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Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,065
11
48,130
Tags: *special, implementation Correct Solution: ``` n, k = map(int, input().split()) arr = list(map(int, input().split())) count = 0 for i in range(len(arr)): if(arr[i] >= arr[k-1] and arr[i] > 0): count += 1 print(count) ```
output
1
24,065
11
48,131
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,066
11
48,132
Tags: *special, implementation Correct Solution: ``` # import sys # sys.stdin = open("test.in","r") # sys.stdout = open("test.out","w") a,b=map(int,input().split()) c=list(map(int,input().split())) d=c[b-1] count=0 for i in range(a): if c[i]>=d and c[i]>0: count+=1 if c[i]<d or c[i]==0: break print(count) ```
output
1
24,066
11
48,133
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,067
11
48,134
Tags: *special, implementation Correct Solution: ``` n,k=map(int,input().split()) arr=list(map(int,input().split())) k=arr[k-1] i=0 if(k<=0): while(i<len(arr) and arr[i]>k and arr[i]>0): i+=1 print(i) else: while(i<len(arr) and arr[i]>=k): i+=1 print(i) ```
output
1
24,067
11
48,135
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,068
11
48,136
Tags: *special, implementation Correct Solution: ``` n, k = [int(i) for i in input().split()] a = [int(i) for i in input().split()] res = 0 if sum(a) == 0: print(0) else: for i in a: if (i >= a[k - 1]) and (i != 0): res += 1 print(res) ```
output
1
24,068
11
48,137
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,069
11
48,138
Tags: *special, implementation Correct Solution: ``` n, k = map(int, input().split()) a = list(map(int, input().split())) ans = 0 for i in a: ans += (i >= a[k-1] and i > 0) print(ans) ```
output
1
24,069
11
48,139
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,070
11
48,140
Tags: *special, implementation Correct Solution: ``` a=input().split() n=int(a[0]) k=int(a[1]) b=input().split() for i in range(0,n): b[i]=int(b[i]) while b[k-1]<=0: k-=1 if k==0:break if k!=n: while b[k-1]==b[k] and k!=0: k+=1 if k==n:break print (k) ```
output
1
24,070
11
48,141
Provide tags and a correct Python 3 solution for this coding contest problem. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score.
instruction
0
24,071
11
48,142
Tags: *special, implementation Correct Solution: ``` n, k = map(int, input().split()) A = list(map(int, input().split())) cnt = 0 for a in A: if a > 0 and a >= A[k-1]: cnt += 1 print(cnt) ```
output
1
24,071
11
48,143
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` n,k=map(int,input().split(' ')) p=input().split(' ') for i in range(0,n): if int(p[i])<int(p[k-1]) or p[i]=='0': print(i) break else: print(i+1) ```
instruction
0
24,072
11
48,144
Yes
output
1
24,072
11
48,145
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` n,k=map(int,input().split()) arr=[int(x) for x in input().split()] max_score=arr[k-1] count=0 for i in range(0,len(arr),1): if arr[i]>0 and arr[i]>=max_score: count+=1 print(count) ```
instruction
0
24,073
11
48,146
Yes
output
1
24,073
11
48,147
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` a = input().split() b = input().split() y = b[int(a[1])-1] n = 0 for i, x in enumerate(b): if int(x) >= int(y) and int(x) > 0: n += 1 print(n) ```
instruction
0
24,074
11
48,148
Yes
output
1
24,074
11
48,149
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` s = input().split() n=int(s[0]) k=int(s[1]) st=input().split()[0:n] res=int() maks=int(st[k-1]) for i in st: i=int(i) if i>=maks and i>0: res+=1 print(res) ```
instruction
0
24,075
11
48,150
Yes
output
1
24,075
11
48,151
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` n,k=map(int,input().split()) a=list(map(int,input().split())) if(k<n): res=a[k] if(res>0): if((a.count(res))==1): tot=a.index(res)+a.count(res)-1 else: tot=a.index(res)+a.count(res) else: tot=0 else: i=n-1 while(a[i]<=0): i=i-1 tot=i+1 print(tot) ```
instruction
0
24,076
11
48,152
No
output
1
24,076
11
48,153
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` size, num = map(int, input().split()) ar = list(map(int, input().split())) tmp = ar[num-1] cnt = 0 for i in ar: if i >= tmp: cnt += 1 print(cnt) ```
instruction
0
24,077
11
48,154
No
output
1
24,077
11
48,155
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` n,k = map(int,input().split()) a = list(map(int,input().split())) c=0 for i in range(n): if a[i]>=a[k-1]: c+=1 print(c) ```
instruction
0
24,078
11
48,156
No
output
1
24,078
11
48,157
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. "Contestant who earns a score equal to or greater than the k-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules. A total of n participants took part in the contest (n ≥ k), and you already know their scores. Calculate how many participants will advance to the next round. Input The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 50) separated by a single space. The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 100), where ai is the score earned by the participant who got the i-th place. The given sequence is non-increasing (that is, for all i from 1 to n - 1 the following condition is fulfilled: ai ≥ ai + 1). Output Output the number of participants who advance to the next round. Examples Input 8 5 10 9 8 7 7 7 5 5 Output 6 Input 4 2 0 0 0 0 Output 0 Note In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers. In the second example nobody got a positive score. Submitted Solution: ``` input1 = input().split() n = int(input1[0]) k = int(input1[1]) total = 0 scores = input().split() #scores = int(scores) if(int(scores[0]) == int(scores[len(scores) - 1]) and int(scores[0]) == 0): total = 0 else: for i in scores: if(int(i) >= int(scores[k-1])): total += 1 print(total) ```
instruction
0
