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
|
|---|---|---|---|---|---|---|---|---|---|---|
p03282 | u139112865 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["#106_C\nS=input()\nK=int(input())\nfor i in range(len(S)):\n if S[i]!='1':\n ans=i\n break\nprint(ans)\nprint('1' if K<=ans else S[ans])", "#106_C\nS=input()\nK=int(input())\nans=0\nfor i in range(len(S)):\n if S[i]!='1':\n ans=i\n break\n\nprint('1' if K<=ans else S[ans])"] | ['Runtime Error', 'Accepted'] | ['s797313032', 's540375181'] | [2940.0, 3060.0] | [17.0, 17.0] | [145, 141] |
p03282 | u140191608 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['ans = 0\nif int(str(S)[0]) == 1 and K == 1:\n ans = 1\nelif int(str(S)[0]) != 1:\n ans = int(str(S)[0])\nelse:\n ans = int(str(S)[1])\nprint(ans)', 'S = int(input())\nK = int(input())\n\nfor i in range(len(str(S))):\n if int(str(S)[i]) == 1:\n if (K-1) == i:\n print(int(str(S)[i]))\n break\n continue\n else:\n print(int(str(S)[i]))\n break'] | ['Runtime Error', 'Accepted'] | ['s302375099', 's213401044'] | [3060.0, 3060.0] | [17.0, 17.0] | [147, 237] |
p03282 | u152671129 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = list(map(int, input().split()))\nk = int(input())\n\nanswer = 1\nfor i in range(min(k, len(s))):\n if s[i] != 1:\n answer = s[i]\n break\n\nprint(answer)\n', "s = input()\nk = int(input())\n\nprint(s)\n\nanswer = 1\nfor i in range(min(k, len(s))):\n if s[i] != '1':\n answer = s[i]\n break\n\nprint(answer)", "s = list(map(int, input().split()))\nk = int(input())\n\nanswer = 1\nfor i in range(min(k, len(s))):\n if s[i] != '1':\n answer = s[i]\n break\n\nprint(answer)", "s = list(input())\nk = int(input())\n\nanswer = 1\nfor i in range(min(k, len(s))):\n if s[i] != '1':\n answer = s[i]\n break\n print(i)\n\nprint(answer)", "s = input()\nk = int(input())\n\nanswer = '1'\nfor i in range(min(k, len(s))):\n if s[i] != '1':\n answer = s[i]\n break\n print(i)\n\nprint(answer)", "n = int(input())\ncnt = 0\n\nfor i in range(n):\n l, r = list(map(int, input().split(' ')))\n cnt += r - l + 1\n\nprint(cnt)", "s = input()\nk = int(input())\n\nanswer = 1\nfor i in range(min(k, len(s))):\n if s[i] != '1':\n answer = s[i]\n break\n\nprint(answer)"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s200437016', 's220502286', 's377457158', 's521343596', 's671417427', 's834575951', 's800438430'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0, 17.0, 17.0, 17.0, 19.0] | [166, 153, 167, 162, 158, 123, 143] |
p03282 | u155828990 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import sys\ns=input()\nk=int(input())\nif(len(s)==1):\n print(s[0])\n sys.exit()\nc=0\nfor i in range(k):\n if(s[i]==1):c+=1\n else:break\nif(c==k):\n print(1)\nelse:\n if(c==0):print(s[0])\n else:print(s[c+1])', "import sys\ns=input()\nk=int(input())\nif(len(s)==1):\n print(s[0])\n sys.exit()\nif(k==1):\n print(s[0])\n sys.exit()\nc=0\nfor i in range(k):\n if(s[i]=='1'):c+=1\n else:break\nif(c==k):\n print(1)\nelse:\n if(c==0):print(s[0])\n else:print(s[c+1])", "import sys\ns=input()\nk=int(input())\nif(len(s)==1):\n print(s[0])\n sys.exit()\nif(k==1):\n print(s[0])\n sys.exit()\nc=0\nfor i in range(k):\n if(s[i]=='1'):c+=1\n else:break\nif(c==k):\n print(1)\nelse:\n print(s[c])"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s553127849', 's821823747', 's856191275'] | [3064.0, 3064.0, 3064.0] | [17.0, 17.0, 17.0] | [217, 260, 228] |
p03282 | u163320134 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=input()\nk=int(input())\nif s[0]!='1':\n print(s[0])\nelse:\n pos=0\n for i in range(len(s)):\n if i!=s[:i].count('1'):\n break\n pos=i\n if k<=pos:\n print('1')\n else:\n print(s[pos+1])", "s=input()\nk=int(input())\nif s[0]!='1':\n print(s[0])\nelse:\n pos=0\n for i in range(len(s)):\n if i!=s[:i].count('1'):\n break\n pos=i\n if k<=pos:\n print('1')\n else:\n print(s[pos])"] | ['Runtime Error', 'Accepted'] | ['s465894810', 's124564823'] | [3060.0, 3060.0] | [17.0, 17.0] | [198, 196] |
p03282 | u163449343 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = list(input())\nK = int(input())\nif K == 1:\n print(S[0])\nelse:\n for x in range(S):\n if int(x) != 1:\n print(S[x])\n break', 'S = list(map(int,input()))\nK = int(input())\nans = 1\nif K == 1 or S.count(S[0]) == len(S) or S.count(1) == len(S):\n ans = S[0]\nelse:\n for x in range(K):\n if S[x] != 1:\n ans = S[x]\n break\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s843912078', 's240448341'] | [2940.0, 2940.0] | [18.0, 17.0] | [138, 214] |
p03282 | u163874353 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nfor i in s:\n if i != "1":\n print(i)\n break\nprint(1)\n ', 's = input()\nk = int(input())\nfor i in range(k):\n if s[i] != "1":\n print(s[i])\n exit()\nprint(1)'] | ['Wrong Answer', 'Accepted'] | ['s163077526', 's710886602'] | [2940.0, 2940.0] | [17.0, 17.0] | [106, 111] |
p03282 | u169138653 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=input()\nk=int(input())\nok=True\nfor i in range(len(s)):\n if i<k:\n if s[i]!='1':\n ok=False\n a=s[i]\n else:\n break\nif ok:\n print('1')\nelse:\n print(a)", "s=input()\nk=int(input())\nok=True\nfor i in range(len(s)):\n if i<k:\n if s[i]!='1':\n ok=False\n a=s[i]\n break\n else:\n break\nif ok:\n print('1')\nelse:\n print(a)\n"] | ['Wrong Answer', 'Accepted'] | ['s351799232', 's403028285'] | [3060.0, 2940.0] | [17.0, 18.0] | [167, 180] |
p03282 | u170324846 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n\nk = -1\nwhile S[k+1] == "1":\n k += 1\n print(k)\n\nif K <= k:\n print(1)\nelse:\n print(S[k+1])', 'import sys\nS = input()\nK = int(input())\n\nfor i in range(K):\n if S[i] != "1":\n print(S[i])\n sys.exit()\nprint(1)'] | ['Runtime Error', 'Accepted'] | ['s949242280', 's457251211'] | [3060.0, 2940.0] | [17.0, 17.0] | [122, 117] |
p03282 | u170765582 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["a=input()\nk=0\nwhile a[k]==1 and len(a)>k:\n k+=1\nb=int(input())\nif b<k or a=='1'*len(a):print(1)\nelse:print(a[k])", "a=input()\nb=int(input())\nif b<101:\n if a[:b]=='1'*b:print(1)\n else:\n k=0\n while a[k]=='1':k+=1\n print(a[k])\nelse:\n k=0\n while a[k]=='1':k+=1\n print(a[k])"] | ['Wrong Answer', 'Accepted'] | ['s819642862', 's025892074'] | [2940.0, 3060.0] | [17.0, 18.0] | [113, 167] |
p03282 | u171065106 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nprint(s[0])', 's = input()\nk = int(input()) - 1\n\n\nfor i in range(len(s)):\n if s[i] != "1" or i == k:\n break\n\nprint(s[i])'] | ['Wrong Answer', 'Accepted'] | ['s172295861', 's854615382'] | [9156.0, 9080.0] | [28.0, 25.0] | [41, 115] |
p03282 | u177756077 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S=str(input())\nK=int(input())\nfor i in range(len(S)) :\n if S[i]!=1 :\n break\nprint(S[i])', 'S=str(input())\nK=int(input())\nfor i in range(K) :\n# print("checking "+S[i])\n if S[i]!="1" :\n break\nprint(S[i])'] | ['Wrong Answer', 'Accepted'] | ['s087823620', 's490935346'] | [2940.0, 2940.0] | [17.0, 17.0] | [97, 123] |
p03282 | u183840468 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['def make_divisors(n):\n divisors = []\n for i in range(1, int(n**0.5)+1):\n if n % i == 0:\n divisors.append(i)\n if i != n // i:\n divisors.append(n//i)\n \n return divisors\n\n\nn = int(input())\nans = 0\nif n < 105:\n print(0)\nelse:\n for i in range(105,n+2,2):\n m = make_divisors(i)\n if len(m) == 8:\n ans += 1\n\nprint(ans)', "s = input()\nk = int(input())\n\nct = 0\nfor i in range(len(s)):\n if s[i] != '1':\n ct += i+1\n break\n\nif ct > k:\n print(1)\nelse:\n print(s[ct-1])"] | ['Wrong Answer', 'Accepted'] | ['s679475707', 's979987641'] | [3316.0, 2940.0] | [2104.0, 17.0] | [394, 162] |
p03282 | u187205913 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s = list(input())\nk = int(input())\nfor i in range(k):\n s_ = s[i]\n if not s_=='1':\n print(s_)\n break\nprint('1')", "s = list(input())\nk = int(input())\nfor i in range(k):\n s_ = s[i]\n if not s_=='1':\n print(s_)\n exit()\nprint('1')\n"] | ['Wrong Answer', 'Accepted'] | ['s755145225', 's209505588'] | [2940.0, 2940.0] | [17.0, 17.0] | [118, 120] |
p03282 | u189326411 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nif len(s)>k and all(s[i]=="1" for i in range(k)):\n print(1)\nelse:\n print(s[0])\n', 's = input()\nk = int(input())\n\nif len(s)>k and all(s[i]=="1" for i in range(k)):\n print(1)\nelse:\n while s[0]=="1":\n s.pop(0)\n print(s[0])\n', 's = input()\nk = int(input())\n\nif all(s[i]=="1" for i in range(min(len(s),k))):\n print(1)\nelse:\n while s[0]=="1":\n s.pop(0)\n print(s[0])\n', 's = input()\nk = int(input())\n\nif all(s[i]=="1" for i in range(min(len(s),k))):\n print(1)\nelse:\n while s[0]=="1":\n s = s[1:]\n print(s[0])\n'] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s201184578', 's713401205', 's805434151', 's495285335'] | [2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 19.0, 20.0, 17.0] | [115, 153, 152, 153] |
