init__(self): self.key = None self.value = None self.left = None self.right = None self.size = 0 self.color = None def is_red(self): return False def is_black(self): return False def __str__(self): return '-' Leaf = LeafNode() class RedBlackBinarySearchTree: """Red Black Binary Search Tree with range, min, max. Originally impremented in the textbook Algorithms, 4th edition by Robert Sedgewick and Kevin Wayne, Addison-Wesley Professional, 2011, ISBN 0-321-57351-X. """ def __init__(self): self.root = Leaf def put(self, key, value=None): def _put(node): if node is Leaf: node = Node(key, value) if node.key > key: node.left = _put(node.left) elif node.key < key: node.right = _put(node.right) else: node.value = value node = self._restore(node) node.size = node.left.size + node.right.size + 1 return node self.root = _put(self.root) self.root.color = Color.BLACK def _rotate_left(self, node): x = node.right node.right = x.left x.left = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _rotate_right(self, node): x = node.left node.left = x.right x.right = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _flip_colors(self, node): node.color = Color.flip(node.color) node.left.color = Color.flip(node.left.color) node.right.color = Color.flip(node.right.color) return node def __contains__(self, key): def _contains(node): if node is Leaf: return False if node.key > key: return _contains(node.left) elif node.key < key: return _contains(node.right) else: return True return _contains(self.root) def get(self, key): def _get(node): if node is Leaf: return None if node.key > key: return _get(node.left) elif node.key < key: return _get(node.right) else: return node.value return _get(self.root) def delete(self, key): def _delete(node): if node is Leaf: return Leaf if node.key > key: if node.left is Leaf: return self._balance(node) if not self._is_red_left(node): node = self._red_left(node) node.left = _delete(node.left) else: if node.left.is_red(): node = self._rotate_right(node) if node.key == key and node.right is Leaf: return Leaf elif node.right is Leaf: return self._balance(node) if not self._is_red_right(node): node = self._red_right(node) if node.key == key: x = self._find_min(node.right) node.key = x.key node.value = x.value node.right = self._delete_min(node.right) else: node.right = _delete(node.right) return self._balance(node) if self.is_empty(): raise ValueError('delete on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = _delete(self.root) if not self.is_empty(): self.root.color = Color.BLACK def delete_max(self): if self.is_empty(): raise ValueError('delete max on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_max(self.root) if not self.is_empty(): self.root.color = Color.BLACK def _delete_max(self, node): if node.left.is_red(): node = self._rotate_right(node) if node.right is Leaf: return Leaf if not self._is_red_right(node): node = self._red_right(node) node.right = self._delete_max(node.right) return self._balance(node) def _red_right(self, node): node = self._flip_colors(node) if node.left.left.is_red(): node = self._rotate_right(node) return node def _is_red_right(self, node): return (node.right.is_red() or (node.right.is_black() and node.right.left.is_red())) def delete_min(self): if self.is_empty(): raise ValueError('delete min on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_min(self.root) if not self.is_empty(): self.root.color = Color.BLACK def _delete_min(self, node): if node.left is Leaf: return Leaf if not self._is_red_left(node): node = self._red_left(node) node.left = self._delete_min(node.left) return self._balance(node) def _red_left(self, node): node = self._flip_colors(node) if node.right.left.is_red(): node.right = self._rotate_right(node.right) node = self._rotate_left(node) return node def _is_red_left(self, node): return (node.left.is_red() or (node.left.is_black() and node.left.left.is_red())) def _balance(self, node): if node.right.is_red(): node = self._rotate_left(node) return self._restore(node) def _restore(self, node): if node.right.is_red() and not node.left.is_red(): node = self._rotate_left(node) if node.left.is_red() and node.left.left.is_red(): node = self._rotate_right(node) if node.left.is_red() and node.right.is_red(): node = self._flip_colors(node) node.size = node.left.size + node.right.size + 1 return node def is_empty(self): return self.root is Leaf def is_balanced(self): if self.is_empty(): return True try: left = self._depth(self.root.left) right = self._depth(self.root.right) return left == right except Exception: return False @property def depth(self): return self._depth(self.root) def _depth(self, node): if node is Leaf: return 0 if node.right.is_red(): raise Exception('right red') left = self._depth(node.left) right = self._depth(node.right) if left != right: raise Exception('unbalanced') if node.is_red(): return