problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 262k
1.05M
| problem_description
stringlengths 48
1.55k
| codes
stringlengths 35
98.9k
| status
stringlengths 28
1.7k
| submission_ids
stringlengths 28
1.41k
| memories
stringlengths 13
808
| cpu_times
stringlengths 11
610
| code_sizes
stringlengths 7
505
|
|---|---|---|---|---|---|---|---|---|---|---|
p03424 | u839873388 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['S = list(input().split())\nc = "kara"\nif S == "Y":\n\tc = "Four"\nelse :\n\tc = "Three"\n\t\nprint(c)', 'n = int(input())\nS = list(input().split())\nc = "kara"\nfor i in range(n):\n\tif S[i] == "Y":\n\t\tc = "Four"\n\telse :\n\t\tc = "Three"\n\t\nprint(c)', 'n = int(input())\nS = list(input().split())\nc = "Three"\nfor i in range(n):\n\tif S[i] == "Y":\n\t\tc = "Four"\n\nprint(c)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s176389082', 's548394155', 's020004580'] | [2940.0, 2940.0, 2940.0] | [18.0, 17.0, 17.0] | [92, 135, 113] |
p03424 | u842118372 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n=input()\ns={str(x) for x in input().split()}\nprint(len(s))', 'n=input()\ns={str(x) for x in input().split()}\nif len(s) == 4:\n print("Four")\nelse:\n print("Three")'] | ['Wrong Answer', 'Accepted'] | ['s981261545', 's093501130'] | [2940.0, 2940.0] | [17.0, 17.0] | [59, 104] |
p03424 | u845427284 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n = int(input())\ns = list(map(int, input().split()))\nprint("Four" if "Y" in s else "Three")', 'n = int(input())\ns = list(input().split())\nprint("Four" if "Y" in s else "Three")'] | ['Runtime Error', 'Accepted'] | ['s086269831', 's927736375'] | [2940.0, 2940.0] | [17.0, 20.0] | [91, 81] |
p03424 | u856555908 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['x = input().split(" ")\nt = True\nfor i in x:\n if i == \'Y\':\n t = False\n\nif t:\n print("Three")\nelse:\n print("Four")', 'x = input().split(" ")\nt = True\nfor i in x:\n if i == \'Y\':\n t = False\n continue\n\nif t:\n print("Three")\nelse:\n print("Four")', 'y = input()\nx = input().split(" ")\nt = True\nfor i in x:\n if i == \'Y\':\n t = False\n continue\n\nif t:\n print("Three")\nelse:\n print("Four")'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s424480473', 's898974839', 's274540697'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 19.0] | [128, 145, 157] |
p03424 | u870286225 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["# -*- coding: utf-8 -*-\n\nN = input()\n\nS = raw_input().split()\n\nif 'Y' in S:\n\tprint('Four')\nelse:\n\tprint('Three')", '# -*- coding: utf-8 -*-\n\nN = input()\n\nS = raw_input().split()\n\nif "Y" in S:\n\tprint(\'Four\')\nelse:\n\tprint(\'Three\')', '# -*- coding: utf-8 -*-\n\nN = input()\n\nS = input().split()\n\nif "Y" in S:\n\tprint(\'Four\')\nelse:\n\tprint(\'Three\')'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s921820664', 's925061858', 's122204821'] | [2940.0, 2940.0, 2940.0] | [19.0, 17.0, 17.0] | [175, 175, 171] |
p03424 | u870518235 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\nS = list(map(int, input().split()))\n\nif "Y" in S:\n print("Four")\nelse:\n print("Three")\n', 'N = int(input())\nS = list(map(str, input().split()))\n\nif "Y" in S:\n print("Four")\nelse:\n print("Three")\n'] | ['Runtime Error', 'Accepted'] | ['s702937056', 's140428428'] | [9028.0, 9064.0] | [25.0, 26.0] | [110, 110] |
p03424 | u874333466 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\nS = list(map(str, input().split()))\n\ncolor = []\n\nfor i in S:\n if i in color:\n pass\n else:\n color.append(i)\n \nprint(len(color))', "N = int(input())\nS = list(map(str, input().split()))\n\ncolor = []\n\nfor i in S:\n if i in color:\n pass\n else:\n color.append(i)\n \nif len(color) == 4:\n print('Four')\nelse:\n print('Three')\n\n"] | ['Wrong Answer', 'Accepted'] | ['s448503042', 's021279097'] | [9072.0, 9016.0] | [28.0, 27.0] | [154, 197] |
p03424 | u879870653 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\nS = list(map(int,input().split()))\np = S.count("P")\nw = S.count("W")\ng = S.count("G")\ny = S>count("Y")\nans = p+w+g+y\nif ans == 3 :\n print("Three")\nelse :\n print("Four")\n ', 'N = int(input())\nS = list(map(str,input().split()))\np = S.count("P")\nw = S.count("W")\ng = S.count("G")\ny = S>count("Y")\nans = p+w+g+y\nif ans == 3 :\n print("Three")\nelse :\n print("Four")\n ', 'N = int(input())\nS = list(map(str,input().split()))\nS = list(set(S))\np = S.count("P")\nw = S.count("W")\ng = S.count("G")\ny = S.count("Y")\nans = p+w+g+y\nif ans == 3 :\n print("Three")\nelse :\n print("Four")\n '] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s501600363', 's569365379', 's954734945'] | [3060.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0] | [196, 196, 213] |
p03424 | u883792993 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["N=int(input())\nS=list(map(int,input().split()))\nbucket=[]\nfor i in range(N):\n if S[i]=='P':\n bucket[0]+=1\n elif S[i]=='W':\n bucket[1]+=1\n elif S[i]=='G':\n bucket[2]+=1\n else:\n bucket[3]+=1\nif 4 - bucket.count(0)==3:\n print('Three')\nelse:\n print('Four')", "N=int(input())\nS=list(map(str,input().split()))\nbucket=[0*i for i in range(4)]\nfor i in range(N):\n if S[i]=='P':\n bucket[0]+=1\n elif S[i]=='W':\n bucket[1]+=1\n elif S[i]=='G':\n bucket[2]+=1\n else:\n bucket[3]+=1\nif 4 - bucket.count(0)==3:\n print('Three')\nelse:\n print('Four')"] | ['Runtime Error', 'Accepted'] | ['s972563215', 's531580642'] | [3060.0, 3064.0] | [18.0, 18.0] | [298, 319] |
p03424 | u893063840 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n = int(input())\ns = list(map(int, input().split()))\n\nprint("Four" if "Y" in s else "Three")\n', 'n = int(input())\ns = list(input().split())\n\nprint("Four" if "Y" in s else "Three")\n'] | ['Runtime Error', 'Accepted'] | ['s597514940', 's510482055'] | [2940.0, 2940.0] | [17.0, 17.0] | [93, 83] |
p03424 | u898058223 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n=int(input())\na=list(map(int,input().split()))\nif len(set(a))==3:\n print("Three")\nelse:\n print("Four")', 'n=int(input())\na=list(input().split())\nif len(set(a))==3:\n print("Three")\nelse:\n print("Four")'] | ['Runtime Error', 'Accepted'] | ['s927995920', 's858608043'] | [9088.0, 8916.0] | [27.0, 29.0] | [105, 96] |
p03424 | u903005414 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["N = int(input())\nA = list(map(int, input().split()))\nfor a in A:\n if a == 'Y':\n print('Four')\n exit()\nprint('Three')", "N = int(input())\nA = input().split()\nfor a in A:\n if a == 'Y':\n print('Four')\n exit()\nprint('Three')\n"] | ['Runtime Error', 'Accepted'] | ['s034683727', 's379380604'] | [2940.0, 2940.0] | [17.0, 17.0] | [123, 108] |
p03424 | u905582793 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n=int(input())\nh=list(map(input().split()))\nif len(list(set(h)))==3:\n print("Three")\nelse:\n print("Four")\n', 'n=int(input())\nh=list(map(int,input().split()))\nif len(list(set(h)))==3:\n print("Three")\nelse:\n print("Four")', 'n=int(input())\nh=list(input().split())\nif len(list(set(h)))==3:\n print("Three")\nelse:\n print("Four")\n'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s536801223', 's928173917', 's273299602'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [108, 111, 103] |
p03424 | u918601425 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N=int(input())\nfor i in range(N):\n x=int(input())\n if x=="Y":\n print("Four")\n break\nelse:\n print("Three")\n ', 'N=int(input())\nls=input().split()\nfor i in range(N):\n x=ls[i]\n if x=="Y":\n print("Four")\n break\nelse:\n print("Three")\n \n'] | ['Runtime Error', 'Accepted'] | ['s249769279', 's579793025'] | [2940.0, 2940.0] | [17.0, 17.0] | [117, 130] |
p03424 | u923279197 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["n=int(input())\ns=list(input().split())\nif 'Y' in s:\n print('four')\nelse:\n print('three')", "n=int(input())\ns=list(input().split())\nif 'Y' in s:\n print('Four')\nelse:\n print('Three')"] | ['Wrong Answer', 'Accepted'] | ['s941294070', 's656184678'] | [2940.0, 3188.0] | [17.0, 22.0] | [94, 94] |
p03424 | u928784113 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['# -*- coding: utf-8 -*-\nN = int(input())\na = []\nfor i in range(1,N+1):\n a.append(input())\n \nif a.count("Y") == 0:\n print("Three")\nelse:\n print("Four")', 'N = int(input())\nS = list(map(int,input().split()))\nif S.count("Y") == 0:\n print("Three")\nelse:\n print("Four")', '# -*- coding: utf-8 -*-\nN = int(input())\na = []\nfor i in range(1,N+1):\n a.append(input())\nif a.count("Y") == 0:\n print("Three")\nelse:\n print("Four")', 'N = int(input())\na = [input() for i in range(1:N)]\nif a.count("Y") == 0:\n print("Three")\nelse:\n print("Four")', 'N = int(input())\nfor i in (list(input().split())):\n if i == "Y":\n print("Four")\n exit()\n\nprint("Three")'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s119836061', 's474514779', 's531647343', 's824786731', 's525302705'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [18.0, 17.0, 18.0, 18.0, 17.0] | [154, 112, 151, 111, 140] |
p03424 | u929618357 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\nstr = input().split()\nprint(str)\nif \'Y\' in str:\n print("Four")\nelse:\n print("Three")\n', 'N = int(input())\nstr = input().split()\nif \'Y\' in str:\n print("Four")\nelse:\n print("Three")'] | ['Wrong Answer', 'Accepted'] | ['s382633175', 's073612604'] | [2940.0, 2940.0] | [17.0, 17.0] | [108, 96] |
p03424 | u931462344 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n = input()\nprint(len(set(input().split())))', 'n = input()\nprint(len(set(input().split()))', 'n = input()\nif len(set(input().split())) == 3:\n\tprint("Three")\nelse:\n\tprint("Four") '] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s567501438', 's758852222', 's658228905'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 18.0] | [44, 43, 85] |
p03424 | u931938233 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["input()\nprint('Three' if 'Y' in input().split() else 'Four')", "input()\nprint('Four' if 'Y' in input().split() else 'Three')"] | ['Wrong Answer', 'Accepted'] | ['s626265391', 's582906629'] | [9028.0, 8992.0] | [29.0, 30.0] | [60, 60] |
p03424 | u936985471 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n=input()\na=input().split()\nprint(("Three","Four")[len(set(a))==4]', 'n=input()\na=input().split()\nprint(("Three","Four")[len(set(a))==4])\n'] | ['Runtime Error', 'Accepted'] | ['s710351336', 's447202396'] | [2940.0, 2940.0] | [17.0, 18.0] | [66, 68] |
p03424 | u940102677 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n = int(input())\na = list(input())\nflag = False\nfor i in range(n):\n if a[i] == "Y":\n flag = True\n\nif flag == Ture:\n print("Four")\nelse:\n print("Three")', 'n = int(input())\na = list(input())\nfor i in range(n):\n if a[i] == "Y":\n print("Four")\n exit()\n \nprint("Three")'] | ['Runtime Error', 'Accepted'] | ['s566305442', 's331412867'] | [2940.0, 2940.0] | [17.0, 17.0] | [157, 120] |
p03424 | u941753895 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["\nN=int(input())\nl=list(map(int,input().split()))\n\nln=len(set(l))\n\nif ln==3:\n print('Three')\nelse:\n print('Four')", "\nN=int(input())\nl=list(map(str,input().split()))\n\nln=len(set(l))\n\nif ln==3:\n print('Three')\nelse:\n print('Four')"] | ['Runtime Error', 'Accepted'] | ['s313497515', 's244554914'] | [2940.0, 2940.0] | [17.0, 17.0] | [122, 122] |