p03282 | u189575640 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import sys\n# N,M = [int(n) for n in input().split()]\n# S = str(input())\n# T = str(input())\n\nS = str(input())\nK = int(input())\nnumplace = 0\n\nfor i in range(len(S)):\n if (S[i] == "1"):\n numplace+=1\n else:\n numplace+= int(S[i])**5000000000000000\n if(numplace>=K):\n print(S[i])\n sys.exit()\n', 'import sys\n# N,M = [int(n) for n in input().split()]\n# S = str(input())\n# T = str(input())\nS = str(input())\nK = int(input())\n\nfor i in range(len(S)):\n if(S[i] == "1" and i+1 == K:\n print("1")\n sys.exit()\n elif(S[i]!="1"):\n print(S[i])\n sys.exit()\n', 'import sys\n# N,M = [int(n) for n in input().split()]\n# S = str(input())\n# T = str(input())\nS = str(input())\nK = int(input())\n\nfor i in range(len(S)):\n if(S[i] == "1" and i+1 == K):\n print("1")\n sys.exit()\n elif(S[i]!="1"):\n print(S[i])\n sys.exit()\n'] | ['Time Limit Exceeded', 'Runtime Error', 'Accepted'] | ['s401442890', 's687637918', 's534085303'] | [197060.0, 2940.0, 2940.0] | [2106.0, 18.0, 18.0] | [323, 281, 282] |
p03282 | u191635495 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import math\nfrom collections import deque\n\ns = input()\nk = int(input())\ns_list = list(map(int, list(s)))\n\nif len(s) == 1:\n print(int(s))\nelif k == 1:\n print(s_list[0])\nelse:\n print(s_list[0])', 'S = input()\nK = int(input())\ns_list = list(map(int, list(S)))\n\nflag = False\nfor k in range( min(len(S), K) ):\n if s_list[k] != 1:\n flag = True\n break\n\nif flag:\n print(s_list[k])\nelse:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s120280647', 's563203266'] | [3316.0, 3060.0] | [21.0, 19.0] | [200, 216] |
p03282 | u192588826 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nonesum = 0\nwhile s[0] == "1" and len(s) > 0\n s = s[1:]\n onesum += 1\nif k <= onesum:\n print(1)\nelse:\n print(s[0])\n', 's = input()\nk = int(input())\nonesum = 0\nwhile s[0] == "1":\n onesum += 1\n if len(s) == 1:\n break\n s = s[1:]\nif k <= onesum:\n print(1)\nelse:\n print(s[0])\n'] | ['Runtime Error', 'Accepted'] | ['s295538884', 's178902047'] | [2940.0, 2940.0] | [17.0, 18.0] | [154, 174] |
p03282 | u193927973 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\nif s[0]=="1":\n print(1)\nelse:\n print(s[1])', 's=input()\nk=int(input())\nif k==1 or len(s)==1:\n print(s[0])\nelse:\n for i in range(k):\n if s[i]!="1":\n print(s[i])\n exit()\n print(1)\n\n '] | ['Runtime Error', 'Accepted'] | ['s307324945', 's584655670'] | [8820.0, 9084.0] | [25.0, 28.0] | [69, 151] |
p03282 | u201928947 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nfor i in range(min(k,len(s))):\n if s[i] != 1:\n print(s[i])\n exit()\nelse:\n print(1)', 's = input()\nk = int(input())\nfor i in range(min(k,len(s))):\n if s[i] != "1":\n print(s[i])\n exit()\nelse:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s856735015', 's101145773'] | [2940.0, 3060.0] | [17.0, 17.0] | [131, 133] |
p03282 | u207185597 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = list(input())\nk = int(input())\nif k >= 2 and s[0] == 1:\n print(s[1])\nif len(s) == 1:\n print(s[0])\nelse:\n print(s[0])\n', 's = list(input())\nk = int(input())\nfor i in range(k):\n if s[i] != 1:\n l = [i for i in s if i != 1]\n print(l[0])\n break\nelse:\n print(1)\n', 's = list(input())\nk = int(input())\ncnt = 0\nif k >= len(s):\n x = len(s)\nelse:\n x = k\n\nfor i in range(x):\n if s[i] == 1:\n cnt += 1\nif cnt == len(s):\n print(1)\nelse:\n for j in s:\n if int(j) >= 1:\n print(int(j))\n break', "s = list(input())\nk = int(input())\ncnt = 0\nx = min(len(s),k)\nfor i in range(x):\n if s[i] != '1':\n print(s[i])\n break\nelse:\n print(1)"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s245899010', 's496875813', 's501865333', 's126786367'] | [2940.0, 2940.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0, 18.0] | [130, 162, 265, 152] |
p03282 | u210827208 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\n\nif s[0]==1:\n print(s[1])\nelse:\n print(s[0])', 's=input()\nk=int(input())\n\nif s[0]==1 and k!=1:\n print(s[1])\nelse:\n print(s[0])\n ', "s=input()\nk=int(input())\ncnt=0\nfor i in range(len(s)):\n if s[i]=='1':\n cnt+=1\n else:\n break\n\nif k>cnt:\n print(s[cnt])\nelse:\n print(1)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s074485164', 's244199039', 's224874180'] | [2940.0, 2940.0, 2940.0] | [18.0, 17.0, 17.0] | [75, 89, 159] |
p03282 | u214434454 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nfor i in (len(s)):\n if s[i] != "1":\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nif k =< len(s):\n if s[:k] == "1" * k:\n ans = 1\n else:\n ans = int(s[k-1])\nelse:\n for i in range(len(s)):\n if s[i] != "1":\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nif k <= len(s):\n if s[:k] == "1" * k:\n ans = 1\n else:\n ans = int(s[k-1])\nelse:\n for i in range(len(s)):\n if s[i] != "1":\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nif k <= len(s):\n if s[:k] == "1" * k:\n ans = 1\n else:\n for i in range(len(s)):\n if s[i] != "1":\n ans = int(s[i])\n break\nelse:\n for i in range(len(s)):\n if s[i] != "1":\n ans = int(s[i])\n break\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s022817305', 's198956958', 's207014620', 's079806258'] | [2940.0, 2940.0, 3060.0, 3060.0] | [17.0, 17.0, 18.0, 18.0] | [116, 244, 244, 332] |
p03282 | u215315599 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nfor s in S:\n if s == "1" and S.index(s) < K-1:\n continue\n else:\n print(s)', 'S = input()\nK = int(input())\ncnt = 0\nfor s in S:\n cnt += 1\n if s == "1" and cnt < K:\n continue\n else:\n print(s)', 'S = input()\nK = int(input())\ncnt = 0\nfor s in S:\n cnt += 1\n if s == "1" and cnt < K:\n continue\n else:\n print(s)\n exit()'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s025548333', 's104689313', 's665351501'] | [3064.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0] | [122, 134, 149] |
p03282 | u215743476 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nif k == 1:\n print(s[0])\nelse:\n if s[0] == 1:\n print(s[1])\n else:\n print(s[0])', "s = input()\nk = int(input())\n\nif len(s) == 1:\n print(s[0])\n exit()\n\nif k == 1:\n print(s[0])\nelse:\n i == 0\n while i <= k:\n if s[i] != '1':\n print(s[i])\n break\n elif i == k:\n print(s[i])\n break\n i += 1\n", "s = input()\nk = int(input())\n\ni = 1\nwhile i <= k:\n if s[i-1] != '1':\n print(s[i-1])\n break\n elif i == k:\n print(s[i-1])\n break\n i += 1\n"] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s777917122', 's996022589', 's348008449'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 18.0] | [130, 280, 172] |
p03282 | u216289806 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['data = input()\nnum = int(input())\n\ncnt = 0\nfor lop in range(len(data)):\n cnt = cnt + (int(data[lop])*1000000000000000000)**5\n if cnt > num:\n print(data[lop])\n break', 'data = input()\nnum = int(input())\n\ncnt = 0\nfor lop in range(len(data)):\n if int(data[lop]) == 1:\n cnt = cnt + 1\n if num == cnt:\n print(data[lop])\n break\n else : \n cnt = cnt + (int(data[lop])*1000000000000000000)**5\n if cnt > num:\n print(data[lop])\n break'] | ['Wrong Answer', 'Accepted'] | ['s176360347', 's019307949'] | [2940.0, 3060.0] | [20.0, 17.0] | [184, 335] |
p03282 | u225388820 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=input()\nk=int(input())\ni=0\nwhile True:\n if s[i]=='1' or i<len(s):\n i+=1\n else:\n break\nif i<k-1:\n print(s[1])\nelse:\n print(1)", "s=input()\nk=int(input())\ni=0\nwhile True:\n if i<len(s) and s[i]=='1':\n i+=1\n else:\n break\nif i>=k and len(s)>=2:\n print(s[1])\nelse:\n print(s[0])\n", "s=input()\nk=int(input())\ni=0\nwhile True:\n if i<len(s) and s[i]=='1':\n i+=1\n else:\n break\nif i==0:\n print(s[0])\n exit()\nif i<k:\n print(s[i])\nelse:\n print(s[0])\n"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s641157096', 's724910312', 's445477815'] | [2940.0, 2940.0, 2940.0] | [18.0, 17.0, 17.0] | [136, 154, 171] |
p03282 | u226108478 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["# -*- coding: utf-8 -*-\n\n\ndef main():\n s = input()\n k = int(input())\n\n if s[0] == 1:\n print(s[1])\n else:\n print(s[0])\n\n\nif __name__ == '__main__':\n main()\n", "# -*- coding: utf-8 -*-\n\n\ndef main():\n s = input()\n k = int(input())\n one_count = 0\n\n for si in s:\n if si == '1':\n one_count += 1\n else:\n break\n\n if s[0] == '1':\n if k == 1:\n print(s[0])\n elif one_count >= k:\n print(1)\n elif one_count < k:\n print(s[k - 1])\n else:\n print(s[1])\n else:\n print(s[0])\n\n\nif __name__ == '__main__':\n main()\n", "# -*- coding: utf-8 -*-\n\n\ndef main():\n s = input()\n k = int(input())\n count = 0\n\n for si in s:\n if si == '1':\n count += 1\n else:\n break\n\n if s[0] != '1':\n print(s[0])\n else:\n if k <= count:\n print(1)\n else:\n print(s[count])\n\n\nif __name__ == '__main__':\n main()\n"] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s173562836', 's502736642', 's679343437'] | [2940.0, 3060.0, 2940.0] | [17.0, 17.0, 19.0] | [184, 470, 361] |
p03282 | u228223940 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nfor i in range(k):\n if s[i] == 1:\n continue\n else:\n print(s[i])\n exit()\n \nprint(1)', 's = input()\nk = int(input())\n\nfor i in range(k):\n if int(s[i]) == 1:\n continue\n else:\n print(s[i])\n exit()\n \nprint(1)'] | ['Wrong Answer', 'Accepted'] | ['s954102838', 's665483355'] | [2940.0, 2940.0] | [17.0, 17.0] | [146, 151] |