left else: return 1 + left def __len__(self): return self.root.size @property def max(self): if self.is_empty(): raise ValueError('max on empty tree') return self._max(self.root) def _max(self, node): x = self._find_max(node) return x.key def _find_max(self, node): if node.right is Leaf: return node else: return self._find_max(node.right) @property def min(self): if self.is_empty(): raise ValueError('min on empty tree') return self._min(self.root) def _min(self, node): x = self._find_min(node) return x.key def _find_min(self, node): if node.left is Leaf: return node else: return self._find_min(node.left) def __iter__(self): def inorder(node): if node is Leaf: return yield from inorder(node.left) yield node.value yield from inorder(node.right) yield from inorder(self.root) def range(self, min_, max_): def _range(node): if node is Leaf: return if node.key > max_: yield from _range(node.left) elif node.key < min_: yield from _range(node.right) else: yield from _range(node.left) yield (node.key, node.value) yield from _range(node.right) if min_ > max_: return yield from _range(self.root) class Map: def __init__(self): self.tree = RedBlackBinarySearchTree() def __getitem__(self, key): if key in self.tree: return self.tree.get(key) else: raise IndexError('key {} not found in map'.format(key)) def __setitem__(self, key, value): self.tree.put(key, value) def __delitem__(self, key): if key in self.tree: self.tree.delete(key) else: raise IndexError('key {} not found in map'.format(key)) def __len__(self): return len(self.tree) def items(self): for k, v in self.tree: yield (k, v) def range(self, min_, max_): for k, v in self.tree.range(min_, max_): yield (k, v) def run(): q = int(input()) m = Map() for _ in range(q): command, *args = input().split() if command == '0': key = args[0] value = int(args[1]) m[key] = value elif command == '1': key = args[0] try: print(m[key]) except IndexError: print(0) elif command == '2': key = args[0] try: del m[key] except IndexError: pass elif command == '3': low, hi = args for k, v in m.range(low, hi): print(k, v) else: raise ValueError('invalid command') if __name__ == '__main__': run() ```
instruction
0
93,522
5
187,044
Yes
output
1
93,522
5
187,045
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from bisect import bisect_left, bisect_right, insort_left q = int(input()) M = {} tbl = [] for i in range(q): query, *x = input().split() if query == "0": x[1] = int(x[1]) if x[0] not in M: insort_left(tbl, x[0]) M[x[0]] = x[1] elif query == "1": print(M[x[0]] if x[0] in M and M[x[0]] != 0 else 0) elif query == "2": M[x[0]] = 0 else: L = bisect_left( tbl, x[0]) R = bisect_right(tbl, x[1]) for i in range(L, R): if M[tbl[i]] != 0: print(tbl[i], M[tbl[i]]) ```
instruction
0
93,523
5
187,046
Yes
output
1
93,523
5
187,047
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from enum import Enum class Color(Enum): BLACK = 0 RED = 1 @staticmethod def flip(c): return [Color.RED, Color.BLACK][c.value] class Node: __slots__ = ('key', 'left', 'right', 'size', 'color', 'value') def __init__(self, key, value): self.key = key self.value = value self.left = Leaf self.right = Leaf self.size = 1 self.color = Color.RED def is_red(self): return self.color == Color.RED def is_black(self): return self.color == Color.BLACK def __str__(self): if self.color == Color.RED: key = '*{}'.format(self.key) else: key = '{}'.format(self.key) return "{}[{}] ({}, {})".format(key, self.size, self.left, self.right) class LeafNode(Node): def __init__(self): self.key = None self.value = None self.left = None self.right = None self.size = 0 self.color = None def is_red(self): return False def is_black(self): return False def __str__(self): return '-' Leaf = LeafNode() class RedBlackBinarySearchTree: """Red Black Binary Search Tree with range, min, max. Originally impremented in the textbook Algorithms, 4th edition by Robert Sedgewick and Kevin Wayne, Addison-Wesley Professional, 2011, ISBN 0-321-57351-X. """ def __init__(self): self.root = Leaf def put(self, key, value=None): def _put(node): if node is Leaf: node = Node(key, value) if node.key > key: node.left = _put(node.left) elif node.key < key: node.right = _put(node.right) else: node.value = value node = self._restore(node) node.size = node.left.size + node.right.size + 1 return node self.root = _put(self.root) self.root.color = Color.BLACK def _rotate_left(self, node): assert node.right.is_red() x = node.right node.right = x.left x.left = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _rotate_right(self, node): assert node.left.is_red() x = node.left node.left = x.right x.right = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _flip_colors(self, node): node.color = Color.flip(node.color) node.left.color = Color.flip(node.left.color) node.right.color = Color.flip(node.right.color) return node def __contains__(self, key): def _contains(node): if node is Leaf: return False if node.key > key: return _contains(node.left) elif node.key < key: return _contains(node.right) else: return True return _contains(self.root) def get(self, key): def _get(node): if node is Leaf: return None if node.key > key: return _get(node.left) elif node.key < key: return _get(node.right) else: return node.value return _get(self.root) def delete(self, key): def _delete(node): if node is Leaf: return Leaf if node.key > key: if node.left is Leaf: return self._balance(node) if not self._is_red_left(node): node = self._red_left(node) node.left = _delete(node.left) else: if node.left.is_red(): node = self._rotate_right(node) if node.key == key and node.right is Leaf: return Leaf elif node.right is Leaf: return self._balance(node) if not self._is_red_right(node): node = self._red_right(node) if node.key == key: x = self._find_min(node.right) node.key = x.key node.value = x.value node.right = self._delete_min(node.right) else: node.right = _delete(node.right) if self.is_empty(): raise ValueError('delete on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = _delete(self.root) if not self.is_empty(): self.root.color = Color.BLACK def delete_max(self): if self.is_empty(): raise ValueError('delete max on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_max(self.root) if not self.is_empty(): self.root.color = Color.BLACK def _delete_max(self, node): if node.left.is_red(): node = self._rotate_right(node) if node.right is Leaf: return Leaf if not self._is_red_right(node): node = self._red_right(node) node.right = self._delete_max(node.right) def _red_right(self, node): node = self._flip_colors(node) if node.left.left.is_red(): node = self._rotate_right(node) return node def _is_red_right(self, node): return (node.right.is_red() or (node.right.is_black() and node.right.left.is_red())) def delete_min(self): if self.is_empty(): raise ValueError('delete min on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_min(self.root) if not self.is_empty(): self.root.color = Color.BLACK def _delete_min(self, node): if node.left is Leaf: return Leaf if not self._is_red_left(node): node = self._red_left(node) node.left = self._delete_min(node.left) return self._balance(node) def _red_left(self, node): node = self._flip_colors(node) if node.right.left.is_red(): node.right = self._rotate_right(node.right) node = self._rotate_left(node) return node def _is_red_left(self, node): return (node.left.is_red() or (node.left.is_black() and node.left.left.is_red())) def _balance(self, node): if node.right.is_red(): node = self._rotate_left(node) return self._restore(node) def _restore(self, node): if node.right.is_red() and not node.left.is_red(): node = self._rotate_left(node) if node.left.is_red() and node.left.left.is_red(): node = self._rotate_right(node) if node.left.is_red() and node.right.is_red(): node = self._flip_colors(node) node.size = node.left.size + node.right.size + 1 return node def is_empty(self): return self.root is Leaf def is_balanced(self): if self.is_empty(): return True try: left = self._depth(self.root.left) right = self._depth(self.root.right) return left == right except Exception: return False @property def depth(self): return self._depth(self.root) def _depth(self, node): if node is Leaf: return 0 if node.right.is_red(): raise Exception('right red') left = self._depth(node.left) right = self._depth(node.right) if left != right: raise Exception('unbalanced') if node.is_red(): return left else: return 1 + left def __len__(self): return self.root.size @property def max(self): if self.is_empty(): raise ValueError('max on empty tree') return self._max(self.root) def _max(self, node): x = self._find_max(node) return x.key def _find_max(self, node): if node.right is Leaf: return node else: return self._find_max(node.right) @property def min(self): if self.is_empty(): raise ValueError('min on empty tree') return self._min(self.root) def _min(self, node): x = self._find_min(node) return x.key def _find_min(self, node): if node.left is Leaf: return node else: return self._find_min(node.left) def __iter__(self): def inorder(node): if node is Leaf: return yield from inorder(node.left) yield node.value yield from inorder(node.right) yield from inorder(self.root) def range(self, min_, max_): def _range(node): if node is Leaf: return if node.key > max_: yield from _range(node.left) elif node.key < min_: yield from _range(node.right) else: yield from _range(node.left) yield (node.key, node.value) yield from _range(node.right) if min_ > max_: return yield from _range(self.root) class Map: def __init__(self): self.tree = RedBlackBinarySearchTree() def __getitem__(self, key): if key in self.tree: return self.tree.get(key) else: raise IndexError('key {} not found in map'.format(key)) def __setitem__(self, key, value): self.tree.put(key, value) def __delitem__(self, key): if key in self.tree: self.tree.delete(key) else: raise IndexError('key {} not found in map'.format(key)) def __len__(self): return len(self.tree) def items(self): for k, v in self.tree: yield (k, v) def range(self, min_, max_): for k, v in self.tree.range(min_, max_): yield (k, v) def run(): q = int(input()) m = Map() for _ in range(q): command, *args = input().split() if command == '0': key = args[0] value = int(args[1]) m[key] = value elif command == '1': key = args[0] try: print(m[key]) except IndexError: print(0) elif command == '2': key = args[0] try: del m[key] except IndexError: pass elif command == '3': low, hi = args for k, v in m.range(low, hi): print(k, v) else: raise ValueError('invalid command') if __name__ == '__main__': run() ```
instruction
0
93,524
5
187,048
No
output
1
93,524
5
187,049
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from collections import OrderedDict dict = OrderedDict() q = int(input()) for i in range(q): query, *val = input().split(' ') if query == '0': dict[val[0]] = int(val[1]) elif query == '1': print(dict.get(val[0], 0)) elif query == '2': if val[0] in dict: dict.pop(val[0]) else: ans = '' for k, v in dict.items(): if val[0] <= k and k <= val[1]: ans += str(k) + ' ' + str(v) + ' ' print(ans.strip()) ```
instruction
0
93,525
5
187,050
No
output
1
93,525
5
187,051
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from enum import Enum class Color(Enum): BLACK = 0 RED = 1 @staticmethod def flip(c): return [Color.RED, Color.BLACK][c.value] class Node: __slots__ = ('key', 'left', 'right', 'size', 'color', 'value') def __init__(self, key, value): self.key = key self.value = value self.left = Leaf self.right = Leaf self.size = 1 self.color = Color.RED def is_red(self): return self.color == Color.RED def is_black(self): return self.color == Color.BLACK def __str__(self): if self.color == Color.RED: key = '*{}'.format(self.key) else: key = '{}'.format(self.key) return "{}[{}] ({}, {})".format(key, self.size, self.left, self.right) class LeafNode(Node): def __init__(self): self.key = None self.value = None self.left = None self.right = None self.size = 0 self.color = None def is_red(self): return False def is_black(self): return False def __str__(self): return '-' Leaf = LeafNode() class RedBlackBinarySearchTree: """Red Black Binary Search Tree with range, min, max. Originally impremented in the textbook Algorithms, 4th edition by Robert Sedgewick and Kevin Wayne, Addison-Wesley Professional, 2011, ISBN 0-321-57351-X. """ def __init__(self): self.root = Leaf def put(self, key, value=None): def _put(node): if node is Leaf: node = Node(key, value) if node.key > key: node.left = _put(node.left) elif node.key < key: node.right = _put(node.right) else: node.value = value node = self._restore(node) node.size = node.left.size + node.right.size + 1 return node self.root = _put(self.root) self.root.color = Color.BLACK def _rotate_left(self, node): assert node.right.is_red() x = node.right node.right = x.left x.left = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _rotate_right(self, node): assert node.left.is_red() x = node.left node.left = x.right x.right = node x.color = node.color node.color = Color.RED node.size = node.left.size + node.right.size + 1 return x def _flip_colors(self, node): node.color = Color.flip(node.color) node.left.color = Color.flip(node.left.color) node.right.color = Color.flip(node.right.color) return node def __contains__(self, key): def _contains(node): if node is Leaf: return False if node.key > key: return _contains(node.left) elif node.key < key: return _contains(node.right) else: return True return _contains(self.root) def get(self, key): def _get(node): if node is Leaf: return None if node.key > key: return _get(node.left) elif node.key < key: return _get(node.right) else: return node.value return _get(self.root) def delete(self, key): def _delete(node): if node is Leaf: return Leaf if node.key > key: if node.left is Leaf: return self._balance(node) if not self._is_red_left(node): node = self._red_left(node) node.left = _delete(node.left) else: if node.left.is_red(): node = self._rotate_right(node) if node.key == key and node.right is Leaf: return Leaf elif node.right is Leaf: return self._balance(node) if not self._is_red_right(node): node = self._red_right(node) if node.key == key: x = self._find_min(node.right) node.key = x.key node.value = x.value node.right = self._delete_min(node.right) else: node.right = _delete(node.right) return self._balance(node) if self.is_empty(): raise ValueError('delete on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = _delete(self.root) if not self.is_empty(): self.root.color = Color.BLACK assert self.is_balanced() def delete_max(self): if self.is_empty(): raise ValueError('delete max