24,079
11
48,158
No
output
1
24,079
11
48,159
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n,k=map(int,input().split()) l=[input() for i in range(n)] z=[] d={} for i in l: t=len(i) if t in d: d[t]+=1 else: d[t]=1 z.append(t) b,w=0,0 su=0 x=len(input()) z.sort() for i in z: if i==x: break su+=d[i] b=((su)//k)*5+su+1 w=((su+d[x]-1)//k)*5+su+d[x] print(b,w) ```
instruction
0
24,350
11
48,700
Yes
output
1
24,350
11
48,701
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n,k=map(int,input().split()) val=[0 for i in range(105) ] ck=[] ml=[] for i in range(n): s=input() if s not in ck: ck+=[s] ml+=[len(s)] val[len(s)]+=1 s=input() d=len(s) c=0 p=val[d] for i in range(1,d+1): c+=val[i] best=((c-p)//k)*5+c-p+1 worst=((c-1)//k)*5+c print(best,worst) ```
instruction
0
24,351
11
48,702
Yes
output
1
24,351
11
48,703
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n, k = map(int, input().split()) a = sorted([len(input()) for i in range(n)]) b = len(input()) ans1, ans2 = 0, 0 for i in range(n): ans1 += 1 if a[i] == b: break if i % k == k - 1: ans1 += 5 for i in range(n): ans2 += 1 if i == n - 1 or a[i + 1] > b: break if i % k == k - 1: ans2 += 5 print(ans1, ans2) ```
instruction
0
24,352
11
48,704
Yes
output
1
24,352
11
48,705
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n,k=[int(a) for a in input().split()] p=[] for i in range(n): p.append(input()) real=input() l=len(real) cmin,cmax=0,0 for s in p: if len(s)<=l: cmax=cmax+1 if len(s)<l: cmin=cmin+1 def calc(c): if c==0: return 0 return c+5*(c//k) print(calc(cmin)+1,calc(cmax-1)+1) ```
instruction
0
24,353
11
48,706
Yes
output
1
24,353
11
48,707
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n, k = map(int, input().split()) lst = [] if n == 1: s1 = 1 e1 = 1 else: for i in range(n): x = input() lst.append(x) p = input() d = {} lst.sort() for i in lst: if len(i) in d: d[len(i)]+=1 else: d[len(i)]=1 l = len(p) s = 1 e = 0 for key in d.keys(): if key < l: s += d[key] if key <= l: e += d[key] s1 = s+((s//k)*5) e1 = e+((e//k)*5) if k == 1 and s == 1: s1 = 1 if k == 1: e1 = e+(((e-1)//k)*5) print(s1, e1) ```
instruction
0
24,354
11
48,708
No
output
1
24,354
11
48,709
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` from collections import Counter n, k = map(int, input().split()) words = list() for i in range(n): words.append(input()) real = input() # Get only lengths of the passwords. words = [len(i) for i in words if len(i) <= len(real)] # Get all lenths of all passwords and their quantity. c = Counter(words) sorted_c = sorted(c) # Sort by key # Assign a quantity of passwords with the same length as real's, # i.e. retrieve the specific value from the dictionary. real_value = c[len(real)] # Get the quantity of passwords with a length less than real's all_but_real = 0 for i in range(len(sorted_c) - 1): all_but_real += c[sorted_c[i]] tmin = 1 + all_but_real + (all_but_real // k) * 5 if real_value == 1: tmax = tmin if (real_value + all_but_real) % 2 != 0: tmax = (real_value + all_but_real + ((real_value + all_but_real) // k) * 5) print("ODD") else: tmax = (real_value + all_but_real + (((real_value + all_but_real) // k) - 1) * 5) print("EVEN") print(tmin, tmax) ```
instruction
0
24,355
11
48,710
No
output
1
24,355
11
48,711
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` # cook your dish here #import sys #sys.setrecursionlimit(10**9) ll=lambda:map(int,input().split()) t=lambda:int(input()) ss=lambda:input() lx=lambda x:map(int,input().split(x)) yy=lambda:print("Yes") nn=lambda:print("No") from math import log10 ,log2,ceil,factorial as fac,gcd,inf,sqrt,log #from itertools import combinations_with_replacement as cs #from functools import reduce from bisect import bisect_right as br,bisect_left as bl from collections import Counter #from math import inf mod=10**9+7 #for _ in range(t()): def f(): n,k=ll() d={} q=[] for i in range(n): s=ss() if len(s) not in d: d[len(s)]=0 d[len(s)]+=1 q.append(s) p=ss() x,y=0,0 for i in sorted(d.keys()): if i>len(p): break if i!=len(p): x+=d[i] y+=d[i] x+=1 x=x+((x//k)*5 if x>k else 0) y=y-q.count(p)+1 y=y+((y//k)*5 if y>k else 0) print(x,y) f() ''' baca bac 1 2 3 baaccca abbaccccaba ''' ```
instruction
0
24,356
11
48,712
No
output
1
24,356
11
48,713
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration. Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice. Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password k times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that. Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds). Input The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds. The next n lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters. The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his n passwords. Output Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively. Examples Input 5 2 cba abc bb1 abC ABC abc Output 1 15 Input 4 100 11 22 1 2 22 Output 3 4 Note Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds. Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all. Submitted Solution: ``` n,k=map(int,input().split()) pas=[] for i in range(n): pas.append(input()) cp=input(); les=0; eq=0 for item in pas: if len(item)<len(cp): les+=1 elif len(item)==len(cp): eq+=1 add=les//k*5 add2=(eq-1)//k*5 print(les+add+1,les+eq+add2) ```
instruction
0
24,357
11
48,714
No
output
1
24,357
11
48,715