p03282 | u230621983 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=str(input())\nn = int(input())\nfor i in range(n):\n if s[i] != "1":\n print(int(s[i]))\n break\n\telse:\n \tprint("1")', 's = input()\nk = int(input())\nfor i in range(k):\n if s[i] != "1":\n break\nprint(s[i])'] | ['Runtime Error', 'Accepted'] | ['s842012398', 's189474326'] | [2940.0, 2940.0] | [17.0, 17.0] | [132, 93] |
p03282 | u239342230 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = str(input())\nK = int(input())\nif S[0] == 1 and K == 1:\n print(1)\n# print('A')\nelif len(S) <= K and list(map(int,list(S[:K]))).count(1) == K:\n print(1)\n# print('B')\nelif len(S) == 1:\n print(S[0])\n# print('C')\nelif S[0] != 1:\n print(S[0])\nelse :\n print(S[1])", 'S = str(input())\nK = int(input())\nif S[0] == 1 & K == 1:\n print(1)\nelif len(S) != 1:\n print(S[0])\nelse:\n print(S[0])', "S = str(input())\nK = int(input())\ni = 0\nif S[0] == '1' and K == 1:\n print(1)\n# print('A')\nelif list(map(int,list(S[:(K-1)]))).count(1) == K-1:\n print(S[K-1])\n# print('B')\n\nelif len(S) == 1:\n print(S[0])\n# print('C')\nelif S[0] != '1':\n print(S[0])\n# print('D')\nelse :\n while(1):\n if list(S[:(K-1)])[i] == '1':\n i = i + 1\n elif i > K:\n print(S[1])\n# print('E')\n else:\n print(list(S[:(K-1)])[i])\n break\n "] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s427008652', 's768745036', 's119998740'] | [3060.0, 3060.0, 3064.0] | [17.0, 18.0, 19.0] | [287, 125, 512] |
p03282 | u244836567 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['a=input()\nb=int(input())\nc=1\nif a[0]!="1":\n print(a[0])\nif a[0]=="1":\n for i in range(len(a)-1):\n if a[i]=="1" and a[i+1]=="1":\n c=c+1\n else:\n break\nif c>b:\n print(1)\nelse:\n print(a[c-1])\n', 'a=input()\nb=int(input())\nc=0\nif a[0]!="1":\n print(a[0])\nif a[0]=="1":\n for i in range(len(a)-1):\n if a[i]=="1" and a[i+1]=="1":\n c=c+1\n else:\n break\nif c>=b:\n print(1)\nelse:\n print(a[c-1])\n', 'a=input()\nb=int(input())\nc=0\nif a[0]!=1:\n print(a[0])\nelse:\n for i in range(len(a)-1):\n if a[i]==a[i+1]:\n c=c+1\n else:\n break\n if c>=b:\n print(1)\n else:\n print(a[1])', 'a=input()\nb=int(input())\nc=0\nif a[0]!="1":\n print(a[0])\nif a[0]=="1":\n for i in range(len(a)-1):\n if a[i]=="1" and a[i+1]=="1":\n c=c+1\n else:\n break\nif c>b:\n print(1)\nelse:\n print(a[c])\n', 'a=input()\nb=int(input())\nc=1\nif a[0]!="1":\n print(a[0])\nif a[0]=="1":\n for i in range(len(a)-1):\n if a[i]=="1" and a[i+1]=="1":\n c=c+1\n else:\n break\n if c>=b:\n print(1)\n else:\n print(a[c+1])\n', 'a=input()\nb=int(input())\nc=1\nif a[0]!="1":\n print(a[0])\nif a[0]=="1":\n for i in range(len(a)-1):\n if a[i]=="1" and a[i+1]=="1":\n c=c+1\n else:\n break\n if c>=b:\n print(1)\n else:\n print(a[c])'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s094041164', 's154868005', 's475051091', 's592354122', 's897217361', 's944179212'] | [9168.0, 8972.0, 9096.0, 9004.0, 9028.0, 9128.0] | [29.0, 28.0, 24.0, 30.0, 31.0, 28.0] | [208, 209, 191, 206, 217, 214] |
p03282 | u245870380 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = input()\n\nif len(S) == 1:\n print(S)\n\nelif set(S) == 1:\n print(S)\n\nelif int(K) == 1:\n print(S[0])\n\nelif S[0] == 1:\n print(S[1])\n \nelse:\n print(int(S[0]))', 'S = input()\nK = input()\n\nfor i in range(int(K)):\n if S[i] == "1":\n if i+1 == int(K):\n print(1)\n break\n else:\n print(S[i])\n break'] | ['Wrong Answer', 'Accepted'] | ['s787539723', 's373546136'] | [3060.0, 2940.0] | [17.0, 17.0] | [185, 177] |
p03282 | u247830763 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import sys\ns = str(input())\nn = int(input())\ni = 0\nprint(s[i])\nwhile True:\n if int(s[i]) == 1:\n if (n-1) == i:\n print("1")\n sys.exit()\n else:\n i += 1\n else:\n print(s[i])\n sys.exit()\n', 'import sys\ns = str(input())\nn = int(input())\ni = 0\nwhile True:\n if s[i] == 1:\n if n-1 == i:\n print("1")\n sys.exit()\n else:\n i += 1\n else:\n print(s[i])\n sys.exit()\n', 's = str(input())\nn = int(input())\nif s[0] == 1:\n print(s[1])\nelse:\n print(s[0])', 'import sys\ns = str(input())\nn = int(input())\ni = 0\nwhile True:\n if s[i] == 1:\n i +=1\n else:\n print(s[i])\n sys.exit()', 'import sys\ns = str(input())\nn = int(input())\ni = 0\nwhile True:\n if int(s[i]) == 1:\n if (n-1) == i:\n print("1")\n sys.exit()\n else:\n i += 1\n else:\n print(s[i])\n sys.exit()\n'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s111610885', 's117844877', 's439627543', 's819785793', 's114496920'] | [2940.0, 3060.0, 3064.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0, 17.0] | [249, 230, 85, 143, 237] |
p03282 | u252828980 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['n = input()\nk = int(input())\n\nif "1" in n:\n one = n.rfind("1")-n.find("1")+1\n #print(n.rfind("1")+1)\n if one < k:\n print(n[n.rfind("1")+1])\n elif one >= k:\n print(1)\nelif "1" not in n:\n print(n[0])', 's = input()\nk = int(input())\nnum = 0\nfor i in s:\n if i == "1":\n num +=1\n if num == k:\n print(1)\n exit()\n else:\n print(i)\n exit()'] | ['Runtime Error', 'Accepted'] | ['s990660835', 's373206558'] | [2940.0, 3060.0] | [17.0, 19.0] | [226, 184] |
p03282 | u252964975 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S=input()\nK=int(input())\n\nprint(S[0])', "S=input()\nK=int(input())\n\none = True\nfor i in range(K):\n one = one and S[i]=='1'\n if not one:\n print(S[i])\n break\nif one:\n print(1)"] | ['Wrong Answer', 'Accepted'] | ['s633478432', 's070096454'] | [2940.0, 2940.0] | [18.0, 17.0] | [37, 140] |
p03282 | u253209284 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['for i,n in enumerate(int(input())):\n if n!= 1:\n break\nprint(n if i<int(input()) else 1)', 'for i,n in enumerate(int(input())):\n if n!= 1:\n break\nprint(n if i<input() else 1)', 'for i,n in enumerate(input()):\n if n!= 1:\n break\nprint(n if i<int(input()) else 1)', 'S = input()\nK = input()\nfor i,n in enumerate(S):\n if n!= 1:\n break\nprint(n if i<K else 1)', "for i,n in enumerate(input()):\n if n!= '1':\n break\nprint(n if i<int(input()) else 1)"] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s189238426', 's647363712', 's772303238', 's996155361', 's681033076'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0, 17.0, 17.0] | [91, 86, 86, 93, 88] |
p03282 | u253952966 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nday = 5 * (10 ** 15)\n\nfor c in s:\n i = int(c)\n l = i ** day\n if k <= l:\n print(c)\n exit()\n else:\n k -= l', "s = input()\nk = int(input())\n\nfor c in s:\n if c != '1':\n print(c)\n exit()\n else:\n k -= 1\n if k == 0:\n print(1)\n exit()\nprint(1)"] | ['Time Limit Exceeded', 'Accepted'] | ['s640591291', 's006081290'] | [197060.0, 2940.0] | [2105.0, 18.0] | [146, 151] |
p03282 | u254086528 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = list(input())\nk = int(input())\n\nfor w in s:\n if w != 1:\n print(w)\n break', 's = input()\nk = int(input())\n\nfor w in s:\n if w == "1":\n k -= 1\n if k == 0:\n print(1)\n break\n else:\n print(int(w))\n break\n '] | ['Wrong Answer', 'Accepted'] | ['s803980276', 's954530909'] | [2940.0, 2940.0] | [17.0, 17.0] | [93, 186] |
p03282 | u266569040 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = list(input())\nK = int(input())\nS = [int(i) for i in S]\nif K == 1 or len(S)==1:\n print(S[0])\nelse:\n if S[0] == 1:\n print(1)\n else:\n print(S[0])', 'S = input().split()\nK = int(input())\nS = [int(i) for i in S]\nif S[0] == 1:\n print(S[1])\nelse:\n print(S[0])', 'import sys\nS = list(input())\nK = int(input())\nS = [int(i) for i in S]\ncount = 0\nfor i in range(K):\n if S[i] is not 1:\n print(S[i])\n sys.exit()\n else:\n count+= 1\nif count == K:\n print(1)'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s773194014', 's868756688', 's289254753'] | [3060.0, 3316.0, 3060.0] | [17.0, 21.0, 17.0] | [169, 112, 215] |
p03282 | u268318377 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = [int(s) for s in input()]\nK = int(input())\nif len(S) >= K and S[K-1] == 1:\n print(1)\nelse: \n for k, s in enumerate(S, start=1):\n if s == 1:\n print(1)\n break\n else:\n print(s)\n break', 'S = [int(s) for s in input()]\nK = int(input())\nif set(S[:K]) == {1}:\n print(1)\nelse:\n for i in S:\n if i != 1:\n print(i)\n break'] | ['Wrong Answer', 'Accepted'] | ['s335760863', 's237750984'] | [3060.0, 2940.0] | [17.0, 17.0] | [251, 161] |
p03282 | u272557899 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n \ns = list(s)\nv = 5 * 10**15\na = []\nfor i in range(9):\n a.append(i ** v)\n\nnuar = []\nlear = []\nco = 0\nrear = 0\n\nfor i in s:\n for j in range(1,10):\n if int(i) == j:\n nuar.append(j)\n rear += a[j - 1]\n lear.append(rear)\n \nfor p in range(len(s)):\n if p == 0:\n if lear[p] >= k:\n print(nuar[p])\n elif lear[p] >= k and lear[p - 1] < k:\n print(nuar[p])\n \n ', 's = input()\nk = int(input())\n \ns = list(s)\n\nco = 1\n \nfor i in s:\n if int(i) == 1:\n if co == k:\n print(int(i))\n break\n else:\n print(int(i))\n break\n co += 1\n \n \n'] | ['Time Limit Exceeded', 'Accepted'] | ['s871539266', 's538789751'] | [197052.0, 2940.0] | [2106.0, 17.0] | [415, 187] |