p03424 | u957872856 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\nS = input().split()\nprint(S)\nif len(set(S)) == 3:\n print("Three")\nelse:\n print("Four")\n', 'N = int(input())\nS = list(input())\nprint(S)\nif len(set(S)) == 3:\n print("Three")\nelse:\n print("Four")\n', 'N = int(input())\nS = input().split()\nif len(set(S)) == 3:\n print("Three")\nelse:\n print("Four")'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s897790429', 's970479814', 's657917548'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 20.0] | [106, 104, 96] |
p03424 | u958053648 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N=int(input())\nS=list(map(str,input().split()))\nSS=set(S)\nlength=len(SS)\nprint("four" if length==4 else "three")', 'N=int(input())\nS=list(map(str,input().split()))\nSS=set(S)\nlength=len(SS)\nprint("Four" if length==4 else "Three")'] | ['Wrong Answer', 'Accepted'] | ['s650448800', 's053186246'] | [2940.0, 2940.0] | [17.0, 17.0] | [112, 112] |
p03424 | u966891144 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["num = int(input())\ndata = input().rstrip().split(' ')\ncount = 0\nif data.count('P') > 0 or data.count('W') > 0 or data.count('G') > 0 or data.count('Y') > 0:\n count++\nprint(count)", "num = int(input())\ndata = input().rstrip().split(' ')\ncount = 0\nif data.count('P') > 0 or data.count('W') > 0 or data.count('G') > 0 or data.count('Y') > 0:\n count++\n\nif count == 3:\n print('Three')\nelif count == 4:\n print('Four')", 'N = int(input())\ns = input().split()\nif "Y" in s:\n print("Four")\nelse:\n print("Three")'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s391921524', 's752492466', 's789957904'] | [2940.0, 2940.0, 2940.0] | [18.0, 17.0, 17.0] | [179, 232, 92] |
p03424 | u969708690 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N=int(input())\nL=list(map(int,input().split()))\nL=set(L)\nif len(L)==4:\n print("Four")\nelse:\n print("Three")', 'N=int(input())\nL=list(input().split())\nL=set(L)\nif len(L)==4:\n print("Four")\nelse:\n print("Three")'] | ['Runtime Error', 'Accepted'] | ['s061182922', 's456469309'] | [9168.0, 9076.0] | [23.0, 28.0] | [109, 100] |
p03424 | u969848070 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["n = int(input())\nk = len(set(map(str, input().split())))\nif k = 4:\n print('Four')\nelse:\n print('Three')\n", "n = int(input())\nk = len(set(map(str, input().split())))\nif k == 4:\n print('Four')\nelse:\n print('Three')"] | ['Runtime Error', 'Accepted'] | ['s283880314', 's089897590'] | [8940.0, 9156.0] | [25.0, 27.0] | [106, 106] |
p03424 | u970198631 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["N = int(input())\nM = set()\nMM = input().split()\nfor i in MM:\n M.add(i)\nprint(len(M))\nif len(M) == 3:\n print('Three')\nelse:\n print('Four')", "N = int(input())\nM = set()\nMM = input().split()\nfor i in MM:\n M.add(i)\n\nif len(M) == 3:\n print('Three')\nelse:\n print('Four')"] | ['Wrong Answer', 'Accepted'] | ['s723140112', 's198536610'] | [9084.0, 9084.0] | [26.0, 27.0] | [140, 127] |
p03424 | u970809473 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["n = int(input())\na = list(map(int, input().split()))\nif a.count('Y') == 0:\n print('Three')\nelse:\n print('Four')", "n = int(input())\na = list(map(str, input().split()))\nif a.count('Y') == 0:\n print('Three')\nelse:\n print('Four')"] | ['Runtime Error', 'Accepted'] | ['s361603688', 's297361778'] | [2940.0, 2940.0] | [18.0, 17.0] | [113, 113] |
p03424 | u972892985 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['n = int(input())\ns = list(input().split())\n\nif len(set(s)) == 3:\n print("Four")\nelse:\n print("Three")\n', 'n = int(input())\ns = list[input().split()]\n\nif len(set(s)) == "Y":\n print("Four")\nelse:\n print("Three")', 'n = int(input())\nfor i in range(n):\n s = input():\n if s == "Y":\n print("Four")\n break\n else:\n print("Three")', 'n = int(input())\ns = list(input().split())\n\nif len(set(s)) == 3:\n print("Three")\nelse:\n print("Four")\n'] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s518872026', 's602053347', 's945693219', 's484743591'] | [2940.0, 2940.0, 2940.0, 3316.0] | [17.0, 17.0, 17.0, 21.0] | [104, 105, 130, 104] |
p03424 | u975012184 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["n = int(input())\nok = False\nfor i in range(n):\n t = input()\n if(t == 'Y'):\n ok = True\n break\n \nif(ok):\n print('Four')\nelse:\n print('Three')", "n = int(input())\nt = 'Y'\nx = input().split()\n \nif(t in x):\n print('Four')\nelse:\n print('Three')"] | ['Runtime Error', 'Accepted'] | ['s811767624', 's097487923'] | [2940.0, 3064.0] | [17.0, 17.0] | [152, 100] |
p03424 | u977349332 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["N = int(input())\nS = input().split()\n\nfor x in S:\n if x is 'Y'\n \tprint('Four')\n exit()\n else:\n print('Three')", "N = int(input())\nS = list(input().split())\n\nprint('Four' if 'Y' in S else 'Three')"] | ['Runtime Error', 'Accepted'] | ['s327451407', 's280041581'] | [2940.0, 3064.0] | [17.0, 17.0] | [118, 82] |
p03424 | u977642052 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['def main(n, s):\n a = set(s)\n\n if len(a) == 3:\n print(\'Three\')\n elif len(a) == 4:\n print(\'Four\')\n\n\nif __name__ == "__main__":\n n = int(input())\n s = list(map(int, input().split()))\n\n main(n, s)\n', 'def main(n, s):\n a = set(s)\n\n if len(a) == 3:\n print(\'Three\')\n elif len(a) == 4:\n print(\'Four\')\n\n\nif __name__ == "__main__":\n n = int(input())\n s = list(input().split())\n\n main(n, s)\n'] | ['Runtime Error', 'Accepted'] | ['s919772639', 's504486497'] | [2940.0, 2940.0] | [18.0, 17.0] | [225, 215] |
p03424 | u988832865 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["N = int(input())\nS = set(map(int, input().split()))\n\nprint('Three' if len(S) == 3 else 'Four')\n", "N = int(input())\nS = set(input().split())\n\nprint('Three' if len(S) == 3 else 'Four')\n"] | ['Runtime Error', 'Accepted'] | ['s842018768', 's316162435'] | [3316.0, 2940.0] | [19.0, 19.0] | [95, 85] |
p03424 | u993461026 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['N = int(input())\narare = list(map(int, input().rstrip().split()))\ncount = {"P": 0, "W": 0, "G": 0, "Y": 0}\nfor i in range(N):\n cur = arare[i]\n if cur == "P":\n count["P"] += 1\n elif cur == "W":\n count["W"] += 1\n elif cur == "G":\n count["G"] += 1\n else:\n count["Y"] += 1\nprint("Four" if count["Y"] >= 1 else "Three")', 'N = int(input())\narare = input().rstrip().split()\ncount = {"P": 0, "W": 0, "G": 0, "Y": 0}\nfor i in range(N):\n cur = arare[i]\n if cur == "P":\n count["P"] += 1\n elif cur == "W":\n count["W"] += 1\n elif cur == "G":\n count["G"] += 1\n else:\n count["Y"] += 1\nprint("Four" if count["Y"] >= 1 else "Three")'] | ['Runtime Error', 'Accepted'] | ['s390431221', 's230849859'] | [3064.0, 3064.0] | [17.0, 17.0] | [331, 315] |
p03424 | u994988729 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ['from collections import Counter\nN=input()\nc = Counter(list(map(int,input().split())))\nif len(c)==4:\n print("Four")\nelse:\n print("Three")', 'N = int(input())\nA = input().split()\n\nif len(set(A)) <= 3:\n print("Three")\nelse:\n print("Four")\n'] | ['Runtime Error', 'Accepted'] | ['s021019725', 's128369381'] | [3316.0, 2940.0] | [20.0, 17.0] | [138, 102] |
p03424 | u999893056 | 2,000 | 262,144 | In Japan, people make offerings called _hina arare_ , colorful crackers, on March 3. We have a bag that contains N hina arare. (From here, we call them arare.) It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow. We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: `P`, white: `W`, green: `G`, yellow: `Y`. If the number of colors of the arare in the bag was three, print `Three`; if the number of colors was four, print `Four`. | ["n = int(input())\n\ns_list = list(input().split())\ns_set = set(s_list)\nanswer = len(list(s_set))\nprint('Four' if answer == 3 else 'Three')", "n = int(input())\n\ns_list = list(input().split())\ns_set = set(s_list)\nanswer = len(list(s_set))\nprint('Four' if answer == 4 else 'Three')"] | ['Wrong Answer', 'Accepted'] | ['s350085829', 's494938109'] | [3060.0, 2940.0] | [19.0, 17.0] | [136, 136] |
p03425 | u000623733 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\nN = int(input())\n\nnum_dic = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}\nfor i in range(N):\n S = input()\n if S[0] in num_dic.keys():\n num_dic[S[0]] += 1\n\nans = 0\nprint(num_dic)\nfor i, j, k in itertools.combinations(num_dic.keys(), 3):\n ans += num_dic[i] * num_dic[j] * num_dic[k]\nprint(ans)\n", "import itertools\nN = int(input())\n\nnum_dic = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}\nfor i in range(N):\n S = input()\n if S[0] in num_dic.keys():\n num_dic[S[0]] += 1\n\nans = 0\nfor i, j, k in itertools.combinations(num_dic.keys(), 3):\n ans += num_dic[i] * num_dic[j] * num_dic[k]\nprint(ans)\n"] | ['Wrong Answer', 'Accepted'] | ['s301354020', 's576146514'] | [3060.0, 3060.0] | [175.0, 184.0] | [320, 305] |
p03425 | u017415492 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['using System;\nusing static System.Console;\nusing System.Linq;\nusing System.Collections.Generic;\nclass Program\n{\n static void Main(string[] args)\n {\n int n = int.Parse(ReadLine());\n List<string> s = new List<string>();\n \n for (int i=0;i<n;i++)\n {\n s.Add(ReadLine());\n }\n var hashSet = new HashSet<string>(s);\n s = hashSet.ToList();\n List<int> ans = new List<int>();\n \n for (int i=0;i<5;i++)\n {\n ans.Add(0);\n }\n \n for (int i=0;i<s.Count;i++)\n {\n if (s[i].Substring(0,1)=="M")\n {\n ans[0]+=1;\n }\n else if (s[i].Substring(0,1)=="A")\n {\n ans[1]+=1;\n }\n else if (s[i].Substring(0,1)=="R")\n {\n ans[2]+=1;\n }\n else if (s[i].Substring(0,1)=="C")\n {\n ans[3]+=1;\n }\n else if (s[i].Substring(0,1)=="H")\n {\n ans[4]+=1;\n }\n }\n int answ=0;\n for (int i=0;i<5;i++)\n {\n for (int j=i+1;j<5;j++)\n {\n for (int k=j+1;k<5;k++)\n {\n if (i!=j && j!=k && i!=k)\n {\n answ+=ans[i]*ans[j]*ans[k];\n }\n }\n }\n }\n WriteLine(answ);\n }\n}', 'import itertools\nn=int(input())\nm=list("MARCH")\nd=[0]*5\nfor i in range(n):\n s=input()\n for j in range(len(m)):\n if s[0]==m[j]:\n d[j]+=1\nans=0\nfor i in list(itertools.combinations(list(range(5)), 3)):\n k=list(i)\n ans+=d[k[0]]*d[k[1]]*d[k[2]]\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s778935506', 's752734191'] | [8940.0, 9220.0] | [24.0, 218.0] | [1276, 263] |
p03425 | u023540496 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = int(input())\n\nlist_pop = [0] * 5\nfor i in range(N):\n s = input()\n if list(s)[0] == 'M':\n list_pop[0] += 1\n if list(s)[0] == 'A':\n list_pop[1] += 1\n if list(s)[0] == 'R':\n list_pop[2] += 1\n if list(s)[0] == 'C':\n list_pop[3] += 1\n if list(s)[0] == 'H':\n list_pop[4] += 1\n\n\nprint(list_pop)\n\nanswer = 0\nfor i in range(3):\n for j in range(4)[i+1:]:\n for k in range(5)[j+1:]:\n answer += list_pop[i] * list_pop[j] * list_pop[k]\n \n \nprint(int(answer))", "N = int(input())\n\nlist_pop = [0] * 5\nfor i in range(N):\n s = input()\n if list(s)[0] == 'M':\n list_pop[0] += 1\n if list(s)[0] == 'A':\n list_pop[1] += 1\n if list(s)[0] == 'R':\n list_pop[2] += 1\n if list(s)[0] == 'C':\n list_pop[3] += 1\n if list(s)[0] == 'H':\n list_pop[4] += 1\n\n\n\n\nanswer = 0\nfor i in range(3):\n for j in range(4)[i+1:]:\n for k in range(5)[j+1:]:\n answer += list_pop[i] * list_pop[j] * list_pop[k]\n \n \nprint(int(answer))"] | ['Wrong Answer', 'Accepted'] | ['s720111898', 's071878213'] | [3064.0, 3444.0] | [378.0, 362.0] | [528, 513] |