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. Constraints * 0≤A,B,C,D≤9 * All input values are integers. * It is guaranteed that there is a solution. Input Input is given from Standard Input in the following format: ABCD Output Print the formula you made, including the part `=7`. Use the signs `+` and `-`. Do not print a space between a digit and a sign. Examples Input 1222 Output 1+2+2+2=7 Input 0290 Output 0-2+9+0=7 Input 3242 Output 3+2+4-2=7 Submitted Solution: ``` # -*- coding: utf-8 -*- import types #import copy # Input abcd = int(input()) n = len(str(abcd)) - 1 # 閾の数 # Print for i in range( 2**n ): op = n * ["-"] # Create operation list # example: ['+', '-', '+'] for j in range(n): if ( i >> j ) & 1: op[n-1 -j] = "+" formula = "" for p_n, p_o in zip(str(abcd), op+[""]): formula += (p_n + p_o) if eval(formula) == 7: print(formula + "=7") break ```
instruction
0
24,601
11
49,202
No
output
1
24,601
11
49,203
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n = int(input()) s = sorted([int(input()) for _ in range(n)]) s = [0]+s x = sum(s) for i in range(n): if (x-s[i]) % 10: print(x-s[i]) exit() print(0) ```
instruction
0
24,613
11
49,226
Yes
output
1
24,613
11
49,227
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n=int(input()) s=[int(input()) for _ in range(n)] x=sum(s) if x%10: print(x) else: a=[] for i in s: if i%10: a.append(i) if a: print(x-min(a)) else: print(0) ```
instruction
0
24,614
11
49,228
Yes
output
1
24,614
11
49,229
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n=int(input()) a=[int(input()) for _ in range(n)] sumA=sum(a) if sumA%10>0: print(sumA) else: a=sorted(a) for b in a: if b%10>0: print(sumA-b) break else: print(0) ```
instruction
0
24,615
11
49,230
Yes
output
1
24,615
11
49,231
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n=int(input()) s=[int(input()) for _ in range(n)] p=sum(s) if p%10!=0: print(p) else: s.sort() d=0 for i in s: if i%10!=0: d=i break if d==0: print(0) else: print(p-d) ```
instruction
0
24,616
11
49,232
Yes
output
1
24,616
11
49,233
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n = int(input()) a = [int(input()) for _ in range(n)] a.sort() suma = sum(a) if suma % 10 != 0: print(suma) else: for i in a: if i % 10 != 0: print(suma - i) ```
instruction
0
24,617
11
49,234
No
output
1
24,617
11
49,235
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n=int(input()) l=sorted([int(input()) for i in range(n)]) if sum(l)%10!=0: print(sum(l)) else: ans=sum(l) for i in range(n): if l[i]%10==0: ans-=l[i] print(ans) quit() print(0) ```
instruction
0
24,618
11
49,236
No
output
1
24,618
11
49,237
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n=int(input()) s=[int(input()) for i in range(n)] #print(n,s) max=0 t=[0] for j in range(n): for i in range(len(t)): t.append(t[i]+s[j]) if max < t[i]+s[j] and (t[i]+s[j])%10 !=0: max = t[i]+s[j] print(max) ```
instruction
0
24,619
11
49,238
No
output
1
24,619
11
49,239
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade? Constraints * All input values are integers. * 1 ≤ N ≤ 100 * 1 ≤ s_i ≤ 100 Input Input is given from Standard Input in the following format: N s_1 s_2 : s_N Output Print the maximum value that can be displayed as your grade. Examples Input 3 5 10 15 Output 25 Input 3 10 10 15 Output 35 Input 3 10 20 30 Output 0 Submitted Solution: ``` n = int(input()) s = [ int(input()) for i in range(n)] dp = [[0 for _ in range(n+1)] for _ in range(n) ] for i in range(1, n + 1): dp[0][i] = s[0] for i in range(1, n): for j in range(n + 1): if (i == j): dp[i][j] = dp[i - 1][j] else: dp[i][j] = dp[i - 1][j] + s[i] print(max([it for it in dp[n - 1] if (it % 10 != 0)])) ```
instruction
0
24,620
11
49,240
No
output
1
24,620
11
49,241
Provide a correct Python 3 solution for this coding contest problem. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10
instruction
0
24,714
11
49,428
"Correct Solution: ``` class WUnionFind: MAX_N = 0 PAR = [] RANK = [] DIFF_WEIGHT=[] def __init__(self,n): self.MAX_N = n self.PAR = [ i for i in range(n)] self.RANK= [ 0 ] * n self.DIFF_WEIGHT=[ 0 ] * n def root(self,x): if self.PAR[x] == x: return x else: r = self.root(self.PAR[x]) self.DIFF_WEIGHT[x] += self.DIFF_WEIGHT[self.PAR[x]] self.PAR[x] = r return r def weight(self,x): self.root(x) return self.DIFF_WEIGHT[x] def issame(self,x,y): return self.root(x) == self.root(y) def diff(self,x,y): return self.weight(y)-self.weight(x) def merge(self,x,y,w): w += self.weight(x) w -= self.weight(y) x = self.root(x) y = self.root(y) if x == y: return False if self.RANK[x] < self.RANK[y]: x,y = y,x w = -w if self.RANK[x] == self.RANK[y]: self.RANK[x] += 1 self.PAR[y] = x self.DIFF_WEIGHT[y] = w return True N,Q = map(int,input().split()) Ans = [] tree = WUnionFind(N+1) for _ in range(Q): s = list(map(int,input().split())) if s[0] == 0: x = s[1] y = s[2] w = s[3] tree.merge(x,y,w) else: x = s[1] y = s[2] if tree.issame(x,y): a = tree.diff(x,y) else: a = '?' Ans.append(a) for a in Ans: print(a) ```
output
1
24,714
11
49,429
Provide a correct Python 3 solution for this coding contest problem. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10
instruction
0
24,716
11
49,432
"Correct Solution: ``` rank = [] p = [] w = [] def makeSet(x): p.append(x) rank.append(0) w.append(0) def union(x, y): link(findSet(x), findSet(y)) def link(x, y): if rank[x] > rank[y]: p[y] = x else: p[x] = y if rank[x] == rank[y]: rank[y] = rank[y] + 1 def findSet(x): if p[x] == x: return x i = findSet(p[x]) w[x] += w[p[x]] p[x] = i return i def same(x, y): return findSet(x) == findSet(y) def relate (x,y,z): i = findSet(x) j = findSet(y) if rank[i] < rank[j]: p[i] = j w[i] = z - w[x] + w[y] else: p[j] = i w[j] = -z - w[y] + w[x] if rank[i] == rank[j]: rank[i] += 1 n, q = map(int, input().split()) for i in range(n): makeSet(i) for i in range(q): com, *cmd = map(int, input().split()) if com == 0: relate(cmd[0], cmd[1],cmd[2]) else: x, y = cmd if same(x, y): print(w[x] - w[y]) else: print("?") ```
output
1
24,716
11
49,433
Provide a correct Python 3 solution for this coding contest problem. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10
instruction
0
24,717