on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_max(self.root) if not self.is_empty(): self.root.color = Color.BLACK assert self.is_balanced() def _delete_max(self, node): if node.left.is_red(): node = self._rotate_right(node) if node.right is Leaf: return Leaf if not self._is_red_right(node): node = self._red_right(node) node.right = self._delete_max(node.right) return self._balance(node) def _red_right(self, node): node = self._flip_colors(node) if node.left.left.is_red(): node = self._rotate_right(node) return node def _is_red_right(self, node): return (node.right.is_red() or (node.right.is_black() and node.right.left.is_red())) def delete_min(self): if self.is_empty(): raise ValueError('delete min on empty tree') if not self.root.left.is_red() and not self.root.right.is_red(): self.root.color = Color.RED self.root = self._delete_min(self.root) if not self.is_empty(): self.root.color = Color.BLACK assert self.is_balanced() def _delete_min(self, node): if node.left is Leaf: return Leaf if not self._is_red_left(node): node = self._red_left(node) node.left = self._delete_min(node.left) return self._balance(node) def _red_left(self, node): node = self._flip_colors(node) if node.right.left.is_red(): node.right = self._rotate_right(node.right) node = self._rotate_left(node) return node def _is_red_left(self, node): return (node.left.is_red() or (node.left.is_black() and node.left.left.is_red())) def _balance(self, node): if node.right.is_red(): node = self._rotate_left(node) return self._restore(node) def _restore(self, node): if node.right.is_red() and not node.left.is_red(): node = self._rotate_left(node) if node.left.is_red() and node.left.left.is_red(): node = self._rotate_right(node) if node.left.is_red() and node.right.is_red(): node = self._flip_colors(node) node.size = node.left.size + node.right.size + 1 return node def is_empty(self): return self.root is Leaf def is_balanced(self): if self.is_empty(): return True try: left = self._depth(self.root.left) right = self._depth(self.root.right) return left == right except Exception: return False @property def depth(self): return self._depth(self.root) def _depth(self, node): if node is Leaf: return 0 if node.right.is_red(): raise Exception('right red') left = self._depth(node.left) right = self._depth(node.right) if left != right: raise Exception('unbalanced') if node.is_red(): return left else: return 1 + left def __len__(self): return self.root.size @property def max(self): if self.is_empty(): raise ValueError('max on empty tree') return self._max(self.root) def _max(self, node): x = self._find_max(node) return x.key def _find_max(self, node): if node.right is Leaf: return node else: return self._find_max(node.right) @property def min(self): if self.is_empty(): raise ValueError('min on empty tree') return self._min(self.root) def _min(self, node): x = self._find_min(node) return x.key def _find_min(self, node): if node.left is Leaf: return node else: return self._find_min(node.left) def __iter__(self): def inorder(node): if node is Leaf: return yield from inorder(node.left) yield node.value yield from inorder(node.right) yield from inorder(self.root) def range(self, min_, max_): def _range(node): if node is Leaf: return if node.key > max_: yield from _range(node.left) elif node.key < min_: yield from _range(node.right) else: yield from _range(node.left) yield (node.key, node.value) yield from _range(node.right) if min_ > max_: return yield from _range(self.root) class Map: def __init__(self): self.tree = RedBlackBinarySearchTree() def __getitem__(self, key): if key in self.tree: return self.tree.get(key) else: raise IndexError('key {} not found in map'.format(key)) def __setitem__(self, key, value): self.tree.put(key, value) def __delitem__(self, key): if key in self.tree: self.tree.delete(key) else: raise IndexError('key {} not found in map'.format(key)) def __len__(self): return len(self.tree) def items(self): for k, v in self.tree: yield (k, v) def range(self, min_, max_): for k, v in self.tree.range(min_, max_): yield (k, v) def run(): q = int(input()) m = Map() for _ in range(q): command, *args = input().split() if command == '0': key = args[0] value = int(args[1]) m[key] = value elif command == '1': key = args[0] try: print(m[key]) except IndexError: print(0) elif command == '2': key = args[0] try: del m[key] except IndexError: pass elif command == '3': low, hi = args for k, v in m.range(low, hi): print(k, v) else: raise ValueError('invalid command') if __name__ == '__main__': run() ```
