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 | u703890795 | 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()\nfor s in S:\n if s != 1:\n print(s)\n break', 'S = input()\nK = int(input())\nf = True\nfor i in range(K):\n if S[i] != 1:\n print(S[i])\n f = False\n break\nif f:\n print(S[K-1])', 'S = input()\nK = int(input())\nf = True\nfor i in range(K):\n if S[i] != "1":\n print(S[i])\n f = False\n break\nif f:\n print(1)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s392491689', 's944460993', 's587682292'] | [2940.0, 2940.0, 2940.0] | [24.0, 18.0, 17.0] | [71, 134, 131] |
p03282 | u708255304 | 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 == 1:\n print(S[0])\nelif S[K] == "1":\n print(1)\nelse:\n for i in range(len(S)):\n if S[i] != "1":\n print(S[i])\n exit()\n', '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(len(S):\n if S[i] != 1:\n print(S[i])\n exit()', 'S = input()\nK = int(input())\nif K == 1:\n print(S[0])\n exit()\nfor i in range(len(S)):\n if S[i] != 1:\n print(S[i])\n exit()\nprint(1)\n', 'S = input()\nK = int(input())\nfor i in range(len(S)):\n if S[i] != 1:\n print(S[i])\n exit()', 'S = input()\nK = int(input())\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', 'Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s202177962', 's376528992', 's426359386', 's525255131', 's737711608', 's758870541'] | [2940.0, 2940.0, 2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0, 17.0, 17.0, 17.0] | [188, 114, 104, 153, 105, 171] |
p03282 | u714533789 | 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 min(s) == '1' or k <= s.count('1'): print(1)\nelse: print(min(s))", "s=input()\nk=int(input())\nfor i in range(k):\n\tif s[i] != '1':\n\t\tprint(s[i]);exit()\nprint(1)"] | ['Wrong Answer', 'Accepted'] | ['s792717267', 's037198804'] | [2940.0, 2940.0] | [17.0, 17.0] | [92, 90] |
p03282 | u722189950 | 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 and K == 1:\n print(S[0])\nelse:\n print(S[0])', 'S = input()\nK = int(input())\nfor s in S:\n if s == "1":\n if K == 1:\n print(1)\n break\n k -= 1\n continue\n print(s)\n break', 'S = input()\nK = int(input())\nif S[0] == 1 and K > 1:\n print(S[1])\nelse:\n print(S[0])', 'S = input()\nK = int(input())\nfor s in S:\n if s == "1":\n if K == 1:\n print(1)\n break\n K -= 1\n continue\n print(s)\n break'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s060935862', 's494475073', 's927078548', 's987169669'] | [3064.0, 2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0] | [91, 170, 90, 170] |
p03282 | u726055614 | 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_len=len(input())\nK=int(input())\nfor i in range(S_len):\n if(int(a[0][i])==1):\n K-=1\n if(K<=0):\n print("1")\n break\n else:\n print(a[0][i])\n break', 'a=input().split("¥n")\nS_len=len(a[0])\nK=int(a[1])\nfor i in range(S_len):\n if(int(a[0][i])==1):\n K-=1\n if(K<=1):\n print("1")\n break\n else:\n print(a[0][i])\n break\n', ' a=input().split("¥n")\n goal=int(a[1])\n keta=len(a[0])\n for i in range(keta):\n if(int(a[0][i])==1):\n goal-=1\n if(goal<=0):\n print("1")\n break\n else:\n print(a[0][i])\n break', 'a=input().split("¥n")\nK=int(a[1])\nS_len=len(a[0])\nfor i in range(S_len):\n\n if(int(a[0][i])==1):\n K-=1\n if(K<=0):\n print("1")\n break\n\n elif(int(a[0][i])==2):\n K2=format(K,\'b\')\n K2_len=len(K2)\n K2_len-=5000*10000*10000*10000\n if(K2_len<=0):\n print("2")\n break\n