11
49,434
"Correct Solution: ``` class UnionFind: def __init__(self, n): self.tree = [-1] * n self.rank = [0] * n self.diff_weight = [0] * n def unite(self, a, b, w): w += self.weight(a) w -= self.weight(b) a = self.root(a) b = self.root(b) if a == b: return if self.rank[a] < self.rank[b]: a, b = b, a w *= -1 if self.rank[a] == self.rank[b]: self.rank[a] += 1 self.tree[b] = a self.diff_weight[b] = w def root(self, a): if(self.tree[a] < 0): return a else: res = self.root(self.tree[a]) self.diff_weight[a] += self.diff_weight[self.tree[a]] self.tree[a] = res return self.tree[a] def same(self, a, b): return self.root(a) == self.root(b) def weight(self, a): self.root(a) return self.diff_weight[a] def diff(self, a, b): return self.weight(b) - self.weight(a) import sys stdin = sys.stdin na = lambda: map(int, stdin.readline().split()) ns = lambda: stdin.readline().rstrip() ni = lambda: int(ns()) def main(): n, m = na() uf = UnionFind(n) for _ in range(m): t = list(na()) a, b, c = t[0], t[1], t[2] if a == 0: d = t[3] uf.unite(b, c, d) else: if uf.same(b, c): print(uf.diff(b, c)) else: print("?") main() ```
output
1
24,717
11
49,435
Provide a correct Python 3 solution for this coding contest problem. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10
instruction
0
24,720
11
49,440
"Correct Solution: ``` class WeightedUnionFind(): def __init__(self,n): self.n = n self.parents = [i for i in range(n)] self.rank = [0 for _ in range(n)] self.weight = [0 for _ in range(n)] def find(self,x): root = x stack = [root] while self.parents[root] != root: root = self.parents[root] stack.append(root) while stack: s = stack.pop() self.weight[s] += self.weight[self.parents[s]] while self.parents[x] != root: parent = self.parents[x] self.parents[x] = root x = parent return root def unite(self,x,y,w): xroot = self.find(x) yroot = self.find(y) if xroot == yroot: return xrank = self.rank[xroot] yrank = self.rank[yroot] if xrank < yrank: self.parents[xroot] = yroot self.weight[xroot] = w-self.weight[x]+self.weight[y] else: self.parents[yroot] = xroot self.weight[yroot] = -w+self.weight[x]-self.weight[y] if xrank == yrank: self.rank[xroot] += 1 def diff(self,x,y): return self.weight[x]-self.weight[y] N,Q = map(int,input().split()) uf = WeightedUnionFind(N) ans = [] for _ in range(Q): q = list(map(int,input().split())) if q[0] == 0: uf.unite(q[1],q[2],q[3]) else: if uf.find(q[1]) == uf.find(q[2]): ans.append(uf.diff(q[1],q[2])) else: ans.append('?') print('\n'.join(map(str,ans))) ```
output
1
24,720
11
49,441
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` class Node: def __init__(self, root, length, dist): self.root = root self.child = None self.length = length self.dist = dist self.tail = root def relate(x, y, z): X = A[x].root Y = A[y].root if A[X].length > A[Y].length: x, y, z, X, Y = y, x, -z, Y, X T = A[Y].tail A[T].child = X A[Y].tail = A[X].tail A[Y].length += A[X].length tmp = X xd = A[x].dist yd = A[y].dist for _ in range(A[X].length): A[tmp].root = Y A[tmp].dist += yd - z - xd tmp = A[tmp].child n,q = [int(i) for i in input().split()] A = [] for i in range(n): A.append(Node(i, 1, 0)) for _ in range(q): c = [int(i) for i in input().split()] if c[0] == 0 and A[c[1]].root != A[c[2]].root: relate(c[1], c[2], c[3]) elif c[0] == 1: if A[c[1]].root == A[c[2]].root: print(A[c[2]].dist - A[c[1]].dist) else: print("?") ```
instruction
0
24,722
11
49,444
Yes
output
1
24,722
11
49,445
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` class WeightedUnionFind: def __init__(self, n): self.par = [i for i in range(n+1)] self.rank = [0] * (n+1) # 根への距離を管理 self.weight = [0] * (n+1) # 検索 def find(self, x): if self.par[x] == x: return x else: y = self.find(self.par[x]) # 親への重みを追加しながら根まで走査 self.weight[x] += self.weight[self.par[x]] self.par[x] = y return y # 併合 def union(self, x, y, w): rx = self.find(x) ry = self.find(y) # xの木の高さ < yの木の高さ if self.rank[rx] < self.rank[ry]: self.par[rx] = ry self.weight[rx] = w - self.weight[x] + self.weight[y] # xの木の高さ ≧ yの木の高さ else: self.par[ry] = rx self.weight[ry] = -w - self.weight[y] + self.weight[x] # 木の高さが同じだった場合の処理 if self.rank[rx] == self.rank[ry]: self.rank[rx] += 1 # 同じ集合に属するか def same(self, x, y): return self.find(x) == self.find(y) # xからyへのコスト def diff(self, x, y): if self.find(x) != self.find(y): return "?" return self.weight[x] - self.weight[y] n, q = map(int, input().split()) tree = WeightedUnionFind(n) for i in range(q): query = list(map(int, input().split())) if query[0] == 0: x, y, z = query[1:4] tree.union(x, y, z) else: x, y = query[1:3] ans = tree.diff(x, y) print(ans) ```
instruction
0
24,723
11
49,446
Yes
output
1
24,723
11
49,447
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` #! python3 # weighted_union_find_tree.py class WeightedUnionFindTree(): def __init__(self, n): self.par = [i for i in range(n+1)] self.rank = [0] * (n+1) self.weight = [0] * (n+1) def find(self, x): if self.par[x] == x: return x else: y = self.find(self.par[x]) self.weight[x] += self.weight[self.par[x]] self.par[x] = y return y def unite(self, x, y, w): rx = self.find(x) ry = self.find(y) if self.rank[rx] < self.rank[ry]: self.par[rx] = ry self.weight[rx] = w - self.weight[x] + self.weight[y] else: self.par[ry] = rx self.weight[ry] = -w - self.weight[y] + self.weight[x] if self.rank[rx] == self.rank[ry]: self.rank[rx] += 1 def same(self, x, y): return self.find(x) == self.find(y) def diff(self, x, y): return self.weight[x] - self.weight[y] n, q = [int(x) for x in input().split(' ')] wuft = WeightedUnionFindTree(n-1) for i in range(q): query = [int(x) for x in input().split(' ')] if query[0] == 0: x, y, z = query[1:] wuft.unite(x, y, z) elif query[0] == 1: x, y = query[1:] if wuft.same(x, y): print(wuft.diff(x,y)) else: print('?') ```