p03425 | u027622859 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = int(input())\ns = [input()[0] for _ in range(N)]\nt = []\nans = 0\nfor k in 'MARCH':\n t.append(s.count(k))\nfor i in range(32):\n u = []\n for j in range(5):\n u.append(t[j])\n if len(u) == 3:\n ans += u[0]*u[1]*u[2]\nprint(ans)", "N = int(input())\ns = [input()[0] for _ in range(N)]\nt = []\nans = 0\nfor k in 'MARCH':\n t.append(s.count(k))\nfor i in range(32):\n u = []\n for j in range(5):\n if i&(1<<j):\n u.append(t[j])\n if len(u) == 3:\n ans += u[0]*u[1]*u[2]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s380261374', 's941451285'] | [3864.0, 3864.0] | [145.0, 139.0] | [247, 272] |
p03425 | u030726788 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['import sys\nN=int(input())\nS=[]\nfor i in range(N):\n S.append(input()[0])\n \nCount=[]\nCount.append(S.count("M"))\nCount.append(S.count("A"))\nCount.append(S.count("R"))\nCount.append(S.count("C"))\nCount.append(S.count("H"))\n\nZero=Count.count(0)\nif(Zero>=3):print(0)\nelif(Zero==2):\n Count.remove(0)\n Count.remove(0)\n pat=1\n for i in Count:pat*=i\n print(pat)\nelif(Zero==1):\n Count.remove(0)\n patA=1\n for i in Count:patA*=i\n pat=0\n for i in Count:pat+=patA//i\n print(pat)\nelif(Zero==0):\n patA=1\n for i in Count:patA*=i\n pat=0\n for i in range(5):\n for j in range(5):\n if(i==j):continue\n else:\n pat+=patA//Count[i]//Count[j]\n print(pat//2)\nprint(Count)', 'import sys\nN=int(input())\nS=[]\nfor i in range(N):\n S.append(input()[0])\n \nCount=[]\nCount.append(S.count("M"))\nCount.append(S.count("A"))\nCount.append(S.count("R"))\nCount.append(S.count("C"))\nCount.append(S.count("H"))\n\nZero=Count.count(0)\nif(Zero>=3):print(0)\nelif(Zero==2):\n Count.remove(0)\n Count.remove(0)\n pat=1\n for i in Count:pat*=i\n print(pat)\nelif(Zero==1):\n Count.remove(0)\n patA=1\n for i in Count:patA*=i\n pat=0\n for i in Count:pat+=patA//i\n print(pat)\nelif(Zero==0):\n patA=1\n for i in Count:patA*=i\n pat=0\n for i in range(5):\n for j in range(5):\n if(i==j):continue\n else:\n pat+=patA//Count[i]//Count[j]\n print(pat//2)'] | ['Wrong Answer', 'Accepted'] | ['s095885488', 's061545800'] | [4016.0, 4016.0] | [149.0, 145.0] | [739, 726] |
p03425 | u036340997 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n = int(input())\nmarch = {'m': 0, 'a': 0, 'r': 0, 'c': 0, 'h': 0}\nfor i in range(n):\n s = input()[0]\n try:\n march[s] += 1\n except:\n pass\nans = 0\n\nfor key1 in march.keys():\n for key2 in march.keys():\n if key1 != key2:\n for key3 in march.keys():\n if key3 != key1 and key3 != key2:\n ans += key1 * key2 * key3\nprint(ans)", "n = int(input())\nmarch = {'m': 0, 'a': 0, 'r': 0, 'c': 0, 'h': 0}\nfor i in range(n):\n s = input()[0].lower()\n try:\n march[s] += 1\n except:\n pass\nans = 0\n\nfor key1 in march.keys():\n for key2 in march.keys():\n if key1 != key2:\n for key3 in march.keys():\n if key3 != key1 and key3 != key2:\n ans += march[key1] * march[key2] * march[key3]\nprint(ans//6)"] | ['Runtime Error', 'Accepted'] | ['s260690327', 's041562395'] | [3060.0, 3188.0] | [196.0, 241.0] | [350, 382] |
p03425 | u046187684 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n = int(input().strip())\ns = [input().strip() for _ in range(n)]\nnum = [0 for _ in range(5)]\nnum[0] = len([True for _s in s if _s[0] == "M"])\nnum[1] = len([True for _s in s if _s[0] == "A"])\nnum[2] = len([True for _s in s if _s[0] == "R"])\nnum[3] = len([True for _s in s if _s[0] == "C"])\nnum[4] = len([True for _s in s if _s[0] == "H"])\nans = 0\nfor i in range(5):\n for j in range(i, 5):\n for k in range(j, 5):\n ans += num[i] * num[j] * num[k]\nprint(ans', 'def solve(string):\n n, *s = string.split()\n num = [len([True for _s in s if _s[0] == _i]) for _i in "M A R C H".split()]\n ans = 0\n for i in range(5):\n for j in range(i + 1, 5):\n for k in range(j + 1, 5):\n ans += num[i] * num[j] * num[k]\n return str(ans)\n\n\nif __name__ == \'__main__\':\n print(solve(input()))\n', 'n = int(input().strip())\ns = [input().strip() for _ in range(n)]\nnum_m = len([True for _s in s if _s[0] == "M"])\nnum_a = len([True for _s in s if _s[0] == "A"])\nnum_r = len([True for _s in s if _s[0] == "R"])\nnum_c = len([True for _s in s if _s[0] == "C"])\nnum_h = len([True for _s in s if _s[0] == "H"])\nprint(num_m * num_a * num_r + num_m * num_a * num_c + num_m * num_a * num_h + num_m * num_r * num_c\n + num_m * num_r * num_h + num_m * num_c * num_h + num_a * num_r * num_c +\n num_a * num_r * num_h + num_a * num_c * num_h + num_r * num_c * num_h', 'n = int(input().strip())\ns = [input().strip() for _ in range(n)]\nnum = [0 for _ in range(5)]\nnum[0] = len([True for _s in s if _s[0] == "M"])\nnum[1] = len([True for _s in s if _s[0] == "A"])\nnum[2] = len([True for _s in s if _s[0] == "R"])\nnum[3] = len([True for _s in s if _s[0] == "C"])\nnum[4] = len([True for _s in s if _s[0] == "H"])\nans = 0\nfor i in range(5):\n for j in range(i, 5):\n for k in range(j, 5):\n ans += num[i] * num[j] * num[k]\nprint(ans)', 'from functools import reduce\nfrom itertools import combinations\nfrom operator import mul\n\n\ndef solve(string):\n n, *s = string.split()\n # num = [len([True for _s in s if _s[0] == _i]) for _i in "M A R C H".split()]\n initial = "MARCH"\n num = [0 for _ in range(6)]\n for _s in s:\n i = initial.find(_s[0])\n num[i] += 1\n return str(sum([reduce(mul, n) for n in combinations(num[:5], 3)]))\n\n\nif __name__ == \'__main__\':\n n = int(input())\n print(solve(\'{}\\n\'.format(n) + \'\\n\'.join([input() for _ in range(n)])))\n'] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s011305958', 's207168321', 's779161061', 's803889605', 's161963818'] | [3064.0, 3060.0, 3064.0, 10520.0, 12468.0] | [17.0, 17.0, 17.0, 170.0, 188.0] | [474, 357, 560, 475, 541] |
p03425 | u059210959 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['#encoding:utf-8\n\nimport math\nimport itertools\n\nn = int(input())\ns = []\nfor i in range(n):\n s.append(str(input()))\n\nchosen_names = {"m":0,"a":0,"r":0,"c":0,"h":0}\n\nfor name in s:\n if name[0] == "M":\n chosen_names["m"] += 1\n elif name[0] == "A":\n chosen_names["a"] += 1\n elif name[0] == "R":\n chosen_names["r"] += 1\n elif name[0] == "C":\n chosen_names["c"] += 1\n elif name[0] == "H":\n chosen_names["h"] += 1\n\nanswer = 0\ncapitals = list("march")\ncapital_set = itertools.com\nbinations(capitals,3)\nfor capital in capital_set:\n answer += chosen_names[capital[0]] * chosen_names[capital[1]] * chosen_names[capital[2]]\n\n\nprint(answer)\n\n\n\n\n# return math.factorial(n) // (math.factorial(r) * math.factorial(n-r))\n', '#encoding:utf-8\n\nimport math\nimport itertools\n\nn = int(input())\ns = []\nfor i in range(n):\n s.append(str(input()))\n\nchosen_names = {"m":0,"a":0,"r":0,"c":0,"h":0}\n\nfor name in s:\n if name[0] == "M":\n chosen_names["m"] += 1\n elif name[0] == "A":\n chosen_names["a"] += 1\n elif name[0] == "R":\n chosen_names["r"] += 1\n elif name[0] == "C":\n chosen_names["c"] += 1\n elif name[0] == "H":\n chosen_names["h"] += 1\n\nanswer = 0\ncapitals = list("march")\ncapital_set = itertools.combinations(capitals,3)\nfor capital in capital_set:\n answer += chosen_names[capital[0]] * chosen_names[capital[1]] * chosen_names[capital[2]]\n\n\nprint(answer)\n\n\n\n\n# return math.factorial(n) // (math.factorial(r) * math.factorial(n-r))\n'] | ['Runtime Error', 'Accepted'] | ['s237976779', 's568685789'] | [9800.0, 9908.0] | [195.0, 191.0] | [793, 792] |
p03425 | u061982241 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\nS ="MARCH"\nS2 = [0]*len(S)\nn=0\n\nfor i in range(N):\n s = input()\n for k, j in enumerate(S1):\n if s[0] == j:\n S2[k]+=1\n \nfor i in range(len(S2)):\n for j in range(i+1,len(S2)):\n for k in range(j+1,len(S2)):\n n +=S2[i]*S2[j]*S2[k]\n \nprint(n)', 'N = int(input())\nS ="MARCH"\nS2 = [0]*len(S)\nn=0\n\nfor i in range(N):\n s = input()\n for k, j in enumerate(S2):\n if s[0] == j:\n S2[k]+=1\n \nfor i in range(len(S2)):\n for j in range(i+1,len(S2)):\n for k in range(j+1,len(S2)):\n n +=S2[i]*S2[j]*S2[k]\n \nprint(n)\n', 'N = int(input())\nS ="MARCH"\nS2 = [0]*len(S)\nn=0\n\nfor i in range(N):\n s = input()\n for k, j in enumerate(S):\n if s[0] == j:\n S2[k]+=1\n \nfor i in range(len(S2)):\n for j in range(i+1,len(S2)):\n for k in range(j+1,len(S2)):\n n +=S2[i]*S2[j]*S2[k]\n \nprint(n)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s049848839', 's210547522', 's540978844'] | [3060.0, 3188.0, 3060.0] | [17.0, 246.0, 275.0] | [313, 314, 312] |
p03425 | u067983636 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\nS = "ACHMR"\ndic = {s:0 for s in S}\nfor _ in range(N):\n s = input()[0]\n if s in S:\n dic[s] += 1\n \n \nres = 0\nfor s0 in S[:3]:\n for s1 in S[1:4]:\n if s0 >= s1:\n continue\n for s2 in S[2:5]:\n if s2 >= s1:\n continue\n res += dic[s0] * dic[s1] * dic[s2] \nprint(res)', 'N = int(input())\nS = "MARCH"\ndic = {s:0 for s in S}\nfor _ in range(N):\n s = input()[0]\n if s in S:\n dic[s] += 1\n \nprint(sum([dic[a] * dic[b] * dic[c] for a in S for b in S for c in S if a < b < c]))\n'] | ['Wrong Answer', 'Accepted'] | ['s586882297', 's967126098'] | [3064.0, 3060.0] | [170.0, 169.0] | [369, 219] |
p03425 | u076917070 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['import sys\ninput=sys.stdin.readline\n\nh = {}\nfor c in "MARCH":\n h[c] = 0\n\nn = int(input())\nfor i in range(n):\n s = input().strip()\n h[s[0]] += 1\n#print(h)\n\nimport itertools\nans = 0\nfor i,j,k in itertools.combinations("MARCH", 3):\n ans += h[i]*h[j]*h[k]\nprint(ans)\n', 'import sys\ninput=sys.stdin.readline\n\nh = {}\nfor c in "MARCH":\n h[c] = 0\n\nfor i in range(int(input())):\n c = input().strip()[0]\n if c in h:\n h[c] += 1\n\nimport itertools\nans = 0\nfor i,j,k in itertools.combinations("MARCH", 3):\n ans += h[i]*h[j]*h[k]\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s109614976', 's402817562'] | [3060.0, 3060.0] | [62.0, 66.0] | [275, 278] |
p03425 | u077179028 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N=int(input())\nS=list(input().split())\nm,a,r,c,h = 0,0,0,0,0\nfor s in S:\n if 'M' == s[0]:\n m+=1\n elif 'A' == s[0]:\n a+=1\n elif 'R' == s[0]:\n r+=1\n elif 'C' == s[0]:\n c+=1\n elif 'H' == s[0]:\n h+=1\n\nprint(m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h)", "N=int(input())\nS=list(input().split())\nm,a,r,c,h = 0,0,0,0,0\nfor s in S:\n if 'M' == s[0]:\n m+=1\n elif 'A' == s[0]:\n a+=1\n elif 'R' == s[0]:\n r+=1\n elif 'C' == s[0]:\n c+=1\n elif 'H' == s[0]:\n h+=1\n\nprint(m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h)", "N=int(input())\nS=[input() for i in range(N)]\nm,a,r,c,h = 0,0,0,0,0\nfor s in S:\n if 'M' == s[0]:\n m+=1\n elif 'A' == s[0]:\n a+=1\n elif 'R' == s[0]:\n r+=1\n elif 'C' == s[0]:\n c+=1\n elif 'H' == s[0]:\n h+=1\n\nprint(m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s139043871', 's298942362', 's041104620'] | [3064.0, 3064.0, 9772.0] | [17.0, 17.0, 161.0] | [301, 301, 307] |