K=pow(2, K2_len)\n\n else:\n print(a[0][i])\n break', ' a=input().split("¥n")\n goal=int(a[1])\n keta=len(a[0])\n for i in range(keta):\n if(int(a[0][i])==1):\n goal-=1\n if(goal<0):\n print("1")\n break\n else:\n print(a[0][i])\n break', 'a=input().split("¥n")\ngoal=int(a[1])\nketa=len(a[0])\nfor i in range(keta):\n if(int(a[0][i])==1):\n goal-=1\n if(goal<=0):\n print("1")\n break\n else:\n print(a[0][i])\n break', 'S=input()\nS_len=len(S)\nK=int(input())\nfor i in range(S_len):\n if(int(S[i])==1):\n K-=1\n if(K<=0):\n print("1")\n break\n else:\n print(S[i])\n break'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s003678577', 's029288737', 's489256366', 's713406755', 's713484931', 's886127941', 's311345762'] | [3060.0, 3060.0, 2940.0, 3064.0, 2940.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0, 18.0, 17.0, 17.0, 20.0] | [168, 186, 272, 423, 271, 224, 166] |
p03282 | u727787724 | 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 len(s)==1 or k==1:\n print(int(s[0]))\nelse:\n for i in range(len(s)):\n if int(s[i])==1 or i!=len(s)-1 or i<k:\n continue\n else:\n print(int(s[i]))\n break', 's=list(input())\nk=int(input())\nif len(s)==1 or k==1:\n print(int(s[0]))\nelse:\n for i in range(len(s)):\n if int(s[i])==1 and i!=len(s)-1 and (i+1)<k:\n continue\n else:\n print(int(s[i]))\n break'] | ['Wrong Answer', 'Accepted'] | ['s492651318', 's678140928'] | [3060.0, 3060.0] | [17.0, 17.0] | [236, 242] |
p03282 | u730476362 | 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())\na = 10 ** 18\nprint(a)\ncounter = 0\nfor i in S:\n if int(i) != 1:\n log = counter\n num = int(i)\n break\n counter += 1\n\nif K == 1:\n print(S[0])\nelif K <= log:\n print(1)\nelse:\n print(num)\n', 'S = input()\nK = int(input())\ncounter = 0\nnum = 1\nlog = 0\nfor i in S:\n if int(i) != 1:\n log = counter\n num = int(i)\n break\n counter += 1\n\nif K <= log:\n print(1)\nelse:\n print(num)\n'] | ['Runtime Error', 'Accepted'] | ['s555050384', 's165982118'] | [3060.0, 2940.0] | [18.0, 18.0] | [246, 211] |
p03282 | u732285858 | 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())\nit=int(-1)\nsum=int(0)\nfor i in s:\n if i == '1':\n sum+=1\n else:\n it=i\n\nif k<=int(it):\n print(1)\nelif it == int(-1):\n print(1)\nelse:\n print(s[it])", "s=input()\nk=int(input())\nit=int(-1)\nsum=int(0)\nfor i in s:\n if i == '1':\n sum+=1\n else:\n it=i\n\nif k<=it:\n print(1)\nelif it is -1:\n print(1)\nelse:\n print(s[it])", "s=input()\nk=int(input())\nit=int(-1)\nsum=int(0)\nfor i in s:\n if i == '1':\n sum+=1\n else:\n it=sum\n break\n\nif k<=int(it):\n print(1)\nelif it == int(-1):\n print(1)\nelse:\n nex=int(it)+int(1)\n print(s[it:nex])"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s259085458', 's940325448', 's099561201'] | [3064.0, 3060.0, 3064.0] | [17.0, 17.0, 18.0] | [198, 188, 241] |
p03282 | u733608212 | 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\ncount = 0\nfor i in s:\n \tcount += 1\n if i != "1" or (count == n):\n print(i)\n break\nelse:\n print(1)', 's = input()\nn = int(input())\n\ncount = 0\nfor i in s:\n \tcount += 1\n if i != "1" or count == n:\n print(i)\n break\nelse:\n print(1)', 's = input()\nn = int(input())\n \ncount = 0\nfor i in s:\n count+=1\n if i != "1" or count == n:\n print(i)\n break\nelse:\n print(1)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s337148304', 's718661071', 's612028036'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0] | [148, 146, 146] |