instruction
0
24,724
11
49,448
Yes
output
1
24,724
11
49,449
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` import sys input = sys.stdin.readline class WeightedUnionFind: def __init__(self, N): self.parent = list(range(N)) self.rank = [0] * N self.weight = [0] * N def find(self, x): stack = [x] while self.parent[x] != x: x = self.parent[x] stack.append(x) r = stack[-1] while stack: x = stack.pop() self.weight[x] += self.weight[self.parent[x]] self.parent[x] = r return r def unite(self, x, y, w): rx = self.find(x) ry = self.find(y) w += self.weight[x] - self.weight[y] x = rx y = ry if x == y: return False if self.rank[x] < self.rank[y]: x, y = y, x w = -w if self.rank[x] == self.rank[y]: self.rank[x] += 1 self.parent[y] = x self.weight[y] = w return True def same(self, x, y): return self.find(x) == self.find(y) def diff(self, x, y): self.find(x) self.find(y) return self.weight[y] - self.weight[x] n, q = map(int, input().split()) wuf = WeightedUnionFind(n) for _ in range(q): t, *data = map(int, input().split()) if t == 0: x, y, z = data wuf.unite(x, y, z) else: x, y = data if not wuf.same(x, y): print('?') else: print(wuf.diff(x, y)) ```
instruction
0
24,725
11
49,450
Yes
output
1
24,725
11
49,451
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` n, q = map(int, input().split()) nodes = [x for x in range(n)] weight = [0 for x in range(n)] def unite(x, y, w) : if nodes[x] < nodes[y] : nodes[y] = x weight[y] = w else : nodes[x] = y weight[x] = -w def get_root(idx) : if nodes[idx] == idx : return idx else : r = get_root(nodes[idx]) weight[idx] += weight[nodes[idx]] nodes[idx] = r return r for i in range(q) : ip = input() # union if ip[0] == "0" : q, a, b, w = map(int, ip.split()) unite(a, b, w) # evaluate else : q, a, b = map(int, ip.split()) if get_root(a) == get_root(b) : print(weight[b] - weight[a]) else : print("?") ```
instruction
0
24,726
11
49,452
No
output
1
24,726
11
49,453
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` # Disjoint Set: Union Find Trees [p, rank] = [[], []] [n, q] = list(map(int, input().split())) w = [0 for i in range(n)] path_c = [[] for i in range(n)] path_c_len = [0 for i in range(n)] def makeSet(x): global p, rank p.append(x) rank.append(0) def findSet(x): global p if x != p[x]: p[x] = findSet(p[x]) return p[x] def link(x, y): global p, rank if rank[x] > rank[y]: p[y] = x else: p[x] = y if rank[x] == rank[y]: rank[y] += 1 def union(x, y): link(findSet(x), findSet(y)) def same(x, y): if findSet(x) == findSet(y): return 1 else: return 0 def relate(x, y, z): global w, path_c path_c[y].append((x, z)) path_c_len[y] += 1 #print(w) if w[y] == 0: w[y] = w[x] + z else: update(x, y, z) union(x, y) def update(x, y, z): global w, path_c w[x] = w[y] - z for i in range(path_c_len[x]): update(path_c[x][i][0], x, path_c[x][i][1]) def diff(x, y): if same(x, y) == 1: global w return w[y] - w[x] #return abs(w[y] - w[x]) else: return "?" ''' print(p) print(rank) print(w) print("\n") ''' for i in range(n): makeSet(i) for i in range(q): data = list(map(int, input().split())) if data[0] == 0: relate(data[1], data[2], data[3]) else: print(diff(data[1], data[2])) ''' print("\n") print(p) print(rank) print(w) print("\n") ''' ```
instruction
0
24,727
11
49,454
No
output
1
24,727
11
49,455
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` # Disjoint Set: Union Find Trees [p, rank] = [[], []] [n, q] = list(map(int, input().split())) w = [0 for i in range(n)] def makeSet(x): global p, rank p.append(x) rank.append(0) def findSet(x): global p if x != p[x]: p[x] = findSet(p[x]) return p[x] def link(x, y): global p, rank if rank[x] > rank[y]: p[y] = x else: p[x] = y if rank[x] == rank[y]: rank[y] += 1 def union(x, y): link(findSet(x), findSet(y)) def same(x, y): if findSet(x) == findSet(y): return 1 else: return 0 def relate(x, y, z): global w #print(w) if w[y] == 0: w[y] = w[x] + z else: w[x] = w[y] - z #print("cost:", w[y]) union(x, y) ''' def cost(x, y, n, total): global w, d #time += 1 #if time > 5: # return None for i in range(len(w[x])): if d[w[x][i][0]] == 0: #print(x, w[x][i][0]) d[w[x][i][0]] = 1 if y == w[x][i][0]: total += w[x][i][1] return total result = cost(w[x][i][0], y, n, total) if result != None: total = result + w[x][i][1] return total for i in range(n): for j in range(len(w[i])): if w[i][j][0] == x and d[i] == 0: #print(i, x) d[i] = 1 if y == i: total -= w[i][j][1] return total result = cost(i, y, n, total) if result != None: total = result - w[i][j][1] return total ''' def diff(x, y): if same(x, y) == 1: global w return w[y] - w[x] ''' global d total = 0 d = [0 for i in range(n)] d[x] = 1 return cost(x, y, n, total) ''' else: return "?" ''' print(p) print(rank) print(w) print("\n") ''' for i in range(n): makeSet(i) #w = [0 for i in range(n)] for i in range(q): data = list(map(int, input().split())) if data[0] == 0: relate(data[1], data[2], data[3]) else: print(diff(data[1], data[2])) ''' print("\n") print(p) print(rank) print(w) print("\n") ''' ```
instruction
0
24,728
11
49,456
No
output
1
24,728
11
49,457