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that each key in $M$ must be unique. * insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$. * get($key$): Print the value with the specified $key$. Print 0 if there is no such element. * delete($key$): Delete the element with the specified $key$. * dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order. Constraints * $1 \leq q \leq 200,000$ * $1 \leq x \leq 1,000,000,000$ * $1 \leq $ length of $key$ $ \leq 20$ * $key$ consists of lower-case letters * $L \leq R$ in lexicographic order * The total number of elements printed by dump operations does not exceed $1,000,000$ Input The input is given in the following format. $q$ $query_1$ $query_2$ : $query_q$ Each query $query_i$ is given by 0 $key$ $x$ or 1 $key$ or 2 $key$ or 3 $L$ $R$ where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations. Output For each get operation, print the corresponding value. For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements (a pair of key and value separated by a space character) in ascending order of the keys. Example Input 9 0 blue 4 0 red 1 0 white 5 1 red 1 blue 2 red 1 black 1 red 3 w z Output 1 4 0 0 white 5 Submitted Solution: ``` from collections import OrderedDict dict = OrderedDict() q = int(input()) for i in range(q): query, *val = input().split(' ') if query == '0': dict[val[0]] = int(val[1]) elif query == '1': print(dict.get(val[0], 0)) elif query == '2': if val[0] in dict: dict.pop(val[0]) else: ans = [] for k, v in dict.items(): if val[0] <= k and k <= val[1]: print(k, v, end=' ') print() ```
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No
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n = int(input()) arr = list(map(int, input().split())) temp = 0 for i in arr: print(temp + i, end=" ") if(i > 0): temp += i ```
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Yes
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93,652
5
187,305
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n = int(input()) b=[] a=[0]*n x=[0]*n b = [int(i) for i in input().split()] #print(x) for i in range(1,n): if i<=(n-1): a[i-1]=b[i-1]+x[i-1] x[i]=max(a[i-1],x[i-1]) a[n-1]=b[n-1]+x[n-1] for i in range(n): print(a[i], end=" ") ```
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Yes
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5
187,307
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` length = int(input()) b = list(map(int, input().split())) x = 0 for i in range(length): a = b[i] + x print(a, end=' ') x = max(a, x) ```
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Yes
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187,309
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n=int(input()) b=list(map(int,input().split())) a=[b[0],b[1]+b[0]] ma=max(a) for x in range(2,n): num=b[x]+ma a.append(num) if num>ma:ma=num print(*a) ```
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Yes
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` N = int(input()) b = list(map(int,input().split())) largest = 0 for i in range(N): temp = b[i] b[i]+=largest if b[i]>0: largest+=temp for i in b: print(i,end=' ') ```
instruction
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No
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5
187,313
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n=int(input()) a=[int(i) for i in input().split()] b=[] c=[] for i in range(n): if(i==0): b.append(a[i]) c.append(a[i]) else: if(a[i]>0): b.append(a[i]+c[i-1]) c.append(a[i]+c[i-1]) else: b.append(0) c.append(abs(a[i])) print(" ".join(str(i) for i in b)) ```
instruction
0
93,657
5
187,314
No
output
1
93,657
5
187,315
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n = int(input()) b = [int(i) for i in input().split()] a = [b[0]] print(b[0],end=' ') ma = a[0] for i in range(1, n): a.append(b[i] + ma) ma = max(ma, a[-1]) print(b[i]+ma) ```
instruction
0
93,658
5
187,316
No
output
1
93,658
5
187,317
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Alicia has an array, a_1, a_2, …, a_n, of non-negative integers. For each 1 ≤ i ≤ n, she has found a non-negative integer x_i = max(0, a_1, …, a_{i-1}). Note that for i=1, x_i = 0. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, then x = \{0, 0, 1, 2, 2\}. Then, she calculated an array, b_1, b_2, …, b_n: b_i = a_i - x_i. For example, if Alicia had the array a = \{0, 1, 2, 0, 3\}, b = \{0-0, 1-0, 2-1, 0-2, 3-2\} = \{0, 1, 1, -2, 1\}. Alicia gives you the values b_1, b_2, …, b_n and asks you to restore the values a_1, a_2, …, a_n. Can you help her solve the problem? Input The first line contains one integer n (3 ≤ n ≤ 200 000) – the number of elements in Alicia's array. The next line contains n integers, b_1, b_2, …, b_n (-10^9 ≤ b_i ≤ 10^9). It is guaranteed that for the given array b there is a solution a_1, a_2, …, a_n, for all elements of which the following is true: 0 ≤ a_i ≤ 10^9. Output Print n integers, a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9), such that if you calculate x according to the statement, b_1 will be equal to a_1 - x_1, b_2 will be equal to a_2 - x_2, ..., and b_n will be equal to a_n - x_n. It is guaranteed that there exists at least one solution for the given tests. It can be shown that the solution is unique. Examples Input 5 0 1 1 -2 1 Output 0 1 2 0 3 Input 3 1000 999999000 -1000000000 Output 1000 1000000000 0 Input 5 2 1 2 2 3 Output 2 3 5 7 10 Note The first test was described in the problem statement. In the second test, if Alicia had an array a = \{1000, 1000000000, 0\}, then x = \{0, 1000, 1000000000\} and b = \{1000-0, 1000000000-1000, 0-1000000000\} = \{1000, 999999000, -1000000000\}. Submitted Solution: ``` n=int(input()) arr=list(map(int,input().split())) pref=[0]*n pref[0]=arr[0] for i in range(1,n): if pref[i-1]==0: if i!