p03282 | u738835924 | 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\nN = 5000000000000000\nprint(N)\nS = input()\nK = int(input())\nfor i in S:\n num = int(i)\n if N * math.log(num) > math.log(K):\n print(num)\n break\n else:\n K -= num', 'import math\nN = 5000000000000000\nprint(N)\nS = input()\nK = math.log(int(input()))\nfor i in S:\n num = int(i)\n kk = N * math.log(num)\n if kk > K:\n print(num)\n break\n else:\n K -= kk', 'import math\nN = 5000000000000000\nprint(N)\nS = input()\nK = math.log(int(input()))\nfor i in S:\n num = int(i)\n kk = N * math.log(num)\n if kk >= K:\n print(num)\n break\n else:\n K -= kk', 'import math\nN = 5000000000000000\nS = input()\nK = math.log(int(input()))\nfor i in S:\n num = int(i)\n if num == 1:\n if num == int(math.exp(K)):\n print(num)\n break\n else:\n K = math.log(int(math.exp(K)-1))\n continue\n kk = N * math.log(num)\n if kk >= K:\n print(num)\n break\n else:\n K -= kk'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s326379423', 's719091453', 's785451806', 's403574429'] | [3060.0, 3060.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0, 17.0] | [181, 190, 191, 322] |
p03282 | u740284863 | 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(k):\n if s[i] != 1:\n print(s[i])\n exit()\nprint(1)', 's = str(input())\nk = int(input())\nfor i in range(k):\n if s[k] != 1:\n print(s[k])\n exit()\nprint(1)', 's = str(input())\nk = int(input())\nfor i in range(k):\n if s[i] != "1":\n print(s[i])\n exit()\nprint(1)'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s450364990', 's876739561', 's942879737'] | [2940.0, 2940.0, 2940.0] | [18.0, 18.0, 17.0] | [114, 114, 116] |
p03282 | u744920373 | 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\nif len(S) == 1:\n print(S[0])\nelif S[0] == 1:\n print(S[1])\nelse:\n print(S[0])', "S = str(input())\nK = int(input())\n\nif len(S) == 1:\n print(S[0])\n exit(0)\n\ni = 0\nwhile S[i] == '1' and i <= len(S)-2:\n if K <= i + 1:\n print(S[i])\n exit(0)\n i += 1\nprint(S[i])"] | ['Wrong Answer', 'Accepted'] | ['s376328482', 's429218754'] | [3060.0, 3060.0] | [17.0, 23.0] | [132, 232] |
p03282 | u745561510 | 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\ncheck = 0\n\nif len(S) == 1:\n print(S[0])\n\n\nfor i in range(len(S)):\n if S[i] == 1:\n check = 1\n else:\n print(S[i])\n exit()\n check = -1\n \n if K == i and check == 1:\n print("1")\n exit()\n\n\n', 'S = input()\nK = int(input())\n\n\ncheck = 0\n\nif len(S) == 1:\n print(S[0])\n exit()\n\nfor i in range(len(S)):\n if S[i] == 1:\n check = 1\n else:\n check = -1\n \n if K == i and check == 1:\n print("1")\n exit()\n\nprint(S[0])\n \n\n', "S = input()\nK = int(input())\n\n\ncheck = 0\n\nfor i in range(K):\n if int(S[i]) != 1:\n print(S[i])\n exit()\n\n\nprint('1')\n"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s152511916', 's306363522', 's274455538'] | [3060.0, 3060.0, 2940.0] | [17.0, 17.0, 17.0] | [321, 313, 132] |
p03282 | u745906650 | 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\ns=input()\nk=int(input())\nlist=list(s)\nans=1\nfor i in range(min(k,list.len())):\n if not list[i] == 1:\n ans=list[i]\n break\nprint(ans)\n', '# coding:utf-8\ns=input()\nk=int(input())\nlists=list(s)\nans=1\nfor i in range(min(k,len(lists))):\n if not lists[i] == "1":\n ans=lists[i]\n break\nprint(ans)\n'] | ['Runtime Error', 'Accepted'] | ['s368731326', 's543335924'] | [2940.0, 2940.0] | [18.0, 17.0] | [164, 169] |