p03282 | u283220417 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nans = ""\n\nif int(S[0]) > 1:\n ans = S[0]\nelif K>=100:\n for i in range(len(S)):\n if int(S[i]) > 1:\n ans = S[i]\n break\n elif i==S-1:\n ans = 1\nelse:\n for i in range(len(S)):\n if int(S[i]) > 1:\n if i+1 <= K:\n ans = S[i]\n break\n else:\n ans = 1\n break\n elif i==S-1:\n ans = 1\nprint(ans)', 'S = input()\nK = int(input())\nans = 1\n\nfor i in range(len(S)):\n if int(S[i]) == 1:\n continue\n else:\n if i+1 <= K:\n ans = S[i]\n break\n\nprint(ans)'] | ['Runtime Error', 'Accepted'] | ['s360018725', 's605685965'] | [3064.0, 2940.0] | [17.0, 17.0] | [380, 161] |
p03282 | u284102701 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s = input()\nk = int(input())\nidx = 0\nwhile idx < len(s) and s[idx] == '1':\n idx += 1\nif k < idx or k == len(s):\n print(1)\nelse:\n print(s[idx])\n", "s = input()\nk = int(input())\nidx = 0\nwhile idx < len(s) and s[idx] == '1':\n idx += 1\nif k <= idx or idx == len(s):\n print(1)\nelse:\n print(s[idx])\n"] | ['Wrong Answer', 'Accepted'] | ['s268697247', 's477238493'] | [3060.0, 2940.0] | [19.0, 18.0] | [152, 155] |
p03282 | u288430479 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nm = 0\nfor i in s:\n if i==1:\n m += 1\n else:\n break\nif m>=k:\n print(1)\nelse:\n for i in s:\n if i!=1:\n print(i)\n break', 's = input()\nk = int(input())\nm = 0\nfor i in s:\n if i==\'1\':\n m += 1\n else:\n break\nif m>=k:\n print(1)\nelse:\n for i in s:\n if i!="1":\n print(i)\n break\n else:\n continue'] | ['Wrong Answer', 'Accepted'] | ['s511005216', 's944385884'] | [2940.0, 2940.0] | [17.0, 18.0] | [166, 195] |
p03282 | u288948615 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = list(input().split())\n\nif S[0] == '1':\n print(S[1])\nelse:\n print(S[0])", "def func(N, K):\n for n in range(len(N)):\n if N[n] != '1':\n if K > n:\n return N[n]\n return 1\n\nprint(func(input(), int(input())))"] | ['Runtime Error', 'Accepted'] | ['s517268726', 's662842983'] | [2940.0, 2940.0] | [17.0, 17.0] | [76, 166] |
p03282 | u294385082 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nsumtmp = 0\n\nfor i in s:\n sumlen += int(i)**(5 * (10)**15)\n if sumtmp >= k:\n print(i)', "s = list(input())\nk = int(input())\ncoun= 0\n\nfor i in range(k):\n coun += 1\n if s[i] != '1':\n print(i)\n exit()\n if coun == 100:\n print(1)\n exit()\n \n ", "s = list(input())\nk = int(input())\n\nif s[0] == '1' and len(s) == 1:\n print(s[0])\nif s[0] == '1' and s[:k] == ['1']*k and len(s) != 1:\n print(s[1])\nelif s[0]!='1':\n print(s[0])", "s = list(input())\nk = int(input())\n\nif s[0] == '1' and len(s) == 1:\n print(s[0])\nif s[0] == '1' and s[:k] == ['1']*k:\n print(s[1])\nelif s[0]!='1':\n print(s[0])", "s = list(input())\nk = int(input())\ncoun= 0\n\nif k <= 100:\n for i in range(k):\n coun += 1\n if s[i] != '1':\n print(s[i])\n exit()\n if coun == k:\n print(1)\n exit()\nelse:\n for i in range(k):\n coun += 1\n if s[i] != '1':\n print(s[i])\n exit()\n if coun == 100:\n print(1)\n exit()\n "] | ['Runtime Error', 'Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s100533212', 's195622065', 's228387646', 's331529803', 's787639041'] | [2940.0, 2940.0, 3060.0, 3060.0, 3064.0] | [18.0, 17.0, 18.0, 17.0, 17.0] | [118, 166, 178, 162, 331] |
p03282 | u297651868 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=list(input())\nk=int(input())\nfor i in s:\n ans=i\n if i!=2:\n break\nprint(ans)', 's=list(input())\nk=int(input())\ntmp=0\nfor i in s:\n ans=i\n if ans!="1":\n break\n else:\n tmp+=1\nif tmp>=k:\n print("1")\nelse:\n print(ans)'] | ['Wrong Answer', 'Accepted'] | ['s807984141', 's007727263'] | [2940.0, 2940.0] | [17.0, 17.0] | [90, 161] |
p03282 | u305965165 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nfor i in range( min(len(s),k) ):\n if s[i] != "1":\n print(s[i])\n\nprint("1")', 's = input()\nk = int(input())\n\nresult = "1"\nfor i in range( min(len(s),k) ):\n if s[i] != "1":\n result = s[i]\n \nprint(result)', 's = input()\nk = int(input())\n\nresult = "1"\nfor i in range( min(len(s),k) ):\n if s[i] != "1":\n result = s[i]\n break\n \nprint(result)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s109326219', 's497077966', 's641832168'] | [2940.0, 2940.0, 2940.0] | [17.0, 18.0, 18.0] | [114, 140, 154] |
p03282 | u306412379 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nK = int(input())\n\nif S[0] == '1':\n count = 0\n while S[i] == '1':\n count += 1\n if count >= K:\n print(1)\n else:\n print(S[count])\nelse:\n print(S[0])", "S = input()\nK = int(input())\ncount = 0\n\nif S[0] != '1':\n print(S[0])\n exit()\n\nfor i in range(len(S)):\n if S[i] == '1':\n count += 1\n else:\n break\n\nif count >= K:\n print(1)\nelse:\n print(S[count])"] | ['Runtime Error', 'Accepted'] | ['s718727527', 's260790312'] | [9172.0, 8940.0] | [29.0, 28.0] | [172, 213] |
p03282 | u315485238 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['N=int(input())\nK=int(input())\n\nif K==1:\n print(N[0])\n \nelse:\n if N[0]==1:\n print(N[1])\n else:\n print(N[0])', "N=input()\nK=int(input())\n\nfor x in N[:min(len(N), K)]:\n if x=='1':\n pass\n else:\n print(x)\n break\nelse:\n print('1')"] | ['Runtime Error', 'Accepted'] | ['s490087817', 's965225214'] | [2940.0, 2940.0] | [17.0, 17.0] | [116, 126] |
p03282 | u318024671 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['numstr = input()\nnum = int(input())\ndef check():\n for i in range(len(numstr)):\n if numstr[i] != 1:\n return numstr[i]\nprint(check())\n \n', 'numstr = input()\nnum = int(input())\ndef check():\n infinity = ""\n for i in range(num):\n if numstr[i] is not "1":\n infinity = infinity + numstr[i]\n return infinity\n else:\n infinity = infinity + "1"\n return "1"\ninfinity = check()\nprint(infinity[len(infinity)-1])\n \n'] | ['Wrong Answer', 'Accepted'] | ['s925430740', 's717961072'] | [2940.0, 2940.0] | [17.0, 18.0] | [162, 325] |
p03282 | u319684393 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["inp1 = input()\ninp1 = list(inp1)\ninp2 = input()\n\nans = 1\nfor i in inp1:\n if i != '1'\n ans = int(i)\n break\n\nprint(ans)", "inp1 = input()\ninp1 = int(inp1)\ninp2 = input()\ninp2 = list(inp2)\n\nans = 1\nfor i in inp2:\n if i != '1'\n ans = int(i)\n break\n\nprint(ans)", "inp1 = input()\ninp1 = list(inp1)\ninp2 = input()\ninp2 = int(inp2)\n\nans = 1\ncount = 0\nfor i in inp1:\n if i != '1':\n ans = int(i)\n break\n else:\n count += 1\n\nif count >= inp2:\n ans = 1\n\nprint(ans)"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s139713977', 's225029027', 's041587757'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0] | [134, 151, 222] |
p03282 | u325282913 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nK = int(input())\nflg = False\nfor i in range(K):\n if S[i] != '1':\n flg = True\nif flg:\n print('1')\n exit()\nfor i in range(len(S)):\n if S[i] != '1':\n print(S[i])\n exit()\nprint('1')", "S = input()\nK = int(input())\nflg = True\nfor i in range(K):\n if len(S) <= i:\n break\n if S[i] != '1':\n flg = False\nif flg:\n print('1')\n exit()\nfor i in range(len(S)):\n if S[i] != '1':\n print(S[i])\n exit()\nprint('1')"] | ['Runtime Error', 'Accepted'] | ['s163418445', 's994118703'] | [3060.0, 3060.0] | [17.0, 17.0] | [222, 256] |
p03282 | u327668449 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nans_idx = 0\n\nfor i in range(k):\n if (s[ans_idx] != "1"):\n break\nprint(s[ans_idx])', 's = input()\nk = int(input())\n\nans_idx = 0\n\ns = list(s)\nfor i in range(min(k, len(s))):\n if (s[i] != "1"):\n break\nprint(s[i])'] | ['Wrong Answer', 'Accepted'] | ['s178254490', 's410234974'] | [2940.0, 3060.0] | [2104.0, 19.0] | [115, 128] |
p03282 | u328131364 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = list(input())\nK = int(input())\ni = 0\nans = 0\n\nwhile K > 0:\n K = K - S[i]**(5*10**16 + 1)\n ans = S[i]\n i += 1\n\nprint(ans)', 'S = input()\nK = int(input())\n\nans = 0\nn1 = 0\n\nfor i in S:\n if i != "1":\n s = int(i)\n break\n \n else:\n n1 += 1\n\nif n1 >= K:\n print(1)\n\nelse:\n print(s)'] | ['Runtime Error', 'Accepted'] | ['s621806009', 's456903919'] | [2940.0, 2940.0] | [18.0, 18.0] | [133, 184] |
p03282 | u328364772 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nk = input()\n\nif S[0] == '1':\n cnt = 0\n for s in S:\n if s == '1':\n cnt += 1\n else:\n not1 = s\n break\n if k <= cnt:\n print(1)\n else:\n print(not1)\nelse:\n print(S[0])\n", "S = input()\nk = int(input())\n\nif S[0] == '1':\n cnt = 0\n for s in S:\n if s == '1':\n cnt += 1\n else:\n not1 = s\n break\n if k <= cnt:\n print(1)\n else:\n print(not1)\nelse:\n print(S[0])\n"] | ['Runtime Error', 'Accepted'] | ['s819513687', 's198243549'] | [2940.0, 2940.0] | [17.0, 17.0] | [250, 255] |
p03282 | u329709276 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n\nfor i in S:\n if i == 1:\n K -= 1\n if K == 0:\n print(1)\n break\n else:\n print(i)\n break', "S = input()\nK = int(input())\n\nfor i in S:\n if i == '1':\n K -= 1\n if K == 0:\n print(1)\n break\n else:\n print(i)\n break"] | ['Wrong Answer', 'Accepted'] | ['s296247768', 's132263563'] | [2940.0, 2940.0] | [17.0, 17.0] | [170, 172] |