p03425 | u077291787 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['# ABC089C - March\nfrom itertools import combinations as comb\n\n\ndef main():\n N, *S = open(0).read().split()\n cnt = [sum(s.startswith(i) for S) for i in "MARCH"]\n ans = sum(x * y * z for x, y, z in comb(cnt, 3))\n print(ans)\n\n\nif __name__ == "__main__":\n main()', '# ABC089C - March\nfrom itertools import combinations as comb\n\n\ndef main():\n N, *S = open(0).read().split()\n capitals = [s[0] for s in S]\n cnt = [capitals.count(i) for i in "MARCH"]\n ans = sum(x * y * z for x, y, z in comb(cnt, 3))\n print(ans)\n\n\nif __name__ == "__main__":\n main()'] | ['Runtime Error', 'Accepted'] | ['s728572520', 's043537056'] | [2940.0, 10864.0] | [17.0, 39.0] | [273, 297] |
p03425 | u086503932 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\nimport math\n\nN = int(input())\nS_list = []\nfor i in range(N):\n S_list.append(input())\n\ncount_M = 0\ncount_A = 0\ncount_R = 0\ncount_C = 0\ncount_H = 0\n\nfor l in S_list:\n if l.startswith('M'):\n count_M += 1\n elif l.startswith('A'):\n count_A += 1\n elif l.startswith('R'):\n count_R += 1\n elif l.startswith('C'):\n count_C += 1\n elif l.startswith('H'):\n count_H += 1\n\ncount = [count_M,count_A,count_R,count_C,count_H]\nc = list(itertools.combinations(count,3))\nprint(c)\nx = 0\nfor i in c:\n i_list = list(i)\n x += i_list[0]*i_list[1]*i_list[2]\nprint(x)\n", "import itertools\nimport math\n\nN = int(input())\nS_list = []\nfor i in range(N):\n S_list.append(input())\n\ncount_M = 0\ncount_A = 0\ncount_R = 0\ncount_C = 0\ncount_H = 0\n\nfor l in S_list:\n if l.startswith('M'):\n count_M += 1\n elif l.startswith('A'):\n count_A += 1\n elif l.startswith('R'):\n count_R += 1\n elif l.startswith('C'):\n count_C += 1\n elif l.startswith('H'):\n count_H += 1\n\ncount = [count_M,count_A,count_R,count_C,count_H]\nc = list(itertools.combinations(count,3))\nx = 0\nfor i in c:\n i_list = list(i)\n x += i_list[0]*i_list[1]*i_list[2]\nprint(x)\n"] | ['Wrong Answer', 'Accepted'] | ['s712821410', 's682315982'] | [9792.0, 9792.0] | [207.0, 206.0] | [616, 607] |
p03425 | u094191970 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n=int(input())\nd={'M':0,'A':0,'R':0,'C':0,'H':0}\nfor i in range(n):\n s=input()\n if s[0]=='M':\n d['M']+=1\n if s[0]=='A':\n d['A']+=1\n if s[0]=='R':\n d['R']+=1\n if s[0]=='C':\n d['C']+=1\n if s[0]=='H':\n d['H']+=1\n\nans=0\nfor i,j,k in combinations('MARCH',3):\n ans+=d[i]*d[j]*d[k]\n\nprint(ans)", "from itertools import combinations\n\nn=int(input())\nd={'M':0,'A':0,'R':0,'C':0,'H':0}\nfor i in range(n):\n s=input()\n if s[0]=='M':\n d['M']+=1\n if s[0]=='A':\n d['A']+=1\n if s[0]=='R':\n d['R']+=1\n if s[0]=='C':\n d['C']+=1\n if s[0]=='H':\n d['H']+=1\n\nans=0\nfor i,j,k in combinations('MARCH',3):\n ans+=d[i]*d[j]*d[k]\n\nprint(ans)"] | ['Runtime Error', 'Accepted'] | ['s976747534', 's850287730'] | [9028.0, 9136.0] | [162.0, 164.0] | [308, 344] |
p03425 | u105124953 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\nn = int(input())\nlis = [input() for _ in range(n)]\nlis.sort()\nchoice = list(itertools.combinations(range(n), 3))\ncount = 0\nfor ch in choice:\n master = set(['M','A','R','C','H'])\n se = set([lis[ch[0]][0],lis[ch[1]][0],lis[ch[2]][0]])\n if len(se) <=2:\n break\n #print(master,se,master - se)\n if len(master - se) == 2:\n count += 1\nprint(count)", "import collections\nimport itertools\nn = int(input())\nli = []\nfor i in range(n):\n li.append(input()[0])\n\nc = collections.Counter(li)\nc_li = [c['M'],c['A'],c['R'],c['C'],c['H']]\n#print(li,c,c_li)\n\ncount = 0\ncombination = list(itertools.combinations([0,1, 2, 3,4], 3))\nfor co in combination:\n tmp = 1\n for c in co:\n if c_li[c] == 0:\n tmp = 0\n break\n tmp *= c_li[c]\n count += tmp\nprint(count)"] | ['Wrong Answer', 'Accepted'] | ['s962829254', 's926586281'] | [1946612.0, 4224.0] | [2235.0, 148.0] | [385, 436] |
p03425 | u106181248 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n = int(input())\nans = [0]*5\nx = 1\n\nfor i in range(n):\n s = input()\n if s[0] == "M":\n ans[0] = ans[0] + 1\n if s[0] == "A":\n ans[1] = ans[1] + 1\n if s[0] == "R":\n ans[2] = ans[2] + 1\n if s[0] == "C":\n ans[3] = ans[3] + 1\n if s[0] == "H":\n ans[4] = ans[4] + 1\n\nif ans.count(0) == 5:\n print(0)\nelse:\n for i in ans:\n if i != 0:\n x *= i\n print(x)\nprint(ans)', 'import itertools\n\nn = int(input())\nli = [0]*5\nans = 0\n\nfor i in range(n):\n s = input()\n if s[0] == "M":\n li[0] = li[0] + 1\n if s[0] == "A":\n li[1] = li[1] + 1\n if s[0] == "R":\n li[2] = li[2] + 1\n if s[0] == "C":\n li[3] = li[3] + 1\n if s[0] == "H":\n li[4] = li[4] + 1\n\nfor l in itertools.combinations(li, 3):\n ans += l[0] * l[1] * l[2] \n\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s447495930', 's389368526'] | [3064.0, 3064.0] | [187.0, 184.0] | [432, 406] |
p03425 | u109133010 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N=int(input())\nm=0\na=0\nr=0\nc=0\nh=0\nfor _ in range(N):\n s=input()\n if s[0]=="M":\n m+=1\n elif s[0]=="A":\n a+=1\n elif s[0]=="R":\n r+=1\n elif s[0]=="C":\n c+=1\n elif s[0]=="H":\n h+=1\nans=0\nans+=m*a*r\nans+=m*a*c\nans+=m*a*h\nans+=a*r*c\nans+=a*r*h\nans+=r*c*h\nprint(ans)', 'N=int(input())\nm=0\na=0\nr=0\nc=0\nh=0\nfor _ in range(N):\n s=input()\n if s[0]=="M":\n m+=1\n elif s[0]=="A":\n a+=1\n elif s[0]=="R":\n r+=1\n elif s[0]=="C":\n c+=1\n elif s[0]=="H":\n h+=1\nans=0\nans+=m*a*r\nans+=m*a*c\nans+=m*a*h\nans+=m*r*c\nans+=m*r*h\nans+=m*c*h\nans+=a*r*c\nans+=a*r*h\nans+=a*c*h\nans+=r*c*h\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s914846930', 's644821321'] | [3064.0, 3064.0] | [170.0, 170.0] | [281, 325] |
p03425 | u113971909 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N= int(input())\nS = ['M','A','R','C','H']\nC = [0]*5\nfor _ in range(N):\n X = (input())[0]\n if X in S:\n C[S.index(X)] += 1\nprint(C)\n\nfrom itertools import permutations, combinations,combinations_with_replacement,product\nL = list(combinations(range(5), 3)) #nCr\nans = 0\nfor i in L:\n ans += C[i[0]]*C[i[1]]*C[i[2]]\nprint(ans)", "N= int(input())\nS = ['M','A','R','C','H']\nC = [0]*5\nfor _ in range(N):\n X = (input())[0]\n if X in S:\n C[S.index(X)] += 1\nfrom itertools import permutations, combinations,combinations_with_replacement,product\nL = list(combinations(range(5), 3)) #nCr\nans = 0\nfor i in L:\n ans += C[i[0]]*C[i[1]]*C[i[2]]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s741647469', 's736502799'] | [3064.0, 3064.0] | [196.0, 193.0] | [329, 319] |
p03425 | u120810144 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n = int(input())\n\ns = [input() for i in range(n)]\n\norder = "MARCH"\n\ninitial_arary = [name[0] for name in s]\nnums = [initial_array.count(order[i]) for i in range(len(order))]\n\nans = 0\nfor i in range(0, 5):\n for j in range(i+1, 5):\n for k in range(j+1, 5):\n ans += nums[i] * nums[j] * nums[k]\n \nprint(ans)\n\n', 'n = int(input())\n\ns = [input() for i in range(n)]\n\norder = "MARCH"\n\ninitial_array = [name[0] for name in s]\nnums = [initial_array.count(order[i]) for i in range(len(order))]\n\nans = 0\nfor i in range(0, 5):\n for j in range(i+1, 5):\n for k in range(j+1, 5):\n ans += nums[i] * nums[j] * nums[k]\n \nprint(ans)\n\n'] | ['Runtime Error', 'Accepted'] | ['s217564682', 's903131242'] | [10520.0, 10520.0] | [134.0, 150.0] | [337, 337] |
p03425 | u125545880 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['import sys\nimport itertools\n\ninput = sys.stdin.readline\n\ndef main():\n common = common_function()\n N = int(input())\n l = [\'M\', \'A\', \'R\', \'C\', \'H\']\n m = [0]*5\n for _ in range(N):\n S = input()[:-1]\n Shead = S[0]\n for i, s0 in enumerate(l):\n if Shead == s0:\n m[i] += 1\n break\n ll = []\n for i, n in enumerate(m):\n if n >= 1:\n ll.append(l[i])\n if len(ll) <= 2:\n print(0)\n return\n ans = 0\n for i, j, k in itertools.combinations(ll, 3):\n ans += m[l.index(i)] * m[l.index(j)] * m[l.index(k)]\n\n print(ans)\n\nif __name__ == "__main__":\n main()\n', 'import sys\nimport itertools\n\ninput = sys.stdin.readline\n\ndef main():\n N = int(input())\n l = [\'M\', \'A\', \'R\', \'C\', \'H\']\n m = [0]*5\n for _ in range(N):\n S = input()[:-1]\n Shead = S[0]\n for i, s0 in enumerate(l):\n if Shead == s0:\n m[i] += 1\n break\n ll = []\n for i, n in enumerate(m):\n if n >= 1:\n ll.append(l[i])\n if len(ll) <= 2:\n print(0)\n return\n ans = 0\n for i, j, k in itertools.combinations(ll, 3):\n ans += m[l.index(i)] * m[l.index(j)] * m[l.index(k)]\n\n print(ans)\n\nif __name__ == "__main__":\n main()\n'] | ['Runtime Error', 'Accepted'] | ['s865699900', 's564232495'] | [9072.0, 9148.0] | [22.0, 73.0] | [669, 638] |
p03425 | u133936772 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n=int(input())\nd={'M':0,'A':0,'R':0,'C':0,'H':0}\nfor _ in range(n): d[input()[0]]+=1\nimport itertools as it\nprint(sum(d[s]*d[t]*d[r] for s,t,r in it.combinations('MARCH',3)))", "_,*l=open(0);from itertools import*;print(sum(x*y*z for x,y,z in combinations([sum(c==s[0]for s in l)for c in'MARCH'],3)))"] | ['Runtime Error', 'Accepted'] | ['s884432326', 's258018735'] | [3060.0, 9876.0] | [163.0, 77.0] | [174, 122] |
p03425 | u135116520 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N=int(input())\nS=[list(map(int,input().split())) for i in range(N)]\nm=0\na=0\nr=0\nc=0\nh=0\nfor i in range(N):\n if S[i][0]=="M":\n m+=1\n if S[i][0]=="A":\n a+=1\n if S[i][0]=="R":\n r+=1\n if S[i][0]=="C":\n c+=1\n if S[i][0]=="H":\n h+=1\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h*a*c*h+r*c*h)', 'N=int(input())\ns=[]\nfor i in range(N):\n S=input()\n s.append(S)\nm=0\na=0\nr=0\nc=0\nh=0\nfor i in range(N):\n if s[i][0]=="M":\n m+=1\n if s[i][0]=="A":\n a+=1\n if s[i][0]=="R":\n r+=1\n if s[i][0]=="C":\n c+=1\n if s[i][0]=="H":\n h+=1\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h)'] | ['Runtime Error', 'Accepted'] | ['s027660665', 's413587756'] | [3064.0, 9776.0] | [19.0, 196.0] | [313, 310] |
p03425 | u141574039 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['import itertools\nN=int(input())\nS=list(str(input()) for i in range(N))\nm,a,r,c,h=[0]*5\nans=0\nif S[0]=="M":m+=1\nif S[0]=="A":a+=1\nif S[0]=="R":r+=1\nif S[0]=="C":c+=1\nif S[0]=="H":h+=1\n\nx=[m,a,r,c,h]\nfor i in itertools.combinations(x,3):\n ans+=i[0]*i[1]*i[2]\nprint(ans)', 'import itertools\nN=int(input())\nS=list(str(input()) for i in range(N))\nm,a,r,c,h=[0]*5\nans=0\nfor i in range(len(S)):\n if S[i][0]=="M":m+=1\n if S[i][0]=="A":a+=1\n if S[i][0]=="R":r+=1\n if S[i][0]=="C":c+=1\n if S[i][0]=="H":h+=1\n\nx=[m,a,r,c,h]\nfor i in itertools.combinations(x,3):\n ans+=i[0]*i[1]*i[2]\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s634699883', 's639273757'] | [9764.0, 9784.0] | [173.0, 207.0] | [268, 317] |
p03425 | u143212659 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['#!/usr/bin/env python3\n# -*- coding: utf-8 -*-\n\nimport itertools\n\n\ndef main():\n N = int(input())\n S = [input()[0] for _ in range(N)]\n\n result = []\n for comb in list(itertools.combinations(S, 3)):\n if comb[0] != comb[1] and comb[0] != comb[2] and comb[1] != comb[2]:\n result.append(comb)\n\n print(len(result))\n\n\nif __name__ == "__main__":\n main()\n', '#!/usr/bin/env python3\n# -*- coding: utf-8 -*-\n\nfrom itertools import combinations\n\n\ndef main():\n\n N = int(input())\n\n count = {x: 0 for x in "MARCH"}\n\n for _ in range(N):\n s = input()[0]\n if s in count:\n count[s] += 1\n\n result = 0\n for a, b, c in combinations("MARCH", 3):\n result += count[a] * count[b] * count[c]\n\n print(result)\n\n\nif __name__ == "__main__":\n main()\n'] | ['Wrong Answer', 'Accepted'] | ['s185283466', 's352364069'] | [1966288.0, 3060.0] | [2230.0, 160.0] | [381, 421] |