p03282 | u746419473 | 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 _s in s:\n if _s == "1":\n if k == 1:\n print(1)\n break\n else:\n k -= 1\n else:\n print(i)\n break\n', 's = input()\nk = int(input())\n\nfor _s in s:\n if _s == "1":\n if k == 1:\n print(1)\n break\n else:\n k -= 1\n else:\n print(_s)\n break\n'] | ['Runtime Error', 'Accepted'] | ['s904533509', 's420902637'] | [2940.0, 2940.0] | [17.0, 17.0] | [193, 194] |
p03282 | u751717561 | 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(len(s)):\n if s[i] != '1':\n if i > k:\n print(int(s[i]))\n exit()\n else:\n print(1)\n exit()\n\nprint(1)\n", "s = input()\nk = int(input())\n\nfor i in range(len(s)):\n if s[i] != '1':\n if i > k:\n print(int(s[i]))\n exit()\n else:\n print(1)\n\nprint(1)\n", "s = input()\nk = int(input())\n\nfor i in range(len(s)):\n if s[i] != '1':\n if k > i:\n print(int(s[i]))\n else:\n print(1)\n exit()\n\nprint(1)\n"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s669251906', 's805409594', 's612984478'] | [9156.0, 9020.0, 9084.0] | [29.0, 27.0, 27.0] | [204, 185, 181] |
p03282 | u754651673 | 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\nif(k == 1):\n ans = s[0]\nelse:\n ans = s[0]\n \nprint(ans)', 's = input()\nk = int(input())\nans = 0\nif(s[0] == 1 and k != 1):\n ans = s[1]\nelse:\n ans = s[0]\n \nprint(ans)', 's = input()\nk = int(input())\nans = 1\nfor i in range(k):\n if(s[i] != 1):\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = "1"\nfor i in range(min(len(s), k)):\n if(s[i] != "1"):\n ans = s[i]\n break\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s283012623', 's736849379', 's891071956', 's291259852'] | [2940.0, 3060.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0, 18.0] | [100, 114, 123, 135] |
p03282 | u756988562 | 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) == 1:\n print(S)\nelse:\n if K == 1:\n print(1)\n else:\n print(S[2])', 'S = input()\nK = int(input())\njudge = True\nfor i in range(len(S)):\n if S[i] !="1":\n # judge = False\n if (i+1) <= K: \n print(S[i])\n exit()\nprint(1)'] | ['Wrong Answer', 'Accepted'] | ['s390182390', 's462833149'] | [3316.0, 3316.0] | [21.0, 21.0] | [126, 184] |
p03282 | u760771686 | 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())\nans = '1'\nfor i in range(K):\n if i >= len(N):\n break\n if N[i] != '1':\n ans = N[i]\n \tbreak\nprint(ans)\n ", "N = input()\nK = int(input())\nif len(N) == 1:\n print(N[0])\nelif K = 1:\n print(N[0])\nelif N[0] == '1':\n print(N[1])\nelse:\n print(N[0])\n ", "N = input()\nK = int(input())\nans = '1'\nfor i in range(K):\n if i >= len(N):\n break\n if N[i] != '1':\n ans = N[i]\n break\nprint(ans)"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s422313186', 's576183505', 's855343582'] | [8956.0, 8980.0, 9160.0] | [24.0, 26.0, 30.0] | [141, 139, 139] |
p03282 | u762420987 | 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 c in S:\n if c != "1":\n print(c)\n break\nprint("1")\n', 'S = input()\nK = int(input())\nif S[0] == "1":\n print(list(set(S))[1])\nelse:\n print(S[0])\n', 'S = input()\nK = int(input())\nif S[0] == "1":\n if K == 1:\n print(1)\n else:\n S.replace("1", "")\n print(S[0])\nelse:\n print(S[0])\n', 'S = input()\nK =int(input())\nif K == S[:K].count("1"):\n print(1)\nelse:\n for c in S:\n if c != "1":\n print(c)\n break'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s353482968', 's391589472', 's701585414', 's742801723'] | [2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0] | [106, 94, 156, 148] |