p03282 | u331381193 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=str(input())\nk=int(input())\n\nres=''\n\nif len(s)>=k:\n\tfor i in range(k):\n\t\tif s[i]!=1:\n\t\t\tres=s[i]\t\t\t\n\t\t\tbreak\n\t\tif i==k-1:\n\t\t\tres='1'\nelse:\n\tfor i in range(len(s)):\n\t\tif s[i]!='1':\n\t\t\tres=s[i]\n\t\t\tbreak\n\nprint(res)", "s=str(input())\nk=int(input())\n\nres=''\n\nif len(s)>=k:\n\tfor i in range(k):\n\t\tif s[i]!=1:\n\t\t\tres=s[i]\t\t\t\n\t\t\tbreak\n\t\tif i==k:\n\t\t\tres='1'\nelse:\n\tfor i in range(len(s)):\n\t\tif s[i]!='1':\n\t\t\tres=s[i]\n\t\t\tbreak\n\nprint(res)", "s=str(input())\nk=int(input())\n\nres=''\n\nif len(s)>=k:\n\tfor i in range(k):\n\t\tif s[i]!='1':\n\t\t\tres=s[i]\t\t\t\n\t\t\tbreak\n\t\tif i==k-1:\n\t\t\tres='1'\nelse:\n\tfor i in range(len(s)):\n\t\tif s[i]!='1':\n\t\t\tres=s[i]\n\t\t\tbreak\n\nprint(res)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s171660096', 's419601001', 's725028886'] | [3060.0, 3060.0, 3060.0] | [18.0, 17.0, 17.0] | [214, 212, 216] |
p03282 | u332253305 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=input()\nn=int(input())\n\nfor i in range(n):\n if s[i]!='1':\n print(s[i])\n break\n else:\n print(1)\n", "s=input()\nn=int(input())\nans=1\nfor i in range(n):\n if s[i]!='1':\n ans=s[i]\n break\n else:\n continue\nprint(ans)\n \n"] | ['Wrong Answer', 'Accepted'] | ['s896905363', 's731726995'] | [2940.0, 2940.0] | [17.0, 17.0] | [124, 145] |
p03282 | u337626942 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\n\nif k==1:\n print(int(s[0]))\n\nelif int(s[0])==1:\n\n if len(s)<k:\n num=1\n for i in range(1, len(s)):\n if int(s[i])!=1:\n num=int(s[i])\n break\n print(num)\n\n else:\n if s[:k]=="1"*k:\n print(1)\n\nelif int(s[0])!=1:\n print(int(s[0]))', 's=input()\nk=int(input())\n\ncnt=0\nfor i in range(len(s)):\n if s[i]=="1":\n cnt+=1\n else:\n ans_index=i\n break\n\nif k<=cnt:\n print(1)\nelse:\n print(s[ans_index])'] | ['Wrong Answer', 'Accepted'] | ['s546790645', 's425537205'] | [3064.0, 3060.0] | [17.0, 17.0] | [340, 187] |
p03282 | u340010271 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S=input()\nK=int(input())\nif S[0]!='1':\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i]!='1':\n if K<=i:\n print('1'):\n else:\n print(S[i])\n break", "S=input()\nK=int(input())\nif S[0]!='1':\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i]!='1':\n if K<=i:\n print('1')\n else:\n print(S[i])\n break\n print(1)", "S=input()\nK=int(input())\nif S[0]!='1':\n print(S[0])\nelse:\n flag=True\n for i in range(len(S)):\n if S[i]!='1':\n if K<=i:\n print('1')\n else:\n print(S[i])\n flag=False\n break\n if flag:\n print(1)"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s555081592', 's679593420', 's685886139'] | [2940.0, 3060.0, 3060.0] | [17.0, 17.0, 18.0] | [223, 235, 289] |
p03282 | u347184682 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["#include <bits/stdc++.h>\n// \n\n#include <algorithm>\n#include <string>\n#include <math.h>\nusing namespace std;\ntypedef long long ll;\n\n#define rep(i,n) for(int i=0;i<n;i++)\n#define rep2(i,a,b) for(int i=(a);i<(b);i++)\n\ntemplate<class T> inline void chmin(T& a, T b) {if (a > b) a = b;}\ntemplate<class T> inline void chmax(T& a, T b) {if (a < b) a = b;}\n\n\nint main() {\n string s;cin >> s;\n int k;cin >> k;\n rep(i,k+3){\n if (s[i]!='1'){\n cout<<s[i]<<endl;\n return 0;\n }\n }\n cout<<1<<endl; \n}\n", 'import sys\ns=input()\nk=int(input())\nfor i in range(k):\n if s[i]!="1":\n print(s[i])\n sys.exit()\nprint(1)\n'] | ['Runtime Error', 'Accepted'] | ['s780550555', 's194651057'] | [3060.0, 2940.0] | [17.0, 17.0] | [544, 111] |
p03282 | u347545486 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = int(input())\nk = int(input())\nli = [int(k) for k in list(str(s))]\n#print(li)\nif k = 1:\n print(li[0])\nelse:\n print(li[1])', 's = int(input())\nk = int(input())\nli = [int(k) for k in list(str(s))]\n#print(li)\ncount = 0\nfor i in li:\n if i ==1:\n count+=1\n else:\n ans = i\n break\n \nif k <= count:\n print(1)\nelse:\n print(ans)'] | ['Runtime Error', 'Accepted'] | ['s658439044', 's362178507'] | [2940.0, 3060.0] | [17.0, 17.0] | [126, 208] |
p03282 | u348868667 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nif S[0]*len(S) == S and S[0] == "1":\n print("1")\nelse:\n ans = 0\n ind = 0\n for i in range(len(S)):\n if S[i] != "1":\n ans = S[i]\n ind = i+1\n if ind > K:\n print("1")\n else:\n print(ans)', 'S = input()\nK = int(input())\nif S[0]*len(S) == S:\n print("1")\nelse:\n ans = 0\n ind = 0\n for i in range(len(S)):\n if S[i] != "1":\n ans = S[i]\n ind = i+1\n if ind > K:\n print("1")\n else:\n print(ans)', 'S = input()\nK = int(input())\nif S[0]*len(S) == S and S[0] == "1":\n print("1")\nelse:\n ans = 0\n ind = 0\n for i in range(len(S)):\n if S[i] != "1":\n ans = S[i]\n ind = i+1\n break\n if ind > K:\n print("1")\n else:\n print(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s216760854', 's284111299', 's120121295'] | [3060.0, 3060.0, 3060.0] | [18.0, 17.0, 17.0] | [271, 255, 289] |
p03282 | u352676541 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nK = int(input())\n\nans = 1\nfor i in S:\n\tprint(i,K)\n\tif i > '1':\n\t\tans = i\n\tif K==1:\n\t\tans = i\n\t\tbreak\n\tK-=1\nprint(ans)", "S = input()\nK = int(input())\n\nans = 1\nfor i in S:\n\tprint(i,K)\n\tif i > '1':\n\t\tans = i\n\t\tbreak\n\tif K==1:\n\t\tans = i\n\t\tbreak\n\tK-=1\nprint(ans)", "S = input()\nK = int(input())\n\nans = 1\nfor i in S:\n\tprint(i,K)\n\tif i != '1':\n\t\tans = i\n\t\tbreak\n\tif K==1:\n\t\tans = i\n\t\tbreak\n\tK-=1\nprint(ans)", "S = input()\nK = int(input())\n\nans = 1\nfor i in S:\n\tif i != '1':\n\t\tans = i\n\t\tbreak\n\tif K==1:\n\t\tans = i\n\t\tbreak\n\tK-=1\nprint(ans)"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s116758349', 's163180636', 's483601270', 's105693698'] | [9048.0, 9072.0, 9012.0, 9052.0] | [27.0, 25.0, 24.0, 31.0] | [129, 137, 138, 126] |
p03282 | u359474860 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nif S[0] != "1":\n print(S[0])\nelse:\n i = 0\n count = 0\n for i in range(len(S)):\n if S[i] == "1":\n count += 1\n i += 1\n if K <= count:\n print(1)\n else:\n print(S[i])', 'S = input()\nK = int(input())\n\n\nif S[0] != "1":\n print(S[0])\nelse:\n i = 0\n count = 0\n while i < len(S) and S[i] == "1":\n count += 1\n i += 1\n if K <= count:\n print(1)\n else:\n print(S[i])'] | ['Runtime Error', 'Accepted'] | ['s206799237', 's207937397'] | [3060.0, 3060.0] | [18.0, 19.0] | [250, 230] |
p03282 | u359747364 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = str(input())\n k = int(input())\n for i in s:\n if i != "1":\n print(i)\n break', 's = str(input())\nk = int(input())\ncounter = 0\nfor i in range(k):\n if i != "1" and i <= k-1:\n print(i)\n counter += 1\n break\nif counter == 0:\n print(1)', "S = input()\nK = int(input())\n \nif K == 1:\n print(S[0])\n exit()\nelse:\n for i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()\nprint(1)"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s318668938', 's784415287', 's287676574'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [133, 176, 173] |
p03282 | u361826811 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["import sys\nimport itertools\nimport numpy as np\n\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nN = readline().decode().strip()\nK = int(readline())\ncnt = 0\nans = 0\nif K <= len(N):\n ans = N[K - 1]\nelse:\n for i in N:\n if i == '1':\n continue\n else:\n ans = i\n break\n\nprint(ans)\n", "import sys\nimport itertools\nimport numpy as np\n\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nN = readline().decode().strip()\nK = int(readline())\ncnt = 0\nans = 0\nfor i in N:\n if len(N):\n ans = N[0]\n break\n if i == '1':\n cnt += 1\n continue\n else:\n if int(i) * 5000 * 10 ** 12 + cnt >= K:\n ans = i\n break\n else:\n ans = N[N.find(i) + 1]\n break\n\nprint(ans)\n", "import sys\nimport itertools\nimport numpy as np\n\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nN = readline().decode().strip()\nK = int(readline())\nans = 1\nfor i in range(K):\n if i != '1':\n ans = i\n break\n\nprint(ans)\n", "import sys\nimport itertools\nimport numpy as np\n\nread = sys.stdin.buffer.read\nreadline = sys.stdin.buffer.readline\nreadlines = sys.stdin.buffer.readlines\n\nN = readline().decode().strip()\nK = int(readline())\nans = 1\nfor i in range(K):\n if N[i] != '1':\n ans = N[i]\n break\n\nprint(ans)\n"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s416257342', 's443944002', 's757229978', 's845063916'] | [12388.0, 12504.0, 12404.0, 18360.0] | [149.0, 151.0, 149.0, 255.0] | [385, 508, 292, 298] |
p03282 | u362560965 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['# S = input()\n# K = int(input())\n\nS = "9999"\nK = 1\n\nans = 1\n\nfor i in range(len(S)):\n if int(S[i]) == 1:\n pass\n else:\n ans = int(S[i])\n ones = i\n break\n\nif not ones:\n ans = 1\nelif K <= ones:\n ans = 1\n\nprint(ans)\n', 'S = input()\nK = int(input())\n\nans = 1\n\nfor i in S:\n if int(i) == 1:\n pass\n else:\n ans = int(i)\n\nprint(ans)\n', 'S = input()\nK = int(input())\n\nans = 1\n\nfor i in range(K):\n if S[i] == 1:\n pass\n else:\n ans = S[i]\n\nprint(ans)\n', 'S = input()\nK = int(input())\n\nans = 1\n\nfor i in range(K):\n if int(S[i]) == 1:\n pass\n else:\n ans = int(S[i])\n break\n\nprint(ans)\n'] | ['Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s382202799', 's878118010', 's979363206', 's339404847'] | [2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 18.0] | [252, 127, 130, 154] |