p03425 | u143509139 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["one = {'M':0, 'A':0, 'R':0, 'C':0, 'H':0}\nn=int(input())\nfor i in range(n):\n one[input()[0]] += 1\nm,a,r,c,h=one['M'],one['A'],one['R'],one['C'],one['H']\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h)", "one = {'M':0, 'A':0, 'R':0, 'C':0, 'H':0}\nn=int(input())\nfor i in range(n):\n c=input()[0]\n one[c] = one.get(c,0)+1\nm,a,r,c,h=one['M'],one['A'],one['R'],one['C'],one['H']\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h)"] | ['Runtime Error', 'Accepted'] | ['s799719470', 's563672083'] | [3064.0, 3064.0] | [156.0, 176.0] | [220, 238] |
p03425 | u152638361 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['5\nCHOKUDAI\nRNG\nMAKOTO\nAOKI\nRINGO', 'import collections\nN = int(input())\nS = [input() for i in range(N)]\nprint(S)\nSI = [S[i][0] for i in range(N)]\ncnt = collections.Counter(SI)\nn = [cnt["M"],cnt["A"],cnt["R"],cnt["C"],cnt["H"]]\nans = 0\nfor i in range(3):\n for j in range(i+1,5):\n for k in range(j+1,5):\n ans += n[i]*n[j]*n[k]\nprint(ans)', 'import collections\nN = int(input())\nS = [input() for i in range(N)]\nSI = [S[i][0] for i in range(N)]\ncnt = collections.Counter(SI)\nn = [cnt["M"],cnt["A"],cnt["R"],cnt["C"],cnt["H"]]\nans = 0\nfor i in range(3):\n for j in range(i+1,5):\n for k in range(j+1,5):\n ans += n[i]*n[j]*n[k]\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s787989405', 's867516766', 's921754541'] | [2940.0, 13432.0, 10880.0] | [18.0, 166.0, 152.0] | [32, 318, 311] |
p03425 | u163320134 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n=int(input())\ncnt=[0]*5\ncheck='MARCH'\nfor i in range(n):\n s=input()\n for j in range(5):\n if s[0]==check[j]:\n cnt[j]+=1\n break\nans=0\nfor i in range(5):\n for j in range(i+1,5):\n for k in range(j+1,5):\n ans+=cnt[i]*cnt[j]*cnt[k]", "n=int(input())\ncnt=[0]*5\ncheck='MARCH'\nfor i in range(n):\n s=input()\n for j in range(5):\n if s[0]==check[j]:\n cnt[j]+=1\n break\nans=0\nfor i in range(5):\n for j in range(i+1,5):\n for k in range(j+1,5):\n ans+=cnt[i]*cnt[j]*cnt[k]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s522815474', 's487097192'] | [3064.0, 3064.0] | [251.0, 258.0] | [250, 261] |
p03425 | u167681750 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\nS = [input() for i in range(N)]\nmarch = {"M":0, "A":0, "R":0, "C":0, "H":0}\n\nfor i in S:\n if i[0] in march:\n march[i[0]] += 1\n\nans = 0\nfor first, second, third in combinations(march.values(), 3):\n ans += first * second * third\n\nprint(ans)', 'from itertools import combinations\n\nN = int(input())\nS = [input() for i in range(N)]\nmarch = {"M":0, "A":0, "R":0, "C":0, "H":0}\n\nfor i in S:\n if i[0] in march:\n march[i[0]] += 1\n\nans = 0\nfor first, second, third in combinations(march.values(), 3):\n ans += first * second * third\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s497236889', 's107452470'] | [9752.0, 9772.0] | [163.0, 157.0] | [268, 304] |
p03425 | u180058306 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['\nN = int(input())\n\nS = [input() for _ in range(N)]\n\nword = "MARCH"\n\nn = len(word)\n\ndic_initial = dict()\nfor letter in word:\n dic_initial[letter] = 0\n\nfor i in range(N):\n \n if S[i][0] in dic_initial:\n \n dic_initial[S[i][0]] += 1\n\nans = 0\n\nfor p in range(n - 2):\n \n num_p = dic_initial[word[p]]\n \n if num_p == 0:\n continue\n for q in range(p + 1, n - 1):\n num_q = dic_initial[word[q]]\n if num_q == 0:\n continue\n for r in range(q + 1, n):\n ans += num_p * num_q *num_r\n\nprint(ans)', '\nN = int(input())\n\nS = [input() for _ in range(N)]\n\nword = "MARCH"\n\nn = len(word)\n\ndic_initial = dict()\nfor letter in word:\n dic_initial[letter] = 0\n\nfor i in range(N):\n \n if S[i][0] in dic_initial:\n \n dic_initial[S[i][0]] += 1\n\nans = 0\n\nfor p in range(n - 2):\n \n num_p = dic_initial[word[p]]\n \n if num_p == 0:\n continue\n for q in range(p + 1, n - 1):\n num_q = dic_initial[word[q]]\n if num_q == 0:\n continue\n for r in range(q + 1, n):\n num_r = dic_initial[word[r]]\n ans += num_p * num_q * num_r\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s135415430', 's431530066'] | [9772.0, 9776.0] | [174.0, 163.0] | [1254, 1296] |
p03425 | u185896732 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n=int(input())\nm=a=r=c=h=0\nls\nfor i in range(n):\n s=input()\n if s[0]=='M':\n m+=1\n elif a[0]=='A':\n a+=1\n elif a[0]=='R':\n r+=1\n elif a[0]=='C':\n c+=1\n elif a[0]=='H':\n h+=1\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h)", "n=int(input())\nm=a=r=c=h=0\nfor i in range(n):\n s=input()\n if s[0]=='M':\n m+=1\n elif s[0]=='A':\n a+=1\n elif s[0]=='R':\n r+=1\n elif s[0]=='C':\n c+=1\n elif s[0]=='H':\n h+=1\nprint(m*a*r+m*a*c+m*a*h+m*r*c+m*r*h+m*c*h+a*r*c+a*r*h+a*c*h+r*c*h)"] | ['Runtime Error', 'Accepted'] | ['s684791579', 's111255317'] | [3188.0, 3064.0] | [18.0, 167.0] | [292, 289] |
p03425 | u190555868 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = input()\n\nd = {'M':0,'A':0,'R':0,'C':0,'H':0}\n\nfor i in range(N):\n s = input()\n if s[0] in 'MARCH':\n d[s[0]]+=1\n\nv = list(d.values())\nans = 0\n\nfor i in range(5):\n for j in range (i+1,5):\n for k in range (j+1,5):\n ans += v[i]*v[j]*v[k]\n\nprint(ans)", "N = input()\n\nd = {'M':0,'A':0,'R':0,'C':0,'H':0}\n\nfor i in range(N):\n s = input()\n if s[0] in 'MARCH':\n d[s[0]]+=1\n\nv = list(d.values())\nans = 0\n\nfor i in range(5):\n for j in range (i+1,5):\n for k in range (j+1,5):\n ans += v[i]*v[j]*v[k]\n\nprint(ans)", "N = int(input())\n\nd = {'M':0,'A':0,'R':0,'C':0,'H':0}\n\nfor i in range(N):\n s = input()\n if s[0] in 'MARCH':\n d[s[0]]+=1\n\nv = list(d.values())\nans = 0\n\nfor i in range(5):\n for j in range (i+1,5):\n for k in range (j+1,5):\n ans += v[i]*v[j]*v[k]\n\nprint(ans)"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s673087159', 's943439673', 's598071925'] | [3064.0, 3064.0, 3064.0] | [18.0, 18.0, 174.0] | [283, 283, 288] |
p03425 | u212328220 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\nn = int(input())\nsl = []\n\ndic = {'M':0,'A':0,'R':0,'C':0,'H':0}\nfor i in range(n):\n name = input()\n if name[0] in dic:\n dic[name[0]] += 1\n\nlst = []\nfor com in itertools.combinations(dic.keys(),3):\n lst.append(com)\nprint(lst)\n\nans = 0\nfor v in lst:\n ans += dic[v[0]] * dic[v[1]] * dic[v[2]]\nprint(ans)", "import itertools\nn = int(input())\nsl = []\n\ndic = {'M':0,'A':0,'R':0,'C':0,'H':0}\nfor i in range(n):\n name = input()\n if name[0] in dic:\n dic[name[0]] += 1\n\nlst = []\nfor com in itertools.combinations(dic.keys(),3):\n lst.append(com)\n\nans = 0\nfor v in lst:\n ans += dic[v[0]] * dic[v[1]] * dic[v[2]]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s600356998', 's524826846'] | [3064.0, 3064.0] | [169.0, 174.0] | [336, 325] |
p03425 | u222634810 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = int(input())\nM = []\nA = []\nR = []\nC = []\nH = []\n\nfor i in range(N):\n temp = input()\n init_temp = list(temp)\n if 'M' == init_temp[0]:\n M.append(temp)\n elif 'A' == init_temp[0]:\n A.append(temp)\n elif 'R' == init_temp[0]:\n R.append(temp)\n elif 'C' == init_temp[0]:\n C.append(temp)\n elif 'H' == init_temp[0]:\n H.append(temp)\n else:\n pass\n\nM = len(list(set(M)))\nA = len(list(set(A)))\nR = len(list(set(R)))\nC = len(list(set(C)))\nH = len(list(set(H)))\n\nD = [M, A, R, C, H]\n\nA = [0, 0, 0, 0, 0, 0, 1, 1, 1, 2]\nB = [1, 1, 1, 2, 2, 3, 2, 2, 3, 3]\nC = [2, 3, 4, 3, 4, 4, 3, 4, 4, 4]\n\nans = 0.0\n\nfor i in range(10):\n ans += D[A[i]] * D[B[i]] * D[C[i]]\n\nprint(ans)", "N = int(input())\nM = []\nA = []\nR = []\nC = []\nH = []\n\nfor i in range(N):\n temp = input()\n init_temp = list(temp) \n if 'M' == init_temp[0]:\n M.append(temp)\n elif 'A' == init_temp[0]:\n A.append(temp)\n elif 'R' == init_temp[0]:\n R.append(temp)\n elif 'C' == init_temp[0]:\n C.append(temp)\n elif 'H' == init_temp[0]:\n H.append(temp)\n else:\n pass\n\nM = len(list(set(M)))\nA = len(list(set(A)))\nR = len(list(set(R)))\nC = len(list(set(C)))\nH = len(list(set(H)))\n\nD = [M, A, R, C, H]\n\nA = [0, 0, 0, 0, 0, 0, 1, 1, 1, 2]\nB = [1, 1, 1, 2, 2, 3, 2, 2, 3, 3]\nC = [2, 3, 4, 3, 4, 4, 3, 4, 4, 4]\n\nans = 0\n\nfor i in range(10):\n ans += D[A[i]] * D[B[i]] * D[C[i]]\n\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s426747302', 's481008651'] | [16020.0, 16016.0] | [261.0, 248.0] | [745, 746] |
p03425 | u223904637 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n=int(input())\nal=list('MARCH')\nd=[0]*5\nfor i in range(n):\n s=list(input())\n if s[0] in al:\n d[al.index(s[0])]+=1\nop=[[0],[1]]\nfor i in range(4):\n tmp=[]\n for j in op:\n j.append(0)\n tmp.append(j)\n j.pop(-1)\n j.append(1)\n tmp.append(j)\n op=tmp\nk=[]\nfor i in op:\n if sum(i)==3:\n k.append(i)\nans=0\nfor i in k:\n t=1\n for j in range(5):\n if i[j]==1:\n t*=d[j]\n ans+=t\nprint(ans)\n ", "n=int(input())\nal=list('MARCH')\nd=[0]*5\nfor i in range(n):\n s=list(input())\n if s[0] in al:\n d[al.index(s[0])]+=1\nop=['0','1']\nfor i in range(4):\n tmp=[]\n for j in op:\n tmp.append(j+'0')\n tmp.append(j+'1')\n op=tmp\nk=[]\nfor i in op:\n g=list(i)\n c=0\n for j in g:\n if j=='1':\n c+=1\n if c==3:\n k.append(i)\nans=0\nfor i in k:\n t=1\n for j in range(5):\n if i[j]=='1':\n t*=d[j]\n ans+=t\nprint(ans)\n \n"] | ['Wrong Answer', 'Accepted'] | ['s061909641', 's956897979'] | [3064.0, 3064.0] | [261.0, 258.0] | [479, 501] |
p03425 | u226108478 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["if __name__ == '__main__':\n people_count = int(input())\n\n person_name = list()\n\n for person in range(people_count):\n person_name.append(str(input()))\n\n initial_strings = list()\n\n for name in person_name:\n initial_strings.append(name[0])\n\n mod_initial_strings = set(initial_strings)\n\n count = 0\n\n matching_strings = ['M', 'A', 'R', 'C', 'H']\n matching_strings_count = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}\n\n for string in matching_strings:\n if string in mod_initial_strings:\n count += initial_strings.count(string)\n\n for string in initial_strings:\n matching_strings_count[string] += 1\n\n if count >= 3:\n print(matching_strings_count)\n else:\n print(0)\n", "# -*- coding: utf-8 -*-\n\n\n\n\n\ndef count_initial_name(people_count: int) -> dict:\n matching_strings_count = {'M': 0, 'A': 0, 'R': 0, 'C': 0, 'H': 0}\n\n for person in range(people_count):\n initial_name = input()[0]\n\n if initial_name in matching_strings_count.keys():\n matching_strings_count[initial_name] += 1\n\n return matching_strings_count\n\n\ndef count_patterns(matching_strings_count):\n import itertools\n\n patterns = itertools.combinations(matching_strings_count.keys(), 3)\n pattern_count = 0\n\n for a_pattern in patterns:\n pattern_count += matching_strings_count[a_pattern[0]] * \\\n matching_strings_count[a_pattern[1]] * \\\n matching_strings_count[a_pattern[2]]\n return pattern_count\n\n\nif __name__ == '__main__':\n matching_strings_count = count_initial_name(people_count=int(input()))\n\n if sum(matching_strings_count.values()) >= 3:\n print(count_patterns(matching_strings_count))\n else:\n print(0)\n"] | ['Runtime Error', 'Accepted'] | ['s764443138', 's818982608'] | [10520.0, 3064.0] | [195.0, 173.0] | [744, 1032] |