p03282 | u762540523 | 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? | ['print(s=input().replace("1","")[0])', 's=input();k=int(input());print(1 if s[0:k]=="1"*k else input().replace("1","")[0])', 'def main():\n s=input()\n k=int(input())\n for i in s[:k]:\n if i!="1":\n print(i)\n return\n print("1")\nmain()'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s292831288', 's748381850', 's521119807'] | [2940.0, 3060.0, 2940.0] | [17.0, 19.0, 17.0] | [35, 82, 121] |
p03282 | u766566560 | 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[0])\nelse:\n print(1)', 'S = input()\nK = int(input())\n\nif S[0] != 1:\n print(S[0])\nelif S[0] == 1 and K > 1:\n print(S[1])\nelse:\n print(1)', 'S = input()\nK = int(input())\n\nif S[0] != 1 and len(S) != 1:\n print(S[0])\nelif S[0] != 1 and len(S) == 1:\n print(S[0])\nelse:\n print(1)', "# ans\n\nS = input()\nK = int(input())\n\nif S[0:K].count('1') == K:\n print(1)\n exit()\n \nfor i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s367011406', 's864944761', 's866313657', 's187255149'] | [2940.0, 2940.0, 3060.0, 2940.0] | [17.0, 17.0, 17.0, 17.0] | [74, 114, 136, 150] |
p03282 | u767664985 | 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 list(set(S)) == ["1"]:\n print(1)\nelif S[0] == "1" and K > 1:\n ind = 1\n while S[ind] == "1":\n ind += 1\n if ind > K:\n print(S[ind])\n else:\n print(1)\nelse:\n print(S[0])\n', 'S = input()\nK = int(input())\nif S[0] != "1":\n print(S[0])\nelse:\n ind = 0\n while S[ind] == "1" and ind < len(S) - 1:\n ind += 1\n if ind < K:\n print(S[ind])\n else:\n print("1")\n'] | ['Wrong Answer', 'Accepted'] | ['s466676134', 's048609816'] | [3060.0, 3064.0] | [17.0, 17.0] | [237, 209] |
p03282 | u769698512 | 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\ndef ones(s):\n ss = s[0]\n idx = 0\n while ss=="1":\n idx += 1\n ss = s[idx]\n return idx\n\nif s[0]!="1":\n print(s[0])\nelse:\n idx = ones(s)\n if idx>=k:\n print("1")\n # else:\n # print(s[idx])\n', 's = input()\nk = int(input())\n\ndef ones(s):\n ss = s[0]\n idx = 0\n while ss=="1" and idx+1 < len(s):\n idx += 1\n ss = s[idx]\n return idx\n\nif s[0]!="1":\n print(s[0])\nelse:\n idx = ones(s)\n if idx>=k:\n print("1")\n else:\n print(s[idx])\n'] | ['Runtime Error', 'Accepted'] | ['s742106762', 's494743748'] | [3056.0, 3060.0] | [17.0, 17.0] | [265, 280] |
p03282 | u780962115 | 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? | ['\na = [int(c) for c in s]\nk=int(input())\nstrong=0\nfor some in range(k):\n if a[some]==1:\n strong+=1\n if strong==k-1:\n print(some)\n if a[some]!=1:\n print(some)\n break\n', 's=input()\nlists=[int(integer)for integer in s] \nnums=int(input())\n\nnumbers=0 \nstrong=0 \n\n\nfor j in lists:\n if lists[0]!=1:\n break\n else:\n if j==1:\n numbers+=1\n else:\n break\n\nif lists[0]==1:\n if numbers>=nums:\n print(1)\n \n elif numbers<nums:\n print(lists[numbers])\nelse:\n print(lists[0])'] | ['Runtime Error', 'Accepted'] | ['s752831543', 's441063625'] | [2940.0, 3064.0] | [17.0, 17.0] | [179, 571] |
p03282 | u781262926 | 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().strip()\nk = int(input())\nfor i, c in enumerate(S, 1):\n if c != '1' or i == k:\nprint(c)", "S = input().strip()\nk = int(input())\nfor i, c in enumerate(S, 1):\n if c != '1' or i == k:\n break\nprint(c)"] | ['Runtime Error', 'Accepted'] | ['s527801298', 's910611800'] | [2940.0, 2940.0] | [17.0, 18.0] | [99, 109] |