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. There is a sequence $A = a_0, a_1, ..., a_{n-1}$. You are given the following information and questions. * relate$(x, y, z)$: $a_y$ is greater than $a_x$ by $z$ * diff$(x, y)$: report the difference between $a_x$ and $a_y$ $(a_y - a_x)$ Constraints * $2 \leq n \leq 100,000$ * $1 \leq q \leq 200,000$ * $0 \leq x, y < n$ * $x \ne y$ * $0 \leq z \leq 10000$ * There are no inconsistency in the given information Input $n \; q$ $query_1$ $query_2$ : $query_q$ In the first line, $n$ and $q$ are given. Then, $q$ information/questions are given in the following format. 0 $x \; y\; z$ or 1 $x \; y$ where '0' of the first digit denotes the relate information and '1' denotes the diff question. Output For each diff question, print the difference between $a_x$ and $a_y$ $(a_y - a_x)$. Example Input 5 6 0 0 2 5 0 1 2 3 1 0 1 1 1 3 0 1 4 8 1 0 4 Output 2 ? 10 Submitted Solution: ``` MAX = 10003 N, Q = (int(x) for x in input().split()) ws = [0 for i in range(MAX)] par = [-1 for i in range(N)] def weight(x): root(x) return ws[x] def root(x): if par[x] < 0: return x p = root(par[x]) ws[x] = ws[x] + ws[par[x]] par[x] = p return par[x] return y def unite(y, x, z): z = z + weight(x) z = z - weight(y) x = root(x) y = root(y) if x == y: return 0 if par[y] < par[x]: tmp = x x = y y = tmp z = -z par[x] = par[x] + par[y] par[y] = x ws[y] = z return 1 def diff(x,y): if root(x) != root(y) :return -1 ans = ws[x] - ws[y] return ans for q in range(Q): t, *cmd = (int(x) for x in input().split()) if t: x, y = cmd z = diff(x, y) if z==-1:print("?") else :print(z) else: x, y, z = cmd unite(x, y, z) ```
instruction
0
24,729
11
49,458
No
output
1
24,729
11
49,459
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A boy named Gena really wants to get to the "Russian Code Cup" finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him. The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena's friends won't agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena's computer is connected to at least ki monitors, each monitor costs b rubles. Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there's no monitors connected to Gena's computer. Input The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena's friends, the number of problems and the cost of a single monitor. The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m. Output Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved. Examples Input 2 2 1 100 1 1 2 100 2 1 1 Output 202 Input 3 2 5 100 1 1 1 100 1 1 2 200 1 2 1 2 Output 205 Input 1 2 1 1 1 1 1 Output -1 Submitted Solution: ``` def calc(zzz, mons0): if not zzz: return mons0*b z = zzz.pop() minc=10**9 for i in dct[z]: dr=drug[i] price2, mon2, zz2 = dr mons=max(mon2, mons0) cost2 = calc(zzz-set(zz2), mons) c=cost2+price2 if c<minc: minc = c return minc def input_row(): return list(map(int, input().strip().split(' '))) if __name__=='__main__': zzz=set() dct={} drug={} xxx = input_row() n, m, b = xxx for i in range(n): x,k,_ = input_row() zz = input_row() drug[i] = x,k, zz for z in zz: if z not in dct: dct[z]=[] dct[z].append(i) zzz=zzz.union(zz) if len(zzz)<m: print (-1) else: solved=set() cost=0 mons=0 cost1 =calc(zzz, mons) print (cost+cost1) ```
instruction
0
25,094
11
50,188
No
output
1
25,094
11
50,189
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A boy named Gena really wants to get to the "Russian Code Cup" finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him. The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena's friends won't agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena's computer is connected to at least ki monitors, each monitor costs b rubles. Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there's no monitors connected to Gena's computer. Input The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena's friends, the number of problems and the cost of a single monitor. The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m. Output Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved. Examples Input 2 2 1 100 1 1 2 100 2 1 1 Output 202 Input 3 2 5 100 1 1 1 100 1 1 2 200 1 2 1 2 Output 205 Input 1 2 1 1 1 1 1 Output -1 Submitted Solution: ``` def calc(zzz, mons0): if not zzz: return mons0*b z = zzz.pop() minc=10**9 for i in dct[z]: dr=drug[i] price2, mon2, zz2 = dr mons=max(mon2, mons0) cost2 = calc(zzz-set(zz2), mons) c=cost2+price2 if c<minc: minc = c return minc def input_row(): return list(map(int, input().strip().split(' '))) if __name__=='__main__': zzz=set() dct={} drug={} xxx = input_row() n, m, b = xxx for i in range(n): x,k,_ = input_row() zz = input_row() drug[i] = x,k, zz for z in zz: if z not in dct: dct[z]=[] dct[z].append(i) zzz=zzz.union(zz) if len(zzz)<m: print (False) else: solved=set() cost=0 mons=0 cost1 =calc(zzz, mons) print (cost+cost1) ```
instruction
0
25,095
11
50,190
No
output
1
25,095
11
50,191
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A boy named Gena really wants to get to the "Russian Code Cup" finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him. The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena's friends won't agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena's computer is connected to at least ki monitors, each monitor costs b rubles. Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there's no monitors connected to Gena's computer. Input The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena's friends, the number of problems and the cost of a single monitor. The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m. Output Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved. Examples Input 2 2 1 100 1 1 2 100 2 1 1 Output 202 Input 3 2 5 100 1 1 1 100 1 1 2 200 1 2 1 2 Output 205 Input 1 2 1 1 1 1 1 Output -1 Submitted Solution: ``` def calc(zzz, mons0): if not zzz: return mons0*b z = zzz.pop() minc=10**9 for i in dct[z]: dr=drug[i] price2, mon2, zz2 = dr mons=max(mon2, mons0) cost2 = calc(zzz-set(zz2), mons) c=cost2+price2 if c<minc: minc = c return minc def input_row(): return list(map(int, input().strip().split(' '))) zzz=set() dct={} drug={} xxx = input_row() n, m, b = xxx #print( list(range(1,m+1))) for i in range(n): x,k,_ = input_row() zz = input_row() drug[i] = x,k, zz for z in zz: if z not in dct: dct[z]=[] dct[z].append(i) zzz=zzz.union(zz) #print(sorted(dct.keys(), key=lambda x:len(dct[x]))) if len(zzz)<m: print (False) else: solved=set() cost=0 mons=0 cost1 =calc(zzz, mons) print (cost+cost1) ```