p03282 | u365993347 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = input()\n\nlst = []\ni = int(S)\nwhile i > 0:\n lst.append(i%10)\n i //= 10 \nlst.reverse()\nprint(lst[0])\nif K == 1:\n print(lst[0])\nelse:\n if lst[0] == 1:\n print(lst[1])\n else:\n print(lst[0])', "S = int(input())\nK = int(input())\n\nlst = []\nwhile S > 0:\n lst.append(S%10)\n S //= 10 \nlst.reverse()\n#print(lst[0])\nj = 0\nn = 0\nwhile lst[j] == 1:\n n = j+1\n j += 1\nprint('n =',n)\nif K <= n:\n print(1)\nelse:\n print(lst[n])", "S = int(input())\nK = int(input())\n\nlst = [100]\nwhile S > 0:\n lst.append(S%10)\n S //= 10 \nlst.reverse()\nprint(lst)\nj = 0\nn = 0\nwhile lst[j] == 1:\n n = j+1\n j += 1\n#print('n =',n)\nif K <= n:\n print(1)\nelse:\n print(lst[n])", "S = int(input())\nK = int(input())\n\nlst = [0]*100\nwhile S > 0:\n lst.append(S%10)\n S //= 10 \nlst.reverse()\n#print(lst)\nj = 0\nn = 0\nwhile lst[j] == 1:\n n = j+1\n j += 1\n#print('n =',n)\nif K <= n:\n print(1)\nelse:\n print(lst[n])"] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s225800117', 's405886957', 's817427346', 's735930637'] | [3060.0, 3064.0, 3064.0, 3064.0] | [18.0, 18.0, 17.0, 17.0] | [237, 245, 245, 248] |
p03282 | u366185462 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['{"nbformat":4,"nbformat_minor":0,"metadata":{"colab":{"name":"ABC106C-ToInfinity.ipynb","provenance":[],"authorship_tag":"ABX9TyNX8Ir2ZzuscRLajj/o6IUg"},"kernelspec":{"name":"python3","display_name":"Python 3"}},"cells":[{"cell_type":"code","metadata":{"id":"3ke4UMpfQZna","colab_type":"code","outputId":"b7f58493-9947-42d9-8953-9dfad87c6dad","executionInfo":{"status":"ok","timestamp":1589608511500,"user_tz":-540,"elapsed":5594,"user":{"displayName":"Taishi","photoUrl":"","userId":"11736711679866118393"}},"colab":{"base_uri":"https://localhost:8080/","height":72}},"source":["s = str(input())\\n","k = int(input())\\n","if int(s[0]) == 1:\\n"," print(s[1])\\n","else:\\n"," print(s[0])"],"execution_count":3,"outputs":[{"output_type":"stream","text":["3\\n","157\\n","3\\n"],"name":"stdout"}]},{"cell_type":"code","metadata":{"id":"G05wyz9VTQI9","colab_type":"code","colab":{"base_uri":"https://localhost:8080/","height":35},"outputId":"8368a8c5-f413-4a10-c257-1f3791af6cf0","executionInfo":{"status":"ok","timestamp":1589608502261,"user_tz":-540,"elapsed":575,"user":{"displayName":"Taishi","photoUrl":"","userId":"11736711679866118393"}}},"source":["3\\n","157"],"execution_count":2,"outputs":[{"output_type":"execute_result","data":{"text/plain":["157"]},"metadata":{"tags":[]},"execution_count":2}]},{"cell_type":"code","metadata":{"id":"TmcZoyerDNFs","colab_type":"code","colab":{}},"source":[""],"execution_count":0,"outputs":[]}]}', '{"nbformat":4,"nbformat_minor":0,"metadata":{"colab":{"name":"ABC106C-ToInfinity.ipynb","provenance":[],"authorship_tag":"ABX9TyNRoFVf941vEvQ2RLhFUphc"},"kernelspec":{"name":"python3","display_name":"Python 3"}},"cells":[{"cell_type":"code","metadata":{"id":"3ke4UMpfQZna","colab_type":"code","colab":{"base_uri":"https://localhost:8080/","height":72},"outputId":"99180d89-276e-4fd8-b2c6-02e5a651ee14","executionInfo":{"status":"ok","timestamp":1589596953669,"user_tz":-540,"elapsed":15115,"user":{"displayName":"Taishi","photoUrl":"","userId":"11736711679866118393"}}},"source":["s = str(input())\\n","k = int(input())\\n","i = 0\\n","n = s[0]\\n","while (i < len(s)):\\n"," if int(s[i]) ** 2 >= k:\\n"," n = s[i]\\n"," break\\n"," else:\\n"," k -= int(s[i])\\n"," i += 1\\n","print(n)"],"execution_count":17,"outputs":[{"output_type":"stream","text":["299792458\\n","9460730472580800\\n","2\\n"],"name":"stdout"}]},{"cell_type":"code","metadata":{"id":"G05wyz9VTQI9","colab_type":"code","colab":{}},"source":["299792458\\n","9460730472580800"],"execution_count":0,"outputs":[]}]}', "s = str(input())\nk = int(input())\ni, count = 0, 0\nwhile (i < len(s)):\n if int(s[i]) == 1:\n if (count >= k - 1):\n print('1')\n break\n count += 1\n else:\n print(s[i])\n break\n i += 1"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s791404722', 's913725096', 's937492919'] | [3064.0, 3064.0, 3060.0] | [17.0, 17.0, 19.0] | [1434, 1078, 202] |
p03282 | u366959492 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\nc=0\nif k<=len(s):\n print(s[k-1])\n exit()\nfor i in range(len(s)):\n if s[i]=="1":\n continue\n else:\n print(s[i])\n break\n', 's=input()\nk=int(input())\nc=0\nfor i in range(len(s)):\n if s[i]=="1":\n if k-1==i:\n print(1)\n break\n continue\n else:\n print(s[i])\n break\n'] | ['Wrong Answer', 'Accepted'] | ['s740451767', 's601511171'] | [2940.0, 2940.0] | [17.0, 18.0] | [175, 190] |
p03282 | u367130284 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['try:print([i for i in input()[:int(input())]][0])\nexcept:print(1)', 'print(max(1,input()[:int(input())].replace("1","")[:1]))', 's=input()\nk=int(input())\nfor i in range(k):\n if s[i] !="1":\n print(s[i])\n break\nprint("1")', "print(1or input()[:int(input())].replace('1','')[:1])", 's=input()\nk=int(input())\nfor i in s[:k]:\n if i!="1":\n print(i)\n exit()\nprint(1)\n'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s436848269', 's499893301', 's674766256', 's730643885', 's117257589'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 19.0, 18.0, 17.0, 17.0] | [65, 56, 107, 53, 97] |
p03282 | u369752439 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['def reflesh_string(current_answer):\n index = min(K, len(current_answer))\n return current_answer[:index]\n \ndef update_string(string_answer):\n string_answer = string_answer.replace("1", "1")\n string_answer = string_answer.replace("2", "22")\n string_answer = string_answer.replace("3", "333")\n string_answer = string_answer.replace("4", "4444")\n string_answer = string_answer.replace("5", "55555")\n string_answer = string_answer.replace("6", "666666")\n string_answer = string_answer.replace("7", "7777777")\n string_answer = string_answer.replace("8", "88888888")\n string_answer = string_answer.replace("9", "999999999")\n return string_answer\n \ndef check_skip(current_answer):\n for i in range(2, 10):\n i = str(i)\n if(i in current_answer):\n return False\n return True\n \nS = input().strip()\nK = int(input().strip())\nfor c in range(5 * 10**15):\n S = update_string(S)\n S = reflesh_string(S)\n if(check_skip(S)):\n break\nprint(S[K-1])', 'S = input().strip()\nK = int(input().strip())\n\nans = 1\nfor i in range(min(K, len(S))):\n if(S[i] != "1"):\n ans = S[i]\n break\n \nprint(ans)'] | ['Time Limit Exceeded', 'Accepted'] | ['s832429728', 's341384198'] | [686736.0, 2940.0] | [2106.0, 17.0] | [1008, 159] |
p03282 | u371132735 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nans = "1"\nfor i in range(K):\n if S[i] != "1":\n ans = i\n break\nprint(ans)', 'S = input()\nK = int(input())\nans = "1"\nfor i in range(len(S)):\n if i >= K:\n break\n if S[i] != "1":\n ans = S[i]\n break\n\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s365199836', 's180612142'] | [2940.0, 2940.0] | [17.0, 18.0] | [108, 140] |
p03282 | u371595076 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nK = input()\n\nfor i, k in enumerate(range(int(K))):\n if S[k] != '1':\n print(S[k])\n break\n if i == len(K)-1:\n print('1')", 'import math\n\nS = input()\nK = int(input())\n\nps = []\n\nfor s in S:\n i = int(s)\n if(i == 1):\n p == -1\n else:\n p = 5*(10**15) * math.log2(i)\n ps.append(p)\n\nsum = 0\nindex = 0\nfor p in ps:\n if not p == -1:\n sum += p\n else:\n K -= 1\n \n if math.log2(K) < sum:\n print(S[index])\n break\n \n index += 1', '# import math\n\nS = input()\nK = int(input())\n\nps = []\n\nfor s in S:\n i = int(s)\n if(i == 1):\n p = -1\n else:\n p = 5*(10**15) * math.log2(i)\n ps.append(p)\n\nsum = 0\nindex = 0\n \nif K == 1:\n print(S[0])\nelse:\n for p in ps:\n\n if not p == -1:\n sum += p\n else:\n K -= 1\n\n if math.log2(K) < sum:\n #print(index)\n print(S[index])\n break\n index += 1\n\n', "S = input()\nK = int(input())\n\nfor i, k in enumerate(range(K)):\n if S[k] != '1':\n print(S[k])\n break\n if i == K - 1:\n print('1')"] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s036356918', 's558009362', 's959192287', 's768658426'] | [2940.0, 3064.0, 3064.0, 2940.0] | [19.0, 19.0, 17.0, 22.0] | [157, 367, 454, 154] |
p03282 | u371763408 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\nif s[0]=="1":\n try:\n if s[k-1]=="1":\n print(\'1\')\n else:\n print(s[s.rfind("1")+1])\n except IndexError:\n try:\n print(s[s.rfind("1")+1])\n except IndexError:\n print(s[-1])\nelse:\n print(s[0])\n\n\n', 's=input()\nk=int(input())\nif s[0]=="1":\n if k>len(s):\n for i in s:\n if i!="1":\n print(i)\n else:\n print(s[k-1])\nelse:\n print(s[0])', 's=input()\nk=int(input())\nif s[0]=="1":\n try:\n if s[k-1]=="1":\n print(\'1\')\n else:\n print(s[s.rfind("1")+1])\n except IndexError:\n try:\n print(s[s.rfind("1")+1])\n except IndexError:\n print(s[s.rfind("1")])\nelse:\n print(s[0])\n\n\n', 's=input()\nk=int(input())\nif s[0]=="1":\n try:\n if s[k-1]=="1":\n print(\'1\')\n else:\n print(s[s.rfind("1")+1])\n except IndexError:\n print(s[s.rfind("1")+1])\nelse:\n print(s[0])\n', 's=input()\nk=int(input())\nfor i in range(k):\n if s[i]!="1":\n print(s[i])\n exit()\nprint("1")\n'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s159154459', 's362417547', 's488459828', 's590965033', 's716379404'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [18.0, 18.0, 18.0, 18.0, 17.0] | [249, 149, 259, 194, 98] |