p03282 | u787562674 | 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\nprint(S[1])\n\nif S[0] != "1":\n print(S[0])\nelse:\n if S[1] != "1":\n print(S[1])\n else:\n if S[2] != "1":\n print(S[2])\n else:\n if S[3] != "1":\n print(S[3])\n else:\n print(1)\n\n\n', 'S = input()\nK = input()\n\nif S[0] == 1 and S[1] == 1 and S[2] == 1 and S[3] == 1:\n print(1)\nelif S[0] == 1 and S[1] == 1 and S[2] == 1:\n print(S[3])\nelif S[0] == 1 and S[1] == 1:\n print(S[2])\nelif S[0] == 1:\n print(S[1])\nelse:\n print(S[0])', 'S = input()\nK = input()\n\nfor i in range(int(K)):\n if S[i] != "1":\n print(S[i])\n break\nelse:\n print(1)\n\n\n\n'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s331455710', 's648351859', 's712593325'] | [3060.0, 3064.0, 2940.0] | [17.0, 17.0, 17.0] | [289, 253, 125] |
p03282 | u788023488 | 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 \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nsaisho = 0\nL = len(S)\n \nfor i in range(L):\n N = int(S[i])\n if k <= saisho+1:\n ans = N\n break\n \n else:\n if N == 1:\n Nn == 1\n else:\n Nn = N**n\n saisho += N**n\n \nprint(ans)', 's = int(input())\nk = int(input())\n\nans=0\n\nn = 5000000000000000\n\nS = str(s) \nsaisho = 0\n\nfor i in range(len(S)):\n N = int(S[i])\n if saisho <= k <= saisho + N**n:\n ans = N\n break\n \n else:\n saisho += N**n\n \nprint(ans)', 's = int(input())\nk = int(input())\n \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nsaisho = 0\n \nfor i in range(len(S)):\n N = int(S[i])\n if k <= saisho+1:\n ans = N\n break\n \n else:\n if N ==1:\n Nn = 1\n else:\n Nn = N**n\n saisho += Nn\n \nprint(ans)', 's = int(input())\nk = int(input())\n\nans=0\n\nn = 5000000000000000\n\nS = str(s) \nsaisho = 0\n\nfor i in range(len(S)):\n N = int(S[i])\n if saisho <= k <= saisho + N**n:\n print(k)\n print(saisho+N**n)\n ans = N\n break\n \n else:\n saisho += N**n\n \nprint(ans)', 's = int(input())\nk = int(input())\n \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nL = len(S)\nsaisho = 0\n \nfor i in range(L):\n N = int(S[i])\n if N == 1:\n Nn == 1\n else:\n Nn = N**n\n \n if saisho <= k <= saisho + Nn:\n ans = N\n break\n \n else:\n if N ==1\n Nn ==\n saisho += Nn\n \nprint(ans)', 's = int(input())\nk = int(input())\n \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nL = len(S)\nsaisho = 0\n \nfor i in range(L):\n N = int(S[i])\n if N == 1:\n Nn == 1\n else:\n Nn = N**n\n \n if saisho <= k <= saisho + Nn:\n ans = N\n break\n \n else:\n saisho += Nn\n \nprint(ans)', 's = int(input())\nk = int(input())\n \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nsaisho = 0\n \nfor i in range(len(S)):\n N = int(S[i])\n if k <= saisho+1:\n ans = N\n break\n \n else:\n saisho += N**n\n \nprint(ans)', 's = int(input())\nk = int(input())\n \nans=0\n \nn = 5000000000000000\n \nS = str(s) \nL = len(S)\nsaisho = 0\n \nfor i in range(L):\n N = int(S[i])\n Nn = N**n\n if saisho <= k <= saisho + Nn:\n ans = N\n break\n \n else:\n saisho += Nn\n \nprint(ans)', 's = int(input())\nk = int(input())\nS = str(s) \n\nans=0\n\nfor i in range(k):\n N = int(S[i])\n \n if N == 1:\n ans = 1\n else:\n ans = N\n break \nprint(ans)'] | ['Runtime Error', 'Time Limit Exceeded', 'Time Limit Exceeded', 'Wrong Answer', 'Runtime Error', 'Runtime Error', 'Time Limit Exceeded', 'Time Limit Exceeded', 'Accepted'] | ['s073682626', 's225716192', 's394198323', 's409189261', 's462588294', 's560824461', 's653829516', 's771489787', 's826231173'] | [197060.0, 197056.0, 197060.0, 166456.0, 2940.0, 197048.0, 197060.0, 197060.0, 2940.0] | [2106.0, 2106.0, 2105.0, 2107.0, 17.0, 2108.0, 2105.0, 2106.0, 18.0] | [327, 277, 317, 313, 360, 335, 266, 295, 178] |