instruction
0
25,096
11
50,192
No
output
1
25,096
11
50,193
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The organizers of a programming contest have decided to present t-shirts to participants. There are six different t-shirts sizes in this problem: S, M, L, XL, XXL, XXXL (sizes are listed in increasing order). The t-shirts are already prepared. For each size from S to XXXL you are given the number of t-shirts of this size. During the registration, the organizers asked each of the n participants about the t-shirt size he wants. If a participant hesitated between two sizes, he could specify two neighboring sizes — this means that any of these two sizes suits him. Write a program that will determine whether it is possible to present a t-shirt to each participant of the competition, or not. Of course, each participant should get a t-shirt of proper size: * the size he wanted, if he specified one size; * any of the two neibouring sizes, if he specified two sizes. If it is possible, the program should find any valid distribution of the t-shirts. Input The first line of the input contains six non-negative integers — the number of t-shirts of each size. The numbers are given for the sizes S, M, L, XL, XXL, XXXL, respectively. The total number of t-shirts doesn't exceed 100 000. The second line contains positive integer n (1 ≤ n ≤ 100 000) — the number of participants. The following n lines contain the sizes specified by the participants, one line per participant. The i-th line contains information provided by the i-th participant: single size or two sizes separated by comma (without any spaces). If there are two sizes, the sizes are written in increasing order. It is guaranteed that two sizes separated by comma are neighboring. Output If it is not possible to present a t-shirt to each participant, print «NO» (without quotes). Otherwise, print n + 1 lines. In the first line print «YES» (without quotes). In the following n lines print the t-shirt sizes the orginizers should give to participants, one per line. The order of the participants should be the same as in the input. If there are multiple solutions, print any of them. Examples Input 0 1 0 1 1 0 3 XL S,M XL,XXL Output YES XL M XXL Input 1 1 2 0 1 1 5 S M S,M XXL,XXXL XL,XXL Output NO Submitted Solution: ``` cs = list(map(int, input().split(" "))) camisas = {} camisas['S'] = cs[0] camisas['M'] = cs[1] camisas['L'] = cs[2] camisas['XL'] = cs[3] camisas['XXL'] = cs[4] camisas['XXXL'] = cs[5] p = int(input()) resposta = {} possivel = True s = [] m = [] l = [] xl = [] xxl = [] for i in range(p): camisa_pessoa = list(input().split(",")) if len(camisa_pessoa) == 1: camisa = camisa_pessoa[0] if camisas[camisa] > 0: resposta[i] = camisa camisas[camisa] = camisas[camisa] - 1 else: possivel = False else: if camisa_pessoa[0] == 'S': s.append(i) elif camisa_pessoa[0] == 'M': m.append(i) elif camisa_pessoa[0] == 'L': l.append(i) elif camisa_pessoa[0] == 'XL': xl.append(i) elif camisa_pessoa[0] == 'XXL': xxl.append(i) if possivel: for i in s: if camisas['S'] > 0: camisas['S'] = camisas['S'] - 1 resposta[i] = 'S' elif camisas['M'] > 0: camisas['M'] = camisas['M'] - 1 resposta[i] = 'M' else: possivel = False break if possivel: for i in m: if camisas['M'] > 0: camisas['M'] = camisas['M'] - 1 resposta[i] = 'M' elif camisas['L'] > 0: camisas['L'] = camisas['L'] - 1 resposta[i] = 'L' else: possivel = False break if possivel: for i in l: if camisas['L'] > 0: camisas['L'] = camisas['L'] - 1 resposta[i] = 'L' elif camisas['XL'] > 0: camisas['XL'] = camisas['XL'] - 1 resposta[i] = 'XL' else: possivel = False break if possivel: for i in xl: if camisas['XL'] > 0: camisas['XL'] = camisas['XL'] - 1 resposta[i] = 'XL' elif camisas['XXL'] > 0: camisas['XXL'] = camisas['XXL'] - 1 resposta[i] = 'XXL' else: possivel = False break if possivel: for i in xxl: if camisas['XXL'] > 0: camisas['XXL'] = camisas['XXL'] - 1 resposta[i] = 'XXL' elif camisas['XXXL'] > 0: camisas['XXXL'] = camisas['XXXL'] - 1 resposta[i] = 'XXXL' else: possivel = False break if not possivel: print('NO') else: print('YES') for k in range(p): print(resposta[k]) ```
instruction
0
25,244
11
50,488
Yes
output
1
25,244
11
50,489
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The organizers of a programming contest have decided to present t-shirts to participants. There are six different t-shirts sizes in this problem: S, M, L, XL, XXL, XXXL (sizes are listed in increasing order). The t-shirts are already prepared. For each size from S to XXXL you are given the number of t-shirts of this size. During the registration, the organizers asked each of the n participants about the t-shirt size he wants. If a participant hesitated between two sizes, he could specify two neighboring sizes — this means that any of these two sizes suits him. Write a program that will determine whether it is possible to present a t-shirt to each participant of the competition, or not. Of course, each participant should get a t-shirt of proper size: * the size he wanted, if he specified one size; * any of the two neibouring sizes, if he specified two sizes. If it is possible, the program should find any valid distribution of the t-shirts. Input The first line of the input contains six non-negative integers — the number of t-shirts of each size. The numbers are given for the sizes S, M, L, XL, XXL, XXXL, respectively. The total number of t-shirts doesn't exceed 100 000. The second line contains positive integer n (1 ≤ n ≤ 100 000) — the number of participants. The following n lines contain the sizes specified by the participants, one line per participant. The i-th line