p03282 | u375616706 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s = input()\nk = (int)(input())\n\nif k <= 5000000000000000:\n for n in s:\n if k == 0:\n print(n)\n elif n == '1':\n k = k-1\n else:\n print(n)\n\nelse:\n for n in s:\n if n == '1':\n k = k-1\n elif n == '2':\n if k >= 2ae0i0:\n else:\n print(n)\n else:\n print(n)\n", "s = input()\nk = (int)(input())\n\nfor n in s:\n if k < 0:\n print(s[s.index(n)-1])\n elif n == '1':\n k = k-1\n else:\n print(n)\n", "s = input()\nk = (int)(input())\n\nif k <= 5000000000000000:\n for n in s:\n if k == 0:\n print(s[s.index(n)-1])\n elif n == '1':\n k = k-1\n else:\n print(n)\n\nelse:\n for n in s:\n if n == '1':\n k = k-1\n elif n == '2':\n if k >= 10**15:\n print()\n else:\n print(n)\n else:\n print(n)\n", "s = input()\nk = (int)(input())\n\nc = 0\nfor ch in s:\n if not ch == '1':\n break\n c += 1\nn = s[c] if c < len(S) else None\na = n if n and c < K else S[0]\nprint(a)\n", "s = input()\nK = (int)(input())\nk = K\nfor n in s:\n if k == 0:\n print(n)\n elif n == '1':\n k = k-1\n else:\n print(n)\nif k == 0:\n print(s[K-1])\n", "s = input()\nk = (int)(input())\n\nfor n in s:\n if k == 0:\n print(s[s.index(n)-1])\n elif n == '1':\n k = k-1\n else:\n print(n)\n", "s = input()\nk = (int)(input())\n\nc = 0\nfor c in s:\n if not c == '1':\n break\n c += 1\nn = s[c] if c < len(S) else None\na = n if n and c < K else S[0]\nprint(a)\n", "s = input()\nk = (int)(input())\n\nc = 0\nfor ch in s:\n if not ch == '1':\n break\n c += 1\nn = s[c] if c < len(s) else None\na = n if n and c < k else s[0]\nprint(a)\n"] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s092106337', 's172045339', 's231445746', 's416359753', 's484234081', 's633992502', 's724433705', 's354655721'] | [2940.0, 2940.0, 3060.0, 2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [18.0, 18.0, 18.0, 17.0, 18.0, 18.0, 18.0, 17.0] | [305, 133, 337, 163, 152, 134, 161, 163] |
p03282 | u375695365 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s=int(input())\nk=int(input())\n\nfor i in range(min(len(str(s),k))):\n if str(s)[i]!='1':\n print(str(s)[i])\n exit()\nprint(1)\n ", "s=int(input())\nk=int(input())\n\nfor i in range(min(len(str(s),k)):\n if str(s)[i]!='1':\n print(str(s)[i])\n exit()\nprint(1)\n ", "s=int(input())\nk=int(input())\n\nfor i in range(min(len(str(s)),k)):\n if str(s)[i]!='1':\n print(str(s)[i])\n exit()\nprint(1)\n "] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s304430073', 's466766607', 's569438073'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [142, 141, 142] |
p03282 | u377834804 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n\nfor i, s in enumerate(S, 1):\n if s == 1:\n if i == K:\n print(s)\n break\n else:\n continue\n else:\n print(s)\n break', "S = input()\nK = int(input())\n \nfor i, s in enumerate(S, 1):\n if s == '1':\n if i == K:\n print(s)\n break\n else:\n continue\n else:\n print(s)\n break"] | ['Wrong Answer', 'Accepted'] | ['s366922113', 's264017985'] | [9020.0, 8760.0] | [35.0, 29.0] | [169, 172] |
p03282 | u382176416 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["n=list(input())\nk=int(input())\nif len(n)==1:\n print('1')\nelse:\n one_list=[]\n while len(n)!=0:\n if n[0]=='1':\n one_list.append(n.pop(0))\n else:\n break\n print(one_list)\n if k<=len(one_list):\n print('1')\n else:\n print(n[0])", "n=list(input())\nk=int(input())\nif len(n)==1:\n print('1')\nelse:\n one_list=[]\n while len(n)!=0:\n if n[0]=='1':\n one_list.append(n.pop(0))\n else:\n break\n if 0<len(one_list)<=k:\n print('1')\n else:\n print(n[0])\n\n", "n=list(input())\nk=int(input())\nif len(n)==1:\n print('1')\nelse:\n one_list=[]\n while len(n)!=0:\n if n[0]=='1':\n one_list.pop(n.pop(0))\n else:\n break\n if 0<len(one_list)<=k:\n print('1')\n else:\n print(n[0])", "n=list(input())\nk=int(input())\nif len(n)==1:\n print('1')\nelse:\n one_list=[]\n while 1==1:\n if n[0]=='1':\n one_list.pop(n.pop(0))\n else:\n break\n if 0<len(one_list)<=k:\n print('1')\n else:\n print(n[0])\n\n", "n=list(input())\nk=int(input())\nif len(n)==1 and n[0]=='1':\n print('1')\nelse:\n one_list=[]\n while len(n)!=0:\n if n[0]=='1':\n one_list.append(n.pop(0))\n else:\n break\n if k<=len(one_list):\n print('1')\n else:\n print(n[0])\n\n"] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s602834804', 's757344996', 's851584012', 's891759433', 's023701995'] | [3060.0, 2940.0, 3060.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0, 17.0] | [248, 234, 229, 226, 246] |
p03282 | u388323466 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = list(map(int, input()))\nk = int(input())\n\nans = 1\n\nfor i in range(len(k)):\n\tif s[i] != 1:\n\t\tans = s[i]\n\t\tbreak\n\nprint(ans)\n\t\t\n\t', 's = list(map(int, input()))\nk = int(input())\n\nans = 1\n\nfor i in range(len(min(s,k))):\n\tif s[i] != 1:\n\t\tans = s[i]\n\t\tbreak\n\nprint(ans)\n\t\t\n\t', 's = list(map(int, input()))\nk = int(input())\n\nans = 1\n\nfor i in range(min(len(s),k)):\n\tif s[i] != 1:\n\t\tans = s[i]\n\t\tbreak\n\nprint(ans)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s352169105', 's686143751', 's278708664'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [131, 138, 133] |
p03282 | u394721319 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["S = input()\nK = int(input())\n\nif K == 1:\n print(S[0])\n exit()\nelse:\n for i in range(K):\n if S[i] != '1':\n print(S{i})\n exit()\nprint(1)\n", "S = input()\nK = int(input())\n\nif K == 1:\n print(S[0])\n exit()\nelse:\n for i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()\nprint(1)\n"] | ['Runtime Error', 'Accepted'] | ['s150207869', 's814255343'] | [2940.0, 2940.0] | [17.0, 17.0] | [173, 173] |
p03282 | u394731058 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["import sys\n\ninput = sys.stdin.readline\n\ndef main():\n ans = 0\n s = input()\n k = int(input())\n for i in range(len(s)):\n if s[i] != '1':\n if k > i+1:\n ans = s[i]\n else:\n ans = '1'\n print(ans)\n\nif __name__ == '__main__':\n main()", "import sys\n\ninput = sys.stdin.readline\n\ndef main():\n ans = 0\n s = input().rstrip('\\n')\n k = int(input())\n for i in range(len(s)):\n if s[i] != '1':\n if k < i+1:\n ans = s[i]\n else:\n ans = '1'\n print(ans)\n\nif __name__ == '__main__':\n main()", "import sys\n\ninput = sys.stdin.readline\n\ndef main():\n ans = ''\n index = 0\n s = input().rstrip('\\n')\n k = int(input())\n for i in range(len(s)):\n if s[i] != '1':\n ans = s[i]\n index = i\n break\n if ans == '' or k < index+1:\n ans = '1'\n print(ans)\n\nif __name__ == '__main__':\n main()"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s023621382', 's457943451', 's815909164'] | [9184.0, 9040.0, 9176.0] | [26.0, 30.0, 28.0] | [301, 314, 348] |
p03282 | u395086545 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\nif int(s[0]) == 1:\n one_count = 0\n i = 0\n while s[i] == 1:\n one_count += 1\n i += 1\n if k <= one_count:\n print(1)\n else:\n print(s[one_count])\nelse:\n print(s[0])\n', 's = input()\nk = int(input())\n\nif int(s[0]) == 1:\n one_count = 0\n i = 0\n while i < len(s) and int(s[i]) == 1:\n one_count += 1\n i += 1\n if k <= one_count:\n print(1)\n else:\n print(s[one_count])\nelse:\n print(s[0])\n'] | ['Wrong Answer', 'Accepted'] | ['s366449209', 's706424933'] | [2940.0, 3060.0] | [17.0, 17.0] | [208, 228] |
p03282 | u396824275 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S=input()\nK=int(input())\nif K<=len(S):\n for s in S[:K]:\n if s!="1":\n break\n else:\n print("1")\nelse:\n\tfor s in S:\n \tif s!="1" :\n \tprint(s)\n \tbreak', 'S=input()\nK=int(input())\nif(K<=len(S)):\n for s in S[:K]:\n if s!="1":\n break\n else:\n print("1")\nelse:\n for s in S:\n if(s!="1"):\n print(s)\n break', 'N=input()\nfor count in range(len(N)): \n if N[count]!="1": break\nif int(input())>count: print(N[count])\nelse: print(1)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s100389950', 's951810612', 's401862785'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 18.0] | [193, 206, 130] |
p03282 | u397531548 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S,K=input().split()\nk=int(K)\nfor i in range(0,k):\n if S[i]!="1":\n print(S[i])\n break\nelse:\n print("1")\n', 'S=input()\nk=int(input())\nfor i in range(0,k):\n if S[i]!="1":\n print(S[i])\n break\nelse:\n print("1")\n'] | ['Runtime Error', 'Accepted'] | ['s632690864', 's145118487'] | [2940.0, 3064.0] | [19.0, 19.0] | [123, 119] |
p03282 | u399973890 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s = input()\nk = int(input())\nfor i in range(5*10**15):\n s = s.replace('2','22')\n s = s.replace('3','333')\n s = s.replace('4','4444')\n s = s.replace('5','55555')\n s = s.replace('6','666666')\n s = s.replace('7','7777777')\n s = s.replace('8','88888888')\n s = s.replace('9','999999999')\n s = s[:k]\nprint(s)", "s = input()\nk = int(input())\ncnt = 0\nwhile s[:1] == '1':\n s = s[1:]\n cnt +=1\nif k <= cnt:\n print('1')\nelse:\n print(s[:1])\n"] | ['Time Limit Exceeded', 'Accepted'] | ['s139645580', 's745966674'] | [493696.0, 3188.0] | [2106.0, 19.0] | [329, 134] |