p03282 | u788137651 | 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 s in S:\n if s == "1":\n K -= 1\n if K <= 0:\n print(s)\n exit()\n else:\n print(s)\n exit()', 'S = input()\nK = int(input())\nfor s in S:\n if int(s)==1:\n\tK-=1\n if K==0:\n print(s)\n exit()\n else:\n print(s)\n exit()', 'S = input()\nK = int(input())\nfor s in S:\n if int(s) == 1:\n K -= 1\n if K == 0:\n print(s)\n exit()\n else:\n print(s)\n exit()'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s332157503', 's723768311', 's665053176'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 18.0] | [162, 135, 176] |
p03282 | u790877102 | 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\nt = 0\nu = s[0]\n\nfor i in range(len(s)):\n\tt +=1\n\tu = s[i]\n\tif s[i] !="1":\n\t\tbreak\nif t<k+1:\n\tprint(u)\nelse:\n\tprint(1', 's =input()\nk = int(input())\n\nt = 0\nu = s[0]\n\nfor i in range(len(s)):\n\tt +=1\n\tu = s[i]\n\tif s[i] !="1":\n\t\tbreak\nif t<k+1:\n\tprint(u)\nelse:\n\tprint(1', 's =input()\nk = int(input())\n\nt = 0\nu = s[0]\n\nfor i in range(len(s)):\n\tt +=1\n\tu = s[i]\n\tif s[i] !="1": \n\t\tbreak\nif t<k+1:\n\tprint(u)\nelse:\n\tprint(1)'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s333707792', 's714453361', 's827400298'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0] | [144, 144, 205] |
p03282 | u794652722 | 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\ndef check_one(sgn,arr):\n cnt = 0\n for i in range(sgn):\n if S[i] == '1':\n cnt += 1\n if cnt == sgn:\n return True\n\ndef first_num(arr):\n for i in range(len(S)):\n if S[i] != '1':\n return int(S[i])\n break\n\nif len(S) >= K:\n if check_one(K,S):\n print(1)\nelse:\n print(first_num(S))\n ", "S = input()\nK = int(input())\n\ndef check_one(sgn,arr):\n cnt = 0\n for i in range(sgn):\n if S[i] == '1':\n cnt += 1\n if cnt == sgn:\n return True\n\ndef first_num(arr):\n list = []\n for i in range(len(arr)):\n if arr[i] != '1':\n list.append(int(arr[i]))\n return list[0]\n \n\nif len(S) >= K and check_one(K,S):\n print(1)\nelse:\n print(first_num(S))\n "] | ['Wrong Answer', 'Accepted'] | ['s565138039', 's364796644'] | [3060.0, 3060.0] | [17.0, 17.0] | [386, 422] |
p03282 | u797550216 | 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()\nprint(S) \n \n', 'S = int(input())\nK = int(input())\n\nlst = []\nwhile S > 0:\n lst.append(S%10)\n S //= 10 \nlst.reverse()\n\nif len(lst) == lst.count(1):\n\tprint(1)\n\nelif K == lst[:K].count(1):\n\tprint(1)\n \nelse:\n\tfor i in range(K):\n\t\tif lst[i] != 1:\n\t\t\tprint(lst[i])\n\t\t\tbreak\n\n\n \n '] | ['Runtime Error', 'Accepted'] | ['s188797456', 's777828106'] | [2940.0, 3060.0] | [17.0, 17.0] | [64, 265] |
p03282 | u801359367 | 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()\ncount = 0\nfor i in range(len(S)):\n if S[i]=="1":\n count+=1\n else:\n break\n \nprint(1) if K <= count else print(S[count])\n', 'S = input()\nK = input()\ncount = 0\nfor i in range(len(S)):\n if S[i]=="1":\n count+=1\n else:\n break\n \nprint(1) if int(K) <= count else print(S[count])\n'] | ['Runtime Error', 'Accepted'] | ['s298121840', 's141870301'] | [2940.0, 2940.0] | [18.0, 17.0] | [170, 175] |