p03425 | u247211039 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N=int(input())\nS=[input() for _ in range(N)]\n\ncnt=[]\nans=0\nfor i in range(N):\n if S[i][0]=='M':\n cnt.append('M')\n elif S[i][0]=='A':\n cnt.append('A')\n elif S[i][0]=='R':\n cnt.append('R')\n elif S[i][0]=='C':\n cnt.append('C')\n elif S[i][0]=='H':\n cnt.append('H')\n \nimport math\nimport itertools\nc_list = list(itertools.combinations(cnt, 3))\n\nfor v in itertools.combinations(cnt, 3):\n print(set(v))\n if len(set(v))==3:\n ans+=1\n\nprint(ans)", "N=int(input())\nS=[input() for _ in range(N)]\ncnt=[0]*N\ncnt2=[]\nflag=0\n\nif len(S)>1:\n for i in range(N):\n if S[i][0]=='M':\n cnt[0]+=1\n elif S[i][0]=='A':\n cnt[1]+=1\n elif S[i][0]=='R':\n cnt[2]+=1\n elif S[i][0]=='C':\n cnt[3]+=1\n elif S[i][0]=='H':\n cnt[4]+=1\n\nfor i in range(N):\n if cnt[i]!=0:\n cnt2.append(cnt[i])\n \nimport math\nimport itertools\nans=0\n\nfor v in itertools.combinations(cnt2, 3):\n ans+=v[0]*v[1]*v[2]\n\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s824196798', 's305623491'] | [1869524.0, 16968.0] | [2271.0, 178.0] | [504, 538] |
p03425 | u257350577 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\nletters = ['M','A','R','C','H']\nN = int(input())\na = []\ncom_s = []\n\nfor i in range(N):\n x = input()\n if x[0] in letters:\n a.append(x)\n\ncoms = list(itertools.combinations(a,3))\nfor com in coms:\n if com[0][0] != com[1][0] and com[1][0] != com[2][0] and com[2][0] != com[0]com[0]:\n com_s.append(com)\n\nprint(len(com_s))", "N = int(input())\n_M = [] \n_A = [] \n_R = [] \n_C = [] \n_H = [] \n\nfor i in range(N):\n x = input()\n if x[0] == 'M':\n _M.append(x)\n elif x[0] == 'A':\n _A.append(x)\n elif x[0] == 'R':\n _R.append(x)\n elif x[0] == 'C':\n _C.append(x)\n elif x[0] == 'H':\n _H.append(x)\n\nm = len(_M)\na = len(_A)\nr = len(_R)\nc = len(_C)\nh = len(_H)\n\nx = m*a*r + m*a*c + m*a*h + m*r*c + m*r*h + m*c*h + a*r*c + a*r*h + a*c*h + r*c*h\n\nprint(x)\n"] | ['Runtime Error', 'Accepted'] | ['s017946293', 's869765064'] | [3060.0, 10208.0] | [17.0, 165.0] | [355, 465] |
p03425 | u267325300 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['from functools import reduce\nfrom itertools import combinations\nN = int(input())\nS = {"M": [], "A": [], "R": [], "C": [], "H": []}\nfor _ in range(N):\n i = input()\n head = i[0]\n if head == "M":\n S["M"].append(i)\n elif head == "A":\n S["A"].append(i)\n elif head == "R":\n S["R"].append(i)\n elif head == "C":\n S["C"].append(i)\n elif head == "H":\n S["H"].append(i)\nresult = [len(S[head]) for head in list("MARCH") if len(S[head]) != 0]\n\nans = sum([reduce(lambda x, y: x * y, list(combinations(result, 3))[i])\n for i in range(len(result))])\nprint(ans)\n', 'from functools import reduce\nfrom itertools import combinations\nN = int(input())\nS = {"M": set([]), "A": set([]), "R": set([]), "C": set([]), "H": set([])}\nfor _ in range(N):\n i = input()\n head = i[0]\n if head == "M":\n S["M"].add(i)\n elif head == "A":\n S["A"].add(i)\n elif head == "R":\n S["R"].add(i)\n elif head == "C":\n S["C"].add(i)\n elif head == "H":\n S["H"].add(i)\nresult = [len(S[head]) for head in list("MARCH") if len(S[head]) != 0]\n\nans = sum([reduce(lambda x, y: x * y, list(combinations(result, 3))[i])\n for i in range(len(result))])\nprint(ans)\n', 'from functools import reduce\nfrom itertools import combinations\nN = int(input())\nS = {"M": [], "A": [], "R": [], "C": [], "H": []}\nfor _ in range(N):\n i = input()\n head = i[0]\n if head == "M":\n S["M"].append(i)\n elif head == "A":\n S["A"].append(i)\n elif head == "R":\n S["R"].append(i)\n elif head == "C":\n S["C"].append(i)\n elif head == "H":\n S["H"].append(i)\ncounter = [len(S[head]) for head in list("MARCH") if len(S[head]) != 0]\ncombs = list(combinations(counter, 3))\nans = sum([reduce(lambda x, y: x * y, e) for e in combs])\nprint(ans)\n'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s122872997', 's908290664', 's924053957'] | [10724.0, 14812.0, 10720.0] | [175.0, 203.0, 176.0] | [611, 621, 595] |
p03425 | u273038590 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n=int(input())\ns=list(input() for i in range(n))\nm=[]\na=[]\nr=[]\nc=[]\nh=[]\nans=1\nfor i in range(n):\n if s[i][0] == "M" and not s[i] in m:\n m.append(s[i])\n elif s[i][0] == "A" and not s[i] in a:\n a.append(s[i])\n elif s[i][0] == "R" and not s[i] in r:\n r.append(s[i])\n elif s[i][0] == "C" and not s[i] in c:\n c.append(s[i])\n elif s[i][0] == "H" and not s[i] in h:\n h.append(s[i])\nnum=len(m)+len(a)+len(r)+len(c)+len(h)\nif num == 0:\n print(0)\nelse:\n lis = list(len(m),len(a),len(r),len(c),len(h))\n for i in range(5):\n if lis[i]>0:\n ans = ans*lis[i] \n print(ans)', 'import math\nn=int(input())\ns=list(input() for i in range(n))\nm=[]\na=[]\nr=[]\nc=[]\nh=[]\nfor i in range(n):\n if s[i][0] == "M" and not s[i] in m:\n m.append(s[i])\n elif s[i][0] == "A" and not s[i] in a:\n a.append(s[i])\n elif s[i][0] == "R" and not s[i] in r:\n r.append(s[i])\n elif s[i][0] == "C" and not s[i] in c:\n c.append(s[i])\n elif s[i][0] == "H" and not s[i] in h:\n h.append(s[i])\nnum = len(m)+len(a)+len(r)+len(c)+len(h)\nprint(math.factorial(num)/(math.factorial(3)*math.factorial(num)))', 'import math\nn=int(input())\ns=list(input() for i in range(n))\nm=[]\na=[]\nr=[]\nc=[]\nh=[]\nans=1\nfor i in range(n):\n if s[i][0] == "M" and not s[i] in m:\n m.append(s[i])\n elif s[i][0] == "A" and not s[i] in a:\n a.append(s[i])\n elif s[i][0] == "R" and not s[i] in r:\n r.append(s[i])\n elif s[i][0] == "C" and not s[i] in c:\n c.append(s[i])\n elif s[i][0] == "H" and not s[i] in h:\n h.append(s[i])\nnum=len(m)+len(a)+len(r)+len(c)+len(h)\nif num == 0:\n print(0)\nelse:\n lis = list(len(m),len(a),len(r),len(c),len(h))\n ans = ans*lis[i] for i in range(5)\n print(ans)', 'm=[]\na=[]\nr=[]\nc=[]\nh=[]\nans=1\nfor i in range(n):\n if s[i][0] == "M" and not s[i] in m:\n m.append(s[i])\n elif s[i][0] == "A" and not s[i] in a:\n a.append(s[i])\n elif s[i][0] == "R" and not s[i] in r:\n r.append(s[i])\n elif s[i][0] == "C" and not s[i] in c:\n c.append(s[i])\n elif s[i][0] == "H" and not s[i] in h:\n h.append(s[i])\nnum=len(m)+len(a)+len(r)+len(c)+len(h)\nif num == 0:\n print(0)\nelse:\n lis = list(len(m),len(a),len(r),len(c),len(h))\n for i in range(5):\n ans = ans*lis[i] \n print(ans)', 'n=int(input())\ns=list(input() for i in range(n))\nmc=ac=rc=cc=hc=0\ncnt=0\nfor i in range(n):\n if s[i][0] == "M":\n mc+=1\n elif s[i][0] == "A":\n ac+=1\n elif s[i][0] == "R":\n rc+=1\n elif s[i][0] == "C":\n cc+=1\n elif s[i][0] == "H":\n hc+=1\nlis = [mc,ac,rc,cc,hc]\nif sum(lis) == 0:\n print(0)\nelse:\n for i in range(5):\n for j in range(i+1,5):\n for k in range(j+1,5):\n cnt+=lis[i]*lis[j]*lis[k]\n print(cnt)'] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s105433474', 's408115642', 's464147436', 's598716921', 's531738043'] | [10164.0, 10012.0, 3064.0, 3064.0, 9764.0] | [2112.0, 2104.0, 17.0, 17.0, 189.0] | [637, 540, 614, 563, 489] |
p03425 | u276204978 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import numpy as np\nfrom itertools\nfrom functools import reduce\nfrom collections import defaultdict\n\nN = int(input())\nd = defaultdict(int)\nfor i in range(N):\n c = input()\n if c[0] in ('M', 'A', 'R', 'C', 'H'):\n d[c[0]] += 1\n\nl = len(d)\nif l < 3:\n print(0)\nelse:\n ans = 0\n for l in list(itertools.combinations(d.values(), 3)):\n ans += reduce(np.multiply, l)\n print(ans)", "import numpy as np\nfrom itertools import combinations\nfrom functools import reduce\nfrom collections import defaultdict\n\nN = int(input())\nd = defaultdict(int)\nfor i in range(N):\n c = input()\n if c[0] in ('M', 'A', 'R', 'C', 'H'):\n d[c[0]] += 1\n\nl = len(d)\nif l < 3:\n print(0)\nelse:\n ans = 0\n for l in list(combinations(d.values(), 3)):\n ans += reduce(np.multiply, l)\n print(ans)"] | ['Runtime Error', 'Accepted'] | ['s759112335', 's911270013'] | [8756.0, 27108.0] | [21.0, 225.0] | [399, 409] |
p03425 | u289288647 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = int(input())\nname = [0]*5\nans = 0\nfor i in range(N):\n a = input()\n if a[0] == 'M': name[0] += 1\n elif a[0] == 'A': name[1] += 1\n elif a[0] == 'R': name[2] += 1\n elif a[0] == 'C': name[3] += 1\n elif a[0] == 'H': name[4] += 1\nif sum(name) < 3:\n print(0)\nelse:\n for i in combinations(name, 3):\n ans += i[0]*i[1]*i[2]\n print(ans)\n\n", "from itertools import combinations\nN = int(input())\nname = [0]*5\nans = 0\nfor i in range(N):\n a = input()\n if a[0] == 'M': name[0] += 1\n elif a[0] == 'A': name[1] += 1\n elif a[0] == 'R': name[2] += 1\n elif a[0] == 'C': name[3] += 1\n elif a[0] == 'H': name[4] += 1\nif sum(name) < 3:\n print(0)\nelse:\n for i in combinations(name, 3):\n ans += i[0]*i[1]*i[2]\n print(ans)\n\n"] | ['Runtime Error', 'Accepted'] | ['s968775339', 's750259471'] | [9132.0, 9200.0] | [153.0, 153.0] | [365, 400] |
p03425 | u301624971 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import itertools\ndef myAnswer(N:int,S:list) -> int:\n ans = 0\n dic = {}\n for s in S:\n if(s[0] in dic.keys()):\n dic[s[0]] += 1\n else:\n dic[s[0]] = 1\n for c in itertools.combinations(dic.values(),3):\n ans += c[0]*c[1]*c[2]\n \n return ans\n\ndef modelAnswer():\n return\ndef main():\n N = int(input())\n S = [input() for _ in range(N)]\n print(myAnswer(N,S[:]))\nif __name__ == '__main__':\n main()", 'import itertools\ndef myAnswer(N:int,S:list) -> int:\n ans = 0\n dic = {"M":0,"A":0,"R":0,"C":0,"H":0}\n for s in S:\n if(s[0] in dic.keys()):\n dic[s[0]] += 1\n for c in itertools.combinations(dic.values(),3):\n ans += c[0]*c[1]*c[2]\n return ans\n\ndef modelAnswer():\n return\ndef main():\n N = int(input())\n S = [input() for _ in range(N)]\n print(myAnswer(N,S[:]))\nif __name__ == \'__main__\':\n main()'] | ['Wrong Answer', 'Accepted'] | ['s940624768', 's578471999'] | [10540.0, 10540.0] | [168.0, 168.0] | [440, 427] |
p03425 | u305366205 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["s = [input() for _ in range(n)]\nprefix = ['M', 'A', 'R', 'C', 'H']\ncnt = [0] * 5\nans = 0\nfor i in range(n):\n for j in range(5):\n if s[i][0] == prefix[j]:\n cnt[j] += 1\nfor i in range(2 ** 5):\n flag = [False] * 5\n for j in range(5):\n if (i >> j) & 1:\n flag[j] = True\n if flag.count(True) == 3:\n tmp = 1\n for k in range(5):\n if flag[k]:\n tmp *= cnt[k]\n ans += tmp\nprint(ans)", "n = int(input())\ns = [input() for _ in range(n)]\nprefix = ['M', 'A', 'R', 'C', 'H']\ncnt = [0] * 5\nans = 0\nfor i in range(n):\n for j in range(5):\n if s[i][0] == prefix[j]:\n cnt[j] += 1\nfor i in range(2 ** 5):\n flag = [False] * 5\n for j in range(5):\n if (i >> j) & 1:\n flag[j] = True\n if flag.count(True) == 3:\n tmp = 1\n for k in range(5):\n if flag[k]:\n tmp *= cnt[k]\n ans += tmp\nprint(ans)"] | ['Runtime Error', 'Accepted'] | ['s833970256', 's649200801'] | [3064.0, 9752.0] | [17.0, 250.0] | [466, 483] |