p03282 | u401452016 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import sys\nimport numpy as np\n\nn = int(sys.stdin.readline())\nA = list(map(int, sys.stdin.readline().split()))\n\nprint(sum(A)-n)', "import sys\n\ns = sys.stdin.readline().rstrip()\nk = int(sys.stdin.readline())\n\nnum=0\ncnt1=0\nfor c in s:\n if c =='1':\n cnt1 +=1\n else:\n num = c\n break\n#print(cnt1, num)\n\nif k<=cnt1:\n print(1)\nelse:\n print(num)"] | ['Wrong Answer', 'Accepted'] | ['s320161305', 's125865296'] | [14440.0, 3060.0] | [151.0, 18.0] | [126, 239] |
p03282 | u403331159 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\nfor i in range(len(s)):\n if s[i]!=1:\n print(s[i])\n exit()\nprint(1)', 's=input()\nk=int(input())\nfor i in range(k):\n if s[i]!="1":\n print(s[i])\n exit()\nprint(1)\n'] | ['Wrong Answer', 'Accepted'] | ['s911739994', 's763959910'] | [2940.0, 2940.0] | [17.0, 18.0] | [98, 96] |
p03282 | u404676457 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["import sys\ns = input()\nk = int(input())\nend = False\nfor i in range(min(len(s), k)):\n if s[i] != '1':\n print(s[i])\n sys.exit\nprint(1)", "s = input()\nk = int(input())\nend = False\nfor i in range(min(len(s), k)):\n if s[i] != '1':\n print(s[i])\n end = True\n break\nif end == False:\n print(1)"] | ['Wrong Answer', 'Accepted'] | ['s656541535', 's277890471'] | [2940.0, 3064.0] | [17.0, 18.0] | [149, 175] |
p03282 | u405256066 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['from sys import stdin\nS = (stdin.readline().rstrip())\nK = int(stdin.readline().rstrip())\ncnt = 0\nfor i in S:\n if i == "1":\n cnt += 1\n else:\n ans = i\nif cnt == K:\n ans = 1\nprint(ans)', 'from sys import stdin\nS = (stdin.readline().rstrip())\nK = int(stdin.readline().rstrip())\ncnt = 0\nfor i in S:\n if i == "1":\n cnt += 1\n else:\n ans = int(i)\nif cnt >= K:\n ans = 1\nprint(ans)', 'from sys import stdin\nS = (stdin.readline().rstrip())\nK = int(stdin.readline().rstrip())\ncnt = 0\nfor i in S:\n if i == "1":\n cnt += 1\n else:\n ans = int(i)\n break\nif cnt >= K:\n ans = 1\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s372656368', 's836413974', 's371874598'] | [3060.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0] | [204, 209, 223] |
p03282 | u408375121 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\ncount = 0\nfor t in S:\n if t == 1:\n count += 1\n else:\n break\nif K <= count:\n print(1)\nelse:\n print(int(S[count]))', 'S = input()\nK = int(input())\ncount = 0\nfor t in S:\n if int(t) == 1:\n count += 1\n else:\n break\nif K <= count:\n print(1)\nelse:\n print(int(S[count]))\n'] | ['Wrong Answer', 'Accepted'] | ['s898697160', 's333026444'] | [2940.0, 2940.0] | [17.0, 18.0] | [151, 157] |
p03282 | u409064224 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\n\n\n\nco = 0\nfor i in s:\n co += 1\n if i == "1":\n if k == co:\n print(1)\n break\n else:\n continue\n else:\n print(int(i))\n', 's = input()\nk = int(input())\n\n\ngosentyo = 5 * (10 ** 15)\n\n\n\nflg = 0\nif k <= 100:\n for i in range(len(s)-1):\n if s[i] != "1":\n print(s[i])\n break\n else:\n if k-1 == i:\n print(1)\n flg = 1\n break\n\nif flg == 0:\n for i in range(len(s)):\n if s[i] == "1":\n continue\n else:\n print(s[i])\n break\n', 's = input()\nk = int(input())\n\n\n\nco = 0\nfor i in s:\n co += 1\n if i == "1":\n if k == co:\n print(1)\n break\n else:\n continue\n else:\n print(str(i))\n', 's = input()\nk = int(input())\n\n\n\nco = 0\nfor i in s:\n co += 1\n if i == "1":\n if k == co:\n print(1)\n break\n else:\n continue\n else:\n print(i)\n', 's = input()\nk = int(input())\n\n\n\n\nco = 0\nfor i in s:\n co += 1\n if i == "1":\n if k == co:\n print(1)\n break\n else:\n continue\n else:\n print(i)\n', 's = input()\nk = int(input())\n\n\ngosentyo = 5 * (10 ** 15)\n\n\n\nflg = 0\nif k <= 100:\n for i in range(len(s)-1):\n if s[i] != "1":\n print(s[i])\n break\n else:\n if k-1 == i:\n print(1)\n flg = 1\n break\n\nif flg == 0:\n for i in range(len(s)):\n if i == len(s)-1:\n print(1)\n if s[i] == "1":\n continue\n else:\n print(s[i])\n break\n', 's = input()\nk = int(input())\n\n\n\nco = 0\nfor i in s:\n co += 1\n if i == "1":\n if k == co:\n print(1)\n break\n else:\n continue\n else:\n print(int(i))\n break\n'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s036862318', 's208912905', 's333539004', 's642927051', 's837980608', 's975187505', 's356481924'] | [2940.0, 3064.0, 2940.0, 2940.0, 3060.0, 3064.0, 2940.0] | [17.0, 18.0, 17.0, 18.0, 17.0, 17.0, 17.0] | [234, 645, 234, 229, 607, 692, 248] |
p03282 | u412116291 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['MOD = 1000000007\ndef fast_power(base, power):\n """\n Returns the result of a^b i.e. a**b\n We assume that a >= 1 and b >= 0\n\n Remember two things!\n - Divide power by 2 and multiply base to itself (if the power is even)\n - Decrement power by 1 to make it even and then follow the first step\n """\n\n result = 1\n while power > 0:\n # If power is odd\n if power % 2 == 1:\n result = (result * base) % MOD\n\n # Divide the power by 2\n power = power / 2\n # Multiply base to itself\n base = (base * base) % MOD\n\n return result\n\n\ns = input()\nk = int(input())\n\ntyo = 5e15\n\nchar = [int(ss) for ss in s]\n\nres = 0\n\n_sum = 0\nok = 0\n\nfor i in range(k):\n _sum = _sum + (fast_power(s[i], tyo))\n if _sum >= k and ok == 0:\n res = s[i]\n ok = 1\n\n# for num in char:\n\n# if _sum >= k and ok == 0:\n# res = num\n# ok = 1\n\nprint(res)\n', "from math import pow\ns = input()\nk = int(input())\n\nchar = [ss for ss in s]\n\n\nres = '1'\n\n_sum = 0\nfor num in range(k):\n if num != '1':\n res = num\n break\n\nprint(res)\n", "from math import pow\ns = input()\nk = int(input())\n\nchar = [ss for ss in s]\n\n\nres = '1'\n\n_sum = 0\nfor i in range(k):\n if s[i] != '1':\n res = s[i]\n break\n\nprint(res)\n"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s810829384', 's983866474', 's444929195'] | [3064.0, 3060.0, 2940.0] | [17.0, 17.0, 17.0] | [965, 181, 181] |
p03282 | u413318755 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["#!/usr/bin/env python\n\ns = input().strip()\nk = input().strip()\n\nflag = True\nfor i in range(k):\n if s[i] == '1':\n print(s[i])\n flag = False\n\nif flag:\n print(s[1])\n", "#!/usr/bin/env python\n\ns = input().strip()\nk = int(input().strip())\n\nflag = True\nfor i in range(k):\n if s[i] == '1':\n print(s[i])\n flag = False\n\nif flag:\n print(s[1])\n", "#!/usr/bin/env python\n\n\ndef main():\n s = input().strip()\n k = int(input().strip())\n\n if s[0] != '1':\n print(s[0])\n return\n\n count_c = 0\n for c in s:\n if c == '1':\n count_c += 1\n\n if k <= count_c:\n print(s[k-1])\n else:\n print(s[count_c])\n\n\nmain()\n", "#!/usr/bin/env python\n\n\ndef main():\n s = input().strip()\n k = int(input().strip())\n\n \n if s[0] != '1':\n print(s[0])\n return\n\n count_one = 0\n for c in s:\n if c == '1':\n count_one += 1\n else:\n break\n\n if k <= count_one:\n print(1)\n else:\n print(s[count_one])\n\n\nmain()\n"] | ['Runtime Error', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s009221611', 's764538404', 's823313764', 's868976771'] | [2940.0, 2940.0, 3060.0, 3060.0] | [17.0, 17.0, 18.0, 17.0] | [182, 187, 313, 406] |
p03282 | u416758623 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s = input()\nk = int(input())\nans = 0\nfor i in range(len(s)):\n if s[i] != 1 or s[i] != 0:\n ans = s[i]\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 0\ncount = 0\nfor i in range(len(s)):\n if k == "1":\n count += 1\n if count == k:\n ans = 1\n break\n else:\n ans = s[i]\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 0\nfor i in range(len(s)):\n if k == 1:\n ans = s[0]\n break\n if s[i] != 1 or s[i] != 0:\n ans = s[i]\n break\nprint(ans)\n', 's = input()\nk = int(input())\ncount = 0\nans = 0\nfor i in range(len(s)):\n if s[i] == "1":\n count += 1\n if count == k:\n ans = 1\n break\n else:\n ans = s[i]\n break\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s428612717', 's609669308', 's877096575', 's059451646'] | [2940.0, 2940.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0, 17.0] | [135, 221, 184, 224] |
p03282 | u426108351 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\nans = 1\nfor i in range(min(len(S), K)):\n if S[i] == 1:\n continue\n else:\n ans = int(S[i])\n break\nprint(ans)\n\n \n \n \n ', "S = input()\nK = int(input())\nans = 1\nfor i in range(min(len(S), K)):\n if S[i] == '1':\n continue\n else:\n ans = int(S[i])\n break\nprint(ans)\n\n \n \n \n "] | ['Wrong Answer', 'Accepted'] | ['s133797140', 's207753631'] | [3060.0, 2940.0] | [21.0, 18.0] | [160, 162] |
p03282 | u439176910 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input().rstrip()\nK = int(input())\nc = 0\nfor item in S:\n if item == "1":\n c+=1\n else:\n break\nn = S[c] if c < len(S) else None\na = n if n and c < K and else S[0]\nprint(a)', 'S = input().rstrip()\nK = int(input())\na = S[1] if K > 1 and S[0] == "1" and len(S) > 1 else S[0]', 'S = input().rstrip()\nK = int(input())\nc = 0\nfor item in S:\n if not item == "1":\n break\n c+=1\nn = S[c] if c < len(S) else None\na = n if n and c < K else S[0]\nprint(a)'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s028157270', 's491318423', 's220936794'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 18.0] | [180, 96, 170] |
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