=1: pref[i]=pref[i-2]+arr[i] elif i==1: pref[i]=pref[i-1]+arr[i] else: pref[i]=pref[i-1]+arr[i] for i in range(n): print(pref[i],end=' ') print('') ```
instruction
0
93,659
5
187,318
No
output
1
93,659
5
187,319
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` def f(arr): val = 0 for x in arr: val ^= x return val def solve(l, r, k): ret = [l] if k >= 2: cand = [l, l + 1] if f(cand) < f(ret): ret = cand if l + 2 <= r: cand = [l + 1, l + 2] if f(cand) < f(ret): ret = cand if k >= 3: x = 1 while x <= l: x *= 2 if x + x // 2 <= r: ret = [x - 1, x + x // 2, x + x // 2 - 1] if k >= 4: cand = [l, l + 1, l + 2, l + 3] if f(cand) < f(ret): ret = cand if l + 4 <= r: cand = [l + 1, l + 2, l + 3, l + 4] if f(cand) < f(ret): ret = cand return ret l, r, k = map(int, input().split()) ans = solve(l, r, k) print(f(ans)) print(len(ans)) print(' '.join(map(str, ans))) ```
instruction
0
93,891
5
187,782
Yes
output
1
93,891
5
187,783
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` import random l, r, k = map(int, input().split(' ')) if k == 1: print(l) print(1) print(l) quit() if k == 2: if r == l+1: a = l b = l^r if a <= b: print(a) print(1) print(l) quit() else: print(b) print(2) print(l, l+1) quit() for i in range(l, r+1): if i%2==0: print(1) print(2) print(i, i+1) quit() if k == 3: if abs(l-r) <= 10000: for i in range(l, r+1): for j in range(i+1, r+1): if l <= i^j <= r: print(0) print(3) print(i, j, i^j) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if abs(l-r) <= 1000000: for x in range(l, r+1): if l <= x^(x+1) <= r: print(0) print(3) print(x, x+1, x^(x+1)) quit() for i in range(500000): if i == 420: x = l y = l+1 elif i == 420420: x = r y = r-1 else: x = random.randint(l, r) y = random.randint(l, r) if x == y: continue if l <= x^y<= r: print(0) print(3) print(x, y, x^y) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if k == 4: if r == l+3: if l%2 == 0: print(0) print(4) print(l, l+1, l+2, l+3) quit() a = l; b = l+1; c = l+2; d = l+3; if a^b^c == 0: print(0) print(3) print(a, b, c) quit() if a^b^d == 0: print(0) print(3) print(a, b, d) quit() if a^c^d == 0: print(0) print(3) print(a, c, d) quit() if b^c^d == 0: print(0) print(3) print(b, c, d) quit() if a^b == 1: print(1) print(2) print(a, b) quit() print(1) print(2) print(b, c) quit() for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() if k >= 5: for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() ```
instruction
0
93,892
5
187,784
Yes
output
1
93,892
5
187,785
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` import random l, r, k = map(int, input().split(' ')) if k == 1: print(l) print(1) print(l) quit() if k == 2: if r == l+1: a = l b = l^r if a <= b: print(a) print(1) print(l) quit() else: print(b) print(2) print(l, l+1) quit() for i in range(l, r+1): if i%2==0: print(1) print(2) print(i, i+1) quit() if k == 3: if abs(l-r) <= 10000: for i in range(l, r+1): for j in range(i+1, r+1): if l <= i^j <= r: print(0) print(3) print(i, j, i^j) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if abs(l-r) <= 1000000: for x in range(l, r+1): if l <= x^(x+1) <= r: print(0) print(3) print(x, x+1, x^(x+1)) quit() for i in range(1000000): x = random.randint(l, r) y = random.randint(l, r) if x == y: continue if l <= x^y<= r: print(0) print(3) print(x, y, x^y) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if k == 4: if r == l+3: if l%2 == 0: print(0) print(4) print(l, l+1, l+2, l+3) quit() a = l; b = l+1; c = l+2; d = l+3; if a^b^c == 0: print(0) print(3) print(a, b, c) quit() if a^b^d == 0: print(0) print(3) print(a, b, d) quit() if a^c^d == 0: print(0) print(3) print(a, c, d) quit() if b^c^d == 0: print(0) print(3) print(b, c, d) quit() if a^b == 1: print(1) print(2) print(a, b) quit() print(1) print(2) print(b, c) quit() for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() if k >= 5: for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() ```
instruction
0
93,893
5
187,786
No
output
1
93,893
5
187,787