p03425 | u308241789 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["N = int(input())\nA = []\nfor i in range(N):\n A.append(input())\nk = 0\nc = [ 0,0,0,0,0]\nfor i in A:\n if i[0] == 'm':\n c[0] += 1\n if i[0] == 'a':\n c[1] += 1\n if i[0] == 'r':\n c[2] += 1\n if i[0] == 'c':\n c[3] += 1\n if i[0] == 'h':\n c[4] += 1\n\nprint(c)\nif not c.count(0) >= 3:\n k += c[0] * c[1] * c[2] + c[0] * c[1] * c[3] + c[0] * c[1] * c[4] + c[0] * c[2] * c[3] + c[0] * c[2] * c[4] + c[0] * c[3] * c[4] + c[1] * c[2] * c[3] + c[1] * c[2] * c[4] + c[1] * c[3] * c[4] + c[2] * c[3] * c[4] \n\nprint(k)", 'N = int(input())\nA = [i for i in range(N) input()]\nk = 0\nc = [ 0,0,0,0,0]\nfor i in A:\n if i[0] == m:\n c[0] += 1\n if i[0] == a:\n c[1] += 1\n if i[0] == r:\n c[2] += 1\n if i[0] == c:\n c[3] += 1\n if i[0] == h:\n c[4] += 1\n\nif c.count(0) >= 3:\n k += c[0] * c[1] * c[2] + c[0] * c[1] * c[3] + c[0] * c[1] * c[4] + c[0] * c[2] * c[3] + c[0] * c[2] * c[4] + c[0] * c[3] * c[4] + c[1] * c[2] * c[3] + c[1] * c[2] * c[4] + c[1] * c[3] * c[4] + c[2] * c[3] * c[4] \n\nprint(k)', "N = int(input())\nA = []\nfor i in range(N):\n A.append(input())\nk = 0\nc = [ 0,0,0,0,0]\nfor i in A:\n if i[0] == 'M':\n c[0] += 1\n if i[0] == 'A':\n c[1] += 1\n if i[0] == 'R':\n c[2] += 1\n if i[0] == 'C':\n c[3] += 1\n if i[0] == 'H':\n c[4] += 1\n\nprint(c)\nif not c.count(0) >= 3:\n k += c[0] * c[1] * c[2] + c[0] * c[1] * c[3] + c[0] * c[1] * c[4] + c[0] * c[2] * c[3] + c[0] * c[2] * c[4] + c[0] * c[3] * c[4] + c[1] * c[2] * c[3] + c[1] * c[2] * c[4] + c[1] * c[3] * c[4] + c[2] * c[3] * c[4] \n\nprint(k)", "N = int(input())\nA = []\nfor i in range(N):\n A.append(input())\nk = 0\nc = [ 0,0,0,0,0]\nfor i in A:\n if i[0] == 'M':\n c[0] += 1\n if i[0] == 'A':\n c[1] += 1\n if i[0] == 'R':\n c[2] += 1\n if i[0] == 'C':\n c[3] += 1\n if i[0] == 'H':\n c[4] += 1\n\n\nif not c.count(0) >= 3:\n k += c[0] * c[1] * c[2] + c[0] * c[1] * c[3] + c[0] * c[1] * c[4] + c[0] * c[2] * c[3] + c[0] * c[2] * c[4] + c[0] * c[3] * c[4] + c[1] * c[2] * c[3] + c[1] * c[2] * c[4] + c[1] * c[3] * c[4] + c[2] * c[3] * c[4] \n\nprint(k)"] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s061975952', 's220385003', 's292189147', 's547471288'] | [9780.0, 2940.0, 9776.0, 9772.0] | [172.0, 17.0, 185.0, 183.0] | [519, 482, 519, 511] |
p03425 | u318029285 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\ns = [0] * N\nd = {"M": 0, "A": 0, "R": 0, "C": 0, "H": 0}\nfor i in range(N):\n s[i] = input()\n if s[i][0] == "M":\n d[\'M\'] += 1\n elif s[i][0] == "A":\n d[\'A\'] += 1\n elif s[i][0] == "R":\n d[\'R\'] += 1\n elif s[i][0] == "C":\n d[\'C\'] += 1\n elif s[i][0] == "H":\n d[\'H\'] += 1\n\ntotal = 1\ncount = 0\ndel_keys = []\nfor k, v in d.items():\n if v == 0:\n del_keys.append(k)\n else:\n total *= v\n count += 1\n\nprint(d)\nfor key in del_keys:\n del d[key]\n\nif count < 3:\n print(0)\n exit()\nelif count == 3:\n print(total)\n exit()\nelif count == 4:\n ans = 0\n for k, v in d.items():\n ans += total / v\n print(int(ans))\n exit()\nelif count == 5:\n ans = 0\n d_lst = list(d)\n for i in range(4):\n for j in range(i+1, 5):\n ans += total / (d_lst[i] * d_lst[j])\n print(int(ans))\n', "from collections import Counter\nfrom itertools import combinations\n\nn = int(input())\ns = [input()[0] for _ in range(n)]\nlc = Counter(s)\nprint(lc)\n\ncon = []\nfor c in 'MARCH':\n if c in lc:\n con.append(lc[c])\n\nans = 0\nfor a, b, c in combinations(con, 3):\n ans += a * b * c\n\nprint(ans)\n", "from collections import Counter\nfrom itertools import combinations\n\nn = int(input())\ns = [input()[0] for _ in range(n)]\nlc = Counter(s)\n\ncon = []\nfor c in 'MARCH':\n if c in lc:\n con.append(lc[c])\n\nans = 0\nfor a, b, c in combinations(con, 3):\n ans += a * b * c\n\nprint(ans)\n"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s667085785', 's980630678', 's326425579'] | [9716.0, 4224.0, 4224.0] | [192.0, 146.0, 140.0] | [898, 295, 285] |
p03425 | u340781749 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["from collections import Counter\nfrom itertools import combinations\n\nn = int(input())\nl = Counter(input()[0] for _ in range(n))\ncon = [l[c] for c in 'MARCH' if c in l]\nprint(sum(a * b * c) for a, b, c in combinations(con, 3))\n", "from collections import Counter\nfrom itertools import combinations\n\nn = int(input())\nl = Counter(input()[0] for _ in range(n))\ncon = [l[c] for c in 'MARCH' if c in l]\nprint(sum(a * b * c for a, b, c in combinations(con, 3)))\n"] | ['Wrong Answer', 'Accepted'] | ['s191668372', 's269363981'] | [3316.0, 3316.0] | [166.0, 161.0] | [225, 225] |
p03425 | u350049649 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N=int(input())\nS=set()\n\nfor i in range(N):\n s=input()\n if s[0] in [M,A,R,C,H]:\n S.add(s)\n\nL=len(S)\n\nprint(L*(L-1)*(L-2)//6)', "N=int(input())\nS=set()\n\nfor i in range(N):\n s=input()\n if s[0] in ['M','A','R','C','H']:\n S.add(s)\n\nL=len(S)\n\nprint(L*(L-1)*(L-2)//6)\n", "N=int(input())\nchars=['M','A','R','C','H']\ndict={}\nfor c in chars:\n dict[c]=0\n \nans=0\n \nfor i in range(N):\n s=input()\n if (s[0] in chars):\n dict[s[0]]+=1\n\nL=list(dict.values())\n \nfor i in range(2**5):\n usedchars=[]\n for j in range(5):\n if i>>j&1==1:\n usedchars.append(L[j])\n \n if len(usedchars)==3:\n ans+=usedchars[0]*usedchars[1]*usedchars[2]\n\nprint(ans)"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s256175522', 's755669638', 's276853698'] | [3060.0, 14316.0, 3064.0] | [18.0, 174.0, 173.0] | [128, 139, 383] |
p03425 | u365858785 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['a=int(input())\nlis=[input()[0] for i in range(a)]\nlis2=[0,0,0,0,0]\ndellist=[]\nfor m in lis:\n if m=="M":\n lis2[0]+=1\n elif m=="A":\n lis2[1]+=1\n elif m=="R":\n lis2[2]+=1\n elif m=="C":\n lis2[3]+=1\n elif m=="H":\n lis2[4]+=1\na=sum(lis2)\nprint(lis2,a)\ndef c3(a):\n return a*(a-1)*(a-2)/6\ndef c2(a):\n return a*(a-1)/2\nans=c3(a)\nfor i in lis2:\n ans=ans-c3(i)-c2(i)*(a-i)\nprint(int(ans))\n ', 'a=int(input())\nlis=[input()[0] for i in range(a)]\nlis2=[0,0,0,0,0]\ndellist=[]\nfor m in lis:\n if m=="M":\n lis2[0]+=1\n elif m=="A":\n lis2[1]+=1\n elif m=="R":\n lis2[2]+=1\n elif m=="C":\n lis2[3]+=1\n elif m=="H":\n lis2[4]+=1\na=sum(lis2)\ndef c3(a):\n return a*(a-1)*(a-2)/6\ndef c2(a):\n return a*(a-1)/2\nans=c3(a)\nfor i in lis2:\n ans=ans-c3(i)-c2(i)*(a-i)\nprint(int(ans))\n '] | ['Wrong Answer', 'Accepted'] | ['s616667859', 's663124775'] | [3992.0, 3888.0] | [156.0, 155.0] | [404, 390] |
p03425 | u367130284 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['from itertools import*;print(sum(p*q*r for p,q,r in combinations([list(s[0]for s in open(0).readlines()).count(s)for s in"MARCH"],3)))', 'import itertools as i,collections as c;print(sum(p*q*r for p,q,r in i.combinations(c.Counter(s[0]*(s[0]in"MARCH")for s in open(0).readlines()[1:]).values(),3)))\n', 'from itertools import*;print(sum(p*q*r for p,q,r in list(combinations([len(list(v))for k,v in groupby(sorted(s[0]for s in open(0).readlines()[1:]))],3))))', 'from itertools import*\nn,*a=open(0).read().split()\nprint(sum(p*q*r for p,q,r in combinations(map([i[0]for i in a].count,"MARCH"),3)))'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s791966041', 's828055897', 's928514879', 's969276548'] | [10748.0, 11004.0, 10664.0, 10992.0] | [36.0, 55.0, 56.0, 41.0] | [134, 161, 154, 133] |
p03425 | u368796742 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n = int(input())\nl = [0]*5\nfor i in range(n):\n if input()[0] == "M":\n l[0] += 1\n elif input()[0] == "A":\n l[1] += 1\n elif input()[0] == "R":\n l[2] += 1\n elif input()[0] == "C":\n l[3] += 1\n elif input()[0] == "H":\n l[4] += 1\nans = 0\nfor i in range(3):\n for j in range(i+1,4):\n for k in range(j+1,5):\n ans += l[i]*l[j]*l[k]\nprint(ans)\n \n ', 'n = int(input())\nl = [0]*5\nfor i in range(n):\n a = input()\n if a[0] == "M":\n l[0] += 1\n elif a[0] == "A":\n l[1] += 1\n elif a[0] == "R":\n l[2] += 1\n elif a[0] == "C":\n l[3] += 1\n elif a[0] == "H":\n l[4] += 1\n \n \nans = 0\nfor i in range(3):\n for j in range(i+1,4):\n for k in range(j+1,5):\n ans += l[i]*l[j]*l[k]\nprint(ans)\n \n \n'] | ['Runtime Error', 'Accepted'] | ['s475500459', 's776933343'] | [3064.0, 3064.0] | [161.0, 168.0] | [367, 406] |
p03425 | u375500286 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['n=int(input())\ns={"M":0,"A":0,"R":0,"C":0,"H":0}\nfor i in range(n):\n v=input()\n if v[0] in s:\n s[v[0]]+=1\nl=s.values()\nsum=0\nfor a in range(5):\n for b in range(a+1,5):\n for c in range(b+1,5):\n sum+=l[a]*l[b]*l[c] \nprint(sum)', 'n=int(input())\ns={"M":0,"A":0,"R":0,"C":0,"H":0}\nfor i in range(n):\n v=input()\n if v[0] in s:\n s[v[0]]+=1\nl=list(s.values())\nprint(l)\nsum=0\nfor a in range(5):\n for b in range(a+1,5):\n for c in range(b+1,5):\n sum+=l[a]*l[b]*l[c] \nprint(sum)\n', 'n=int(input())\n\ns={"M":0,"A":0,"R":0,"C":0,"H":0}\n\n\nfor i in range(n):\n v=input()\n if v[0] in s:\n s[v[0]]+=1\nsum=0\nfor a in s.keys():\n for b in s.keys():\n for c in s.keys():\n if a==b or a==c or b==c:\n continue\n sum+=s[a]*s[b]*s[c]\n \nprint(sum)', 'n=int(input())\ns={"M":0,"A":0,"R":0,"C":0,"H":0}\nfor i in range(n):\n v=input()\n if v[0] in s:\n s[v[0]]+=1\nl=list(s.values())\nans=0\nfor a in range(5):\n for b in range(a+1,5):\n for c in range(b+1,5):\n ans+=l[a]*l[b]*l[c] \nprint(ans)\n'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s024640891', 's122223341', 's876972281', 's002893175'] | [3064.0, 3064.0, 3188.0, 3064.0] | [167.0, 174.0, 173.0, 164.0] | [266, 282, 313, 273] |
p03425 | u388323466 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["# n = int(input())\n# arr = []\n\n# arr += [list(input())]\nn = 5\narr = [['Z','H'],['Z','N'],['Z','A'],['Z','O'],['Z','I']]\n\nMARCH = [0] * 5 \n \nfor j in range(n):\n if arr[j][0] == 'M':\n MARCH[0] += 1\n elif arr[j][0] == 'A':\n MARCH[1] += 1\n elif arr[j][0] == 'R':\n MARCH[2] += 1\n elif arr[j][0] == 'C':\n MARCH[3] += 1\n elif arr[j][0] == 'H':\n MARCH[4] += 1\n\ndef C(a,b):\n aa = 1\n bb = 1\n for k in range(a-b+1, a+1):\n aa *= k\n for l in range(1, b+1):\n bb *= l\n return int(aa/bb)\n \nans = C(sum(MARCH), 3)\n\nfor m in range(5):\n if MARCH[m] == 2:\n ans -= C(sum(MARCH)-MARCH[m],1)\n elif MARCH[m] >= 3:\n ans -= C(MARCH[m],2) * C(sum(MARCH)-MARCH[m],1)\n ans -= C(MARCH[m],3)\n\nprint(ans)\n ", "n = int(input())\narr = []\nfor i in range(n):\n arr += [list(input())]\n\nMARCH = [0] * 5 \n \nfor j in range(n):\n if arr[j][0] == 'M':\n MARCH[0] += 1\n elif arr[j][0] == 'A':\n MARCH[1] += 1\n elif arr[j][0] == 'R':\n MARCH[2] += 1\n elif arr[j][0] == 'C':\n MARCH[3] += 1\n elif arr[j][0] == 'H':\n MARCH[4] += 1\n\ndef C(a,b):\n aa = 1\n bb = 1\n for k in range(a-b+1, a+1):\n aa *= k\n for l in range(1, b+1):\n bb *= l\n return int(aa/bb)\n \nans = C(sum(MARCH), 3)\n\nfor m in range(5):\n if MARCH[m] == 2:\n ans -= C(sum(MARCH)-MARCH[m],1)\n elif MARCH[m] >= 3:\n ans -= C(MARCH[m],2) * C(sum(MARCH)-MARCH[m],1)\n ans -= C(MARCH[m],3)\n\nprint(ans)\n "] | ['Wrong Answer', 'Accepted'] | ['s948716802', 's341056719'] | [3064.0, 20304.0] | [18.0, 316.0] | [820, 748] |
p03425 | u401686269 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["from collections import defaultdict\nfrom itertools import combinations\n\nN=int(input())\nS=[input() for _ in range(N)]\n\nc = defaultdict(int)\n\nfor i in range(N):\n if S[i][0] in {'M','A','R','C','H'}:\n c[S[i][0]] += 1\n\nif len(c) < 3:\n print(0)\n exit()\n\nif len(c)==3:\n ans=0\n for k in c.keys():\n ans *= c[k]\n print(ans)\n exit()\n\nelse:\n ans = 0\n l = len(c)\n keys = list(c.keys())\n for C in combinations(range(l),3):\n tmp = c[keys[C[0]]]*c[keys[C[1]]]*c[keys[C[2]]]\n ans += tmp\n print(ans)", "from collections import defaultdict\nfrom itertools import combinations\n\nN=int(input())\nS=[input() for _ in range(N)]\n\nc = defaultdict(int)\n\nfor i in range(N):\n if S[i][0] in {'M','A','R','C','H'}:\n c[S[i][0]] += 1\n\nif len(c) < 3:\n print(0)\n exit()\n\nif len(c)==3:\n ans=1\n for k in c.keys():\n ans *= c[k]\n print(ans)\n exit()\n\nelse:\n ans = 0\n l = len(c)\n keys = list(c.keys())\n for C in combinations(range(l),3):\n tmp = c[keys[C[0]]]*c[keys[C[1]]]*c[keys[C[2]]]\n ans += tmp\n print(ans)"] | ['Wrong Answer', 'Accepted'] | ['s289957164', 's332755646'] | [16408.0, 16472.0] | [158.0, 149.0] | [506, 506] |