contains information provided by the i-th participant: single size or two sizes separated by comma (without any spaces). If there are two sizes, the sizes are written in increasing order. It is guaranteed that two sizes separated by comma are neighboring. Output If it is not possible to present a t-shirt to each participant, print «NO» (without quotes). Otherwise, print n + 1 lines. In the first line print «YES» (without quotes). In the following n lines print the t-shirt sizes the orginizers should give to participants, one per line. The order of the participants should be the same as in the input. If there are multiple solutions, print any of them. Examples Input 0 1 0 1 1 0 3 XL S,M XL,XXL Output YES XL M XXL Input 1 1 2 0 1 1 5 S M S,M XXL,XXXL XL,XXL Output NO Submitted Solution: ``` import copy sizes = list(map(int, input().split())) participants = {} participants_list = [] N = int(input()) for i in range(N): s = input() size_string = s d = participants.get(size_string, 0) participants[size_string] = d + 1 participants_list.append(size_string) single_sizes = ["S", "M", "L", "XL", "XXL", "XXXL"] multiple_sizes = [ single_sizes[i] + "," + single_sizes[i + 1] for i in range(len(single_sizes) - 1) ] size_to_index = { "S": 0, "M": 1, "L": 2, "XL": 3, "XXL": 4, "XXXL": 5, } # print(sizes) # print(participants) # print(single_sizes) # print(multiple_sizes) multiple_answers = {} sizes_copy = copy.copy(sizes) res = "YES" for single_size in single_sizes: nb_participants = participants.get(single_size, 0) ind = size_to_index[single_size] if nb_participants > sizes_copy[ind]: res = "NO" break sizes_copy[ind] -= nb_participants for multiple_size in multiple_sizes: size_1, size_2 = multiple_size.split(",") nb_participants = participants.get(multiple_size, 0) ind_1 = size_to_index[size_1] first_size = min(sizes_copy[ind_1], nb_participants) sizes_copy[ind_1] -= first_size nb_participants -= first_size ind_2 = size_to_index[size_2] if nb_participants > sizes_copy[ind_2]: res = "NO" break sizes_copy[ind_2] -= nb_participants multiple_answers[multiple_size] = first_size, nb_participants print(res) # print(multiple_answers) if res == "YES": for participant_size in participants_list: if "," in participant_size: size_1, size_2 = participant_size.split(",") r1, r2 = multiple_answers.get(participant_size, 0) if r1 > 0: print(size_1) r1 -= 1 else: print(size_2) r2 -= 1 multiple_answers[participant_size] = r1, r2 else: print(participant_size) ```
instruction
0
25,245
11
50,490
Yes
output
1
25,245
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50,491
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. The organizers of a programming contest have decided to present t-shirts to participants. There are six different t-shirts sizes in this problem: S, M, L, XL, XXL, XXXL (sizes are listed in increasing order). The t-shirts are already prepared. For each size from S to XXXL you are given the number of t-shirts of this size. During the registration, the organizers asked each of the n participants about the t-shirt size he wants. If a participant hesitated between two sizes, he could specify two neighboring sizes — this means that any of these two sizes suits him. Write a program that will determine whether it is possible to present a t-shirt to each participant of the competition, or not. Of course, each participant should get a t-shirt of proper size: * the size he wanted, if he specified one size; * any of the two neibouring sizes, if he specified two sizes. If it is possible, the program should find any valid distribution of the t-shirts. Input The first line of the input contains six non-negative integers — the number of t-shirts of each size. The numbers are given for the sizes S, M, L, XL, XXL, XXXL, respectively. The total number of t-shirts doesn't exceed 100 000. The second line contains positive integer n (1 ≤ n ≤ 100 000) — the number of participants. The following n lines contain the sizes specified by the participants, one line per participant. The i-th line contains information provided by the i-th participant: single size or two sizes separated by comma (without any spaces). If there are two sizes, the sizes are written in increasing order. It is guaranteed that two sizes separated by comma are neighboring. Output If it is not possible to present a t-shirt to each participant, print «NO» (without quotes). Otherwise, print n + 1 lines. In the first line print «YES» (without quotes). In the following n lines print the t-shirt sizes the orginizers should give to participants, one per line. The order of the participants should be the same as in the input. If there are multiple solutions, print any of them. Examples Input 0 1 0 1 1 0 3 XL S,M XL,XXL Output YES XL M XXL Input 1 1 2 0 1 1 5 S M S,M XXL,XXXL XL,XXL Output NO Submitted Solution: ``` # print(list(zip(input().split(), "S M L XL XXL XXXL".split()))) SIZES = "S M L XL XXL XXXL".split() have = {key:int(count) for count,key in zip(input().split(), SIZES)} n = int(input()) possible = [] uses = [] for size in range(n): possible.append(input().split(',')) if len(possible[-1]) > 1: uses.append(None) else: uses.append(possible[-1][0]) have[possible[-1][0]] -= 1 if have[possible[-1][0]] < 0: print("NO") exit(0) possible[-1] = None # print(*uses) for size in SIZES: for i in range(2): for index, poss in enumerate(possible): if poss is not None: if have[size] > 0: if (i == 0 and size == poss[1]) or (i == 1 and size in poss): have[size] -= 1 uses[index] = size possible[index] = None else: break # print(*uses) if None in uses: print("NO") else: print("YES") for i in uses: print(i) ```
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Yes
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