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` import random l, r, k = map(int, input().split(' ')) if k == 1: print(l) print(1) print(l) quit() if k == 2: if r == l+1: a = l b = l^r if a <= b: print(a) print(1) print(l) quit() else: print(b) print(2) print(l, l+1) quit() for i in range(l, r+1): if i%2==0: print(1) print(2) print(i, i+1) quit() if k == 3: if abs(l-r) <= 10000: for i in range(l, r+1): for j in range(i+1, r+1): if l <= i^j <= r: print(0) print(3) print(i, j, i^j) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if abs(l-r) <= 1000000: for x in range(l, r+1): if l <= x^(x+1) <= r: print(0) print(3) print(x, x+1, x^(x+1)) quit() for i in range(1000000): if i == 420: x = l y = l+1 if i == 420420: x = r y = r-1 x = random.randint(l, r) y = random.randint(l, r) if x == y: continue if l <= x^y<= r: print(0) print(3) print(x, y, x^y) quit() print(1) print(2) if (l%2==0): print(l, l+1) quit() else: print(l+1, l+2) quit() if k == 4: if r == l+3: if l%2 == 0: print(0) print(4) print(l, l+1, l+2, l+3) quit() a = l; b = l+1; c = l+2; d = l+3; if a^b^c == 0: print(0) print(3) print(a, b, c) quit() if a^b^d == 0: print(0) print(3) print(a, b, d) quit() if a^c^d == 0: print(0) print(3) print(a, c, d) quit() if b^c^d == 0: print(0) print(3) print(b, c, d) quit() if a^b == 1: print(1) print(2) print(a, b) quit() print(1) print(2) print(b, c) quit() for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() if k >= 5: for i in range(l, r+1): if i%2 == 0: print(0) print(4) print(i, i+1, i+2, i+3) quit() ```
instruction
0
93,894
5
187,788
No
output
1
93,894
5
187,789
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` def main(): from itertools import combinations from functools import reduce from random import randrange l, r, k = [int(i) for i in input().split()] if r - l < 4: f = float("inf") result = None numbers = list(range(l, r + 1)) for mask in range(1, 1 << (r - l + 1)): tmp = [numbers[i] for i in range(r - l + 1) if mask & (1 << i)] if len(tmp) <= k and reduce(lambda x, y: x ^ y, tmp) < f: f = reduce(lambda x, y: x ^ y, tmp) result = tmp elif k >= 4: a = l + l % 2 f = 0 result = [a, a + 1, a + 2, a + 3] elif k == 1: f = l result = [l] elif k == 2: a = l + l % 2 f = 1 result = [a, a + 1] elif k == 3: a = l + l % 2 f = 1 result = [a, a + 1] for a in range(l, min(r + 1, l + 400)): for b in range(max(l, r - 400), r + 1): if l <= a ^ b <= r: f = 0 result = [a, b, a ^ b] break print(f) print(len(result)) print(' '.join(str(i) for i in sorted(result))) main() ```
instruction
0
93,895
5
187,790
No
output
1
93,895
5
187,791
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties: * for all x <image> the following inequality holds l ≤ x ≤ r; * 1 ≤ |S| ≤ k; * lets denote the i-th element of the set S as si; value <image> must be as small as possible. Help Victor find the described set. Input The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)). Output Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order. If there are multiple optimal sets, you can print any of them. Examples Input 8 15 3 Output 1 2 10 11 Input 8 30 7 Output 0 5 14 9 28 11 16 Note Operation <image> represents the operation of bitwise exclusive OR. In other words, it is the XOR operation. Submitted Solution: ``` def solve(L,R,k): if k < 5: if k==1: return [L] elif k==2: if R-L+1==2: if L < (L^R): return [L] else: return [R] else: if L&1: return [L+1,L+2] else: return [L,L+1] else: if L==1: return [1,2,3] elif k==3: b = 60 while (L&(1<<b))==0 and (R&(1<<b))==0: b -= 1 if (L&(1<<b))==(R&(1<<b)): if L&1: return [L+1,L+2] else: return [L,L+1] else: return [L,L^((1<<b)-1),(1<<b)-1] else: if (L^1)&1: return [L,L+1,L+2,L+3] elif L+4 <= R: return [L+1,L+2,L+3,L+4] else: # odd, odd+1, odd+2, odd+3 011 100 101 110 return [L+1,L+2] else: if L&1: return [L+1,L+2,L+3,L+4] else: return [L,L+1,L+2,L+3] ans = solve(*map(int,input().split())) xor = 0 for x in ans: xor ^= x print(xor) print(len(ans)) print(*ans) ```
instruction
0
93,896
5
187,792
No
output
1
93,896
5
187,793
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You're given Q queries of the form (L, R). For each query you have to find the number of such x that L ≤ x ≤ R and there exist integer numbers a > 0, p > 1 such that x = ap. Input The first line contains the number of queries Q (1 ≤ Q ≤ 105). The next Q lines contains two integers L, R each (1 ≤ L ≤ R ≤ 1018). Output Output Q lines — the answers to the queries. Example Input 6 1 4 9 9 5 7 12 29 137 591 1 1000000 Output 2 1 0 3 17 1111 Note In query one the suitable numbers are 1 and 4. Submitted Solution: ``` from math import sqrt def isnps(x): return sqrt(x)%1 != 0 MAX = 10**18 sp = set() for p in range(3, 60): maxx = int(10**(18/p)) for i in range(2, maxx): if isnps(i**p): sp.add(i**p) sp = list(sp) sp.sort() def bs(a,val): low,high=0,len(a)-1 while low<=high: mid=(low+high)//2 if a[mid]>=val: high=mid-1 else: low=mid+1 return high q = int(input()) for t in range(q): l, r = map(int, input().split()) ps = int(sqrt(r)) - int(sqrt(l-1)) print(ps,end=' ') i = bs(sp, r) j = bs(sp, l) ps += i-j print(ps) ```
instruction
0
94,123
5
188,246
No
output
1
94,123
5
188,247