p03425 | u411923565 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["#60 C - March\nimport collections\nN = int(input())\nS = [input() for _ in range(N)]\nS = [s[0] for s in S if s[0] == 'M' or s[0] == 'A' or s[0] == 'R' or s[0] == 'C' or s[0] == 'H']\n\ncnt = collections.Counter(S)\nn = len(cnt)\n\nans = ((n*(n-1)*(n-2))//(3*2))\nif ans == 1:\n for i in cnt.values():\n if i>=2:\n ans * i\nelse:\n for i in cnt.values():\n if i>=2:\n ans += (n- (((n-1)*(n-2)*(n-3))//(3*2)))*(i-1)\nprint(ans)", "#60 C - March\nimport collections\nimport itertools\nN = int(input())\nS = [input() for _ in range(N)]\nS = [s[0] for s in S if s[0] == 'M' or s[0] == 'A' or s[0] == 'R' or s[0] == 'C' or s[0] == 'H']\n\ncnt = collections.Counter(S)\n\ncomb = itertools.combinations(cnt.keys(),3)\n\nans = 0\nfor a,b,c in comb:\n ans += cnt[a]*cnt[b]*cnt[c]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s571880291', 's778441072'] | [16992.0, 17196.0] | [146.0, 148.0] | [450, 341] |
p03425 | u413165887 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['from collections import Counter\nn = int(input())\ns = [input() for _ in range(n)]\nname_count = [0 for _ in range(5)]\nfor i in s:\n if i[0] == "M":\n name_count[0] += 1\n elif i[0] == "A":\n name_count[1] += 1\n elif i[0] == "R":\n name_count[2] += 1\n elif i[0] == "C":\n name_count[3] += 1\n elif i[0] == "H":\n name_count[4] += 1\n\nresult = 0\nfor i in range(5):\n result += name_count[i]*name_count[(i+1)%5]*name_count[(i+2)%5]\nprint(result)', 'from collections import Counter\nn = int(input())\ns = [input() for _ in range(n)]\nname_count = [0 for _ in range(5)]\nfor i in s:\n if i[0] == "M":\n name_count[0] += 1\n elif i[0] == "A":\n name_count[1] += 1\n elif i[0] == "R":\n name_count[2] += 1\n elif i[0] == "C":\n name_count[3] += 1\n elif i[0] == "H":\n name_count[4] += 1\n\nresult = 0\nfor s in range(3):\n for t in range(s+1, 4):\n for u in range(t+1, 5):\n result += name_count[s]*name_count[t]*name_count[u]\nprint(result)'] | ['Wrong Answer', 'Accepted'] | ['s112658041', 's858277645'] | [10112.0, 10108.0] | [175.0, 167.0] | [483, 539] |
p03425 | u417835834 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["import collections\nN = int(input())\na = ['']*N\nfor i in range(N):\n a[i] = list(input())[0]\nprint(a)\nc_list = [0]*5\nMARCH = ['M','A','R','C','H']\nc_list = [a.count(i) for i in MARCH]\nprint(c_list)\nsummation = 0\nfor i in range(3):\n for j in range(i+1,4):\n for k in range(j+1,5):\n summation += c_list[i]*c_list[j]*c_list[k]\nprint(summation)", "import collections\nN = int(input())\na = ['']*N\nfor i in range(N):\n a[i] = list(input())[0]\nc_list = [0]*5\nMARCH = ['M','A','R','C','H']\nc_list = [a.count(i) for i in MARCH]\nsummation = 0\nfor i in range(3):\n for j in range(i+1,4):\n for k in range(j+1,5):\n summation += c_list[i]*c_list[j]*c_list[k]\nprint(summation)"] | ['Wrong Answer', 'Accepted'] | ['s017613072', 's744396944'] | [5616.0, 4084.0] | [222.0, 225.0] | [361, 338] |
p03425 | u426108351 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\nS = []\nfor i in range(N):\n S.append(input())\nmarch = [0, 0, 0, 0, 0]\n\nfor i in range(N):\n if S[i][0] == "M":\n march[0] += 1\n if S[i][0] == "A":\n march[1] += 1\n if S[i][0] == "R":\n march[2] += 1\n if S[i][0] == "C":\n march[3] += 1\n if S[i][0] == "H":\n march[4] += 1\nans = 0\nprint(march)\nfor i in range(5):\n for j in range(i+1, 5):\n for k in range(j+1, 5):\n ans += march[i] * march[j] * march[k]\nprint(ans)\n', 'N = int(input())\nS = []\nfor i in range(N):\n S.append(input())\nmarch = [0, 0, 0, 0, 0]\n\nfor i in range(N):\n if S[i][0] == "M":\n march[0] += 1\n if S[i][0] == "A":\n march[1] += 1\n if S[i][0] == "R":\n march[2] += 1\n if S[i][0] == "C":\n march[3] += 1\n if S[i][0] == "H":\n march[4] += 1\nans = 0\nprint(march)\nfor i in range(5):\n for j in range(i+1, 5):\n for k in range(j+1, 5):\n ans += march[i] * march[j] * march[k]\nprint(ans)\n', 'N = int(input())\nS = []\nfor i in range(N):\n S.append(input())\nmarch = [0, 0, 0, 0, 0]\n\nfor i in range(N):\n if S[i][0] == "M":\n march[0] += 1\n if S[i][0] == "A":\n march[1] += 1\n if S[i][0] == "R":\n march[2] += 1\n if S[i][0] == "C":\n march[3] += 1\n if S[i][0] == "H":\n march[4] += 1\nans = 0\nfor i in range(5):\n for j in range(i+1, 5):\n for k in range(j+1, 5):\n ans += march[i] * march[j] * march[k]\nprint(ans)\n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s790133078', 's942412057', 's732815511'] | [9772.0, 9772.0, 9772.0] | [200.0, 202.0, 193.0] | [495, 495, 482] |
p03425 | u440129511 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["from itertools import combinations as c\nn=int(input())\ns=[input() for _ in range(n)]\nm=[m1 for m1 in s if s.startswith('M')]\na=[a1 for a1 in s if s.startswith('A')]\nr=[r1 for r1 in s if s.startswith('R')]\nc=[c1 for c1 in s if s.startswith('C')]\nh=[h1 for h1 in s if s.startswith('H')]\nmarch=[len(m),len(a),len(r),len(c),len(h)]\nprint(sum(i*j*k for i, j, k in c(march, 3)))", "from itertools import combinations as c\nn=int(input())\ns=[input() for _ in range(n)]\nmarch=[0]*5\nfor i in range(n):\n if s[i][0]=='M':march[0]+=1\n elif s[i][0]=='A':march[1]+=1\n elif s[i][0]=='R':march[2]+=1\n elif s[i][0]=='C':march[3]+=1\n elif s[i][0]=='H':march[4]+=1\nprint(sum(i*j*k for i, j, k in c(march, 3)))"] | ['Runtime Error', 'Accepted'] | ['s500804036', 's204830307'] | [9776.0, 9776.0] | [137.0, 174.0] | [372, 328] |
p03425 | u444856278 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ['N = int(input())\nS = {"M":0,"A":0,"R":0,"C":0,"H":0}\nfor i in range(N):\n s = input()\n if s[0] in ["M","A","R","C","H"]\n S[s[0]] += 1\n\nprint(S["M"] * S["A"] * (S["R"]+S["C"]+S["H"]) + S["M"] * S["R"] * (S["C"] + S["H"]) + S["M"] * S["C"] * S["H"] + S["A"] * S["R"] * (S["C"] + S["H"]) + S["C"] * S["A"] * S["H"] + S["R"] * S["C"] * S["H"])\n', 'N = int(input())\nS = {"M":0,"A":0,"R":0,"C":0,"H":0}\nfor i in range(N):\n s = input()\n if s[0] in ["M","A","R","C","H"]:\n S[s[0]] += 1\n\nprint(S["M"] * S["A"] * (S["R"] + S["C"] + S["H"])\n + S["M"] * S["R"] * (S["C"] + S["H"])\n + S["M"] * S["C"] * S["H"]\n + S["A"] * S["R"] * (S["C"] + S["H"])\n + S["A"] * S["C"] * S["H"]\n + S["R"] * S["C"] * S["H"])\n'] | ['Runtime Error', 'Accepted'] | ['s828931161', 's680705272'] | [2940.0, 3064.0] | [17.0, 175.0] | [353, 387] |
p03425 | u451017206 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n = int(input())\nl = ['M','A','R','C','H']\nm = []\na = []\nr = []\nc = []\nh = []\nfor i in range(n):\n s = input()\n if not s[0] in l: continue\n if s[0] == 'M':\n m.append(s)\n elif s[0] == 'A':\n a.append(s)\n elif s[0] == 'R':\n r.append(s)\n elif s[0] == 'C':\n c.append(s)\n else:\n h.append(s)\n\nn = sum([len(i) for i in [m,a,r,c,h]])\nall = n * (n - 1) * (n - 2) // 6\nprint(r)\nfor t in [m,a,r,c,h]:\n i = len(t)\n if i >= 2:\n k = i * (i -1) // 2\n all -= k*(n-2)\n if i >= 3:\n k = i * (i - 1) * (i - 2) // 6\n all -= k\nprint(all)", "n = int(input())\nl = ['M','A','R','C','H']\nm = []\na = []\nr = []\nc = []\nh = []\nfor i in range(n):\n s = input()\n if not s[0] in l: continue\n if s[0] == 'M':\n m.append(s)\n elif s[0] == 'A':\n a.append(s)\n elif s[0] == 'R':\n r.append(s)\n elif s[0] == 'C':\n c.append(s)\n else:\n h.append(s)\n\nn = sum([len(i) for i in [m,a,r,c,h]])\nall = n * (n - 1) * (n - 2) // 6\nfor t in [m,a,r,c,h]:\n i = len(t)\n if i >= 3:\n k = i * (i - 1) * (i - 2) // 6\n all -= k\n if i >= 2:\n k = i * (i - 1) // 2\n all -= k * (n - i)\nprint(all)\n"] | ['Wrong Answer', 'Accepted'] | ['s433250686', 's569569041'] | [10612.0, 10228.0] | [183.0, 180.0] | [606, 603] |
p03425 | u470542271 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["n = int(input())\ns_ = [input() for i in range(n)]\n\ns_list = [0, 0, 0, 0, 0]\nfor i in range(n):\n if s_[i][0] == 'M':\n s_list[0] += 1\n elif s_[i][0] == 'A':\n s_list[1] += 1\n elif s_[i][0] == 'R':\n s_list[2] += 1\n elif s_[i][0] == 'C':\n s_list[3] += 1\n elif s_[i][0] == 'H':\n s_list[4] += 1\n\nimport math\ncounter = 0\ns_list_nonzero = []\nfor i in range(5):\n if s_list[i] != 0:\n counter += 1\n s_list_nonzero.append(s_list[i])\n\nif counter < 3:\n print(0)\nelif counter == 4:\n print(s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[2] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[3])\nelif counter == 5:\n print(s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[2] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[4] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[4] +\n s_list_nonzero[0] * s_list_nonzero[3] * s_list_nonzero[4] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[4] +\n s_list_nonzero[1] * s_list_nonzero[3] * s_list_nonzero[4] +\n s_list_nonzero[2] * s_list_nonzero[3] * s_list_nonzero[4])\n", "n = int(input())\ns_ = [input() for i in range(n)]\n\ns_list = [0, 0, 0, 0, 0]\nfor i in range(n):\n if s_[i][0] == 'M':\n s_list[0] += 1\n elif s_[i][0] == 'A':\n s_list[1] += 1\n elif s_[i][0] == 'R':\n s_list[2] += 1\n elif s_[i][0] == 'C':\n s_list[3] += 1\n elif s_[i][0] == 'H':\n s_list[4] += 1\n\nimport math\ncounter = 0\ns_list_nonzero = []\nfor i in range(5):\n if s_list[i] != 0:\n counter += 1\n s_list_nonzero.append(s_list[i])\n\nif counter < 3:\n print(0)\nelif counter == 3:\n print(s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[2])\nelif counter == 4:\n print(s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[2] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[3])\nelif counter == 5:\n print(s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[2] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[1] * s_list_nonzero[4] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[0] * s_list_nonzero[2] * s_list_nonzero[4] +\n s_list_nonzero[0] * s_list_nonzero[3] * s_list_nonzero[4] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[3] +\n s_list_nonzero[1] * s_list_nonzero[2] * s_list_nonzero[4] +\n s_list_nonzero[1] * s_list_nonzero[3] * s_list_nonzero[4] +\n s_list_nonzero[2] * s_list_nonzero[3] * s_list_nonzero[4])\n"] | ['Wrong Answer', 'Accepted'] | ['s843270487', 's398871270'] | [9832.0, 9784.0] | [180.0, 174.0] | [1557, 1645] |
p03425 | u475675023 | 2,000 | 262,144 | There are N people. The name of the i-th person is S_i. We would like to choose three people so that the following conditions are met: * The name of every chosen person begins with `M`, `A`, `R`, `C` or `H`. * There are no multiple people whose names begin with the same letter. How many such ways are there to choose three people, disregarding order? Note that the answer may not fit into a 32-bit integer type. | ["from itertools import combinations\nn=int(input())\ns=[input()[0] for _ in range(n)]\na=[]\nfor i in ['M','A','R','C','H']:\n a+=[s.count(i)]\nprint(a)\nans=0\nfor i in combinations(a,3):\n ans+=i[0]*i[1]*i[2]\nprint(ans)", "from itertools import combinations\nn=int(input())\ns=[input()[0] for _ in range(n)]\na=[]\nfor i in ['M','A','R','C','H']:\n a+=[s.count(i)]\nans=0\nfor i in combinations(a,3):\n ans+=i[0]*i[1]*i[2]\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s410119233', 's321227338'] | [3888.0, 3888.0] | [142.0, 141.0] | [213, 204] |
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