p03282 | u801512570 | 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\nD=5000000000000000\n\ntmp=0\nfor s in S:\n tmp+=int(s)*5000000000000000 if s!='1' else 1\n if tmp>K:\n print(s)", 'S = input()\nK = int(input())\n\nD=5000000000000000\n\ntmp=0\nfor s in S:\n t=1\n for _ in range(D):\n t=t*int(s)\n if tmp+t>=K:\n print(s)\n break\n else:\n continue\n break', "S = int(input())\nK = int(input())\n\nD=5000000000000000\n\ntmp=0\nfor s in S:\n tmp+=int(s)*5000000000000000 if s!='1' else 1\n if tmp>K:\n print(s)", 'S = input()\nK = int(input())\n\nD=5000000000000000\n\ntmp=0\nfor s in S:\n t=1\n for _ in range(D):\n t=t*int(s)\n if tmp+t>=K:\n print(s)\n break\n else:\n tmp+=t\n continue\n break', "S = input()\nK = int(input())\n\nD=5000000000000000\n\ntmp=0\nfor s in S:\n if s=='1':\n tmp+=1\n if tmp>=K:\n print(s)\n break\n continue\n \n t=1\n for _ in range(D):\n t=t*int(s)\n if tmp+t>=K:\n print(s)\n break\n else:\n tmp+=t\n continue\n break"] | ['Wrong Answer', 'Time Limit Exceeded', 'Runtime Error', 'Time Limit Exceeded', 'Accepted'] | ['s119802311', 's595210487', 's647893422', 's967154000', 's971612053'] | [3060.0, 2940.0, 3060.0, 2940.0, 3060.0] | [18.0, 2104.0, 18.0, 2104.0, 18.0] | [140, 182, 145, 193, 277] |
p03282 | u802772880 | 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())\non=[int(j) for j in range(2,10)]\ni=0\nif int(s[0]) in on:\n print(s[0])\n\nwhile int(s[i]) not in on:\n if s[i] in on:\n print(s[i])\n break\n elif i==k-1:\n print(s[k-1])\n break\n i+=1', 's=input()\nk=int(input())\non=[int(j) for j in range(2,10)]\ni=0\nif int(s[0]) in on:\n print(s[0])\n\nwhile int(s[i]) not in on:\n if s[i] in on:\n print(s[i])\n break\n elif i==k-1:\n print(s[k-1])\n break\n i+=1', "s=input()\nk=int(input())\non=[int(j) for j in range(2,10)]\nfor i in range(k):\n if int(s[i]) in on:\n print(s[i])\n break\n elif i==k-1:\n print('1')\n else:\n i+=1"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s542999822', 's625227018', 's591774462'] | [3060.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0] | [240, 240, 193] |
p03282 | u809690311 | 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())\nS_list = [int(x) for x in list(str(S))]\n\ndef saiki(i):\n if i == len(S_list):\n print(S_list[i-1])\n exit()\n if S_list[i] > 2:\n print(S_list[i])\n exit()\n if S_list[i] == 2:\n if K > 5*1000000000000000:\n print(2)\n exit()\n else:\n saiki(i+1)\n if S_list[i] == 1:\n saiki(i+1)\n\nprint(saiki(0))', 'S = int(input())\nK = int(input())\nS_list = [int(x) for x in list(str(S))]\n \ndef saiki(i):\n if i == len(S_list):\n print("1")\n exit()\n if K == i:\n print("1")\n exit()\n if S_list[i] > 1:\n print(S_list[i])\n exit()\n if S_list[i] == 1:\n saiki(i+1)\n \nprint(saiki(0))'] | ['Wrong Answer', 'Accepted'] | ['s624251657', 's898120023'] | [3064.0, 3060.0] | [18.0, 18.0] | [357, 283] |
p03282 | u810288681 | 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 i!= '1':\n ans = i\n break\nprint(ans)", "s = input()\nk = int(input())\nans = '1'\nfor i in range(k):\n if s[i]!= '1':\n ans = s[i]\n break\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s159002290', 's301005215'] | [2940.0, 2940.0] | [17.0, 17.0] | [112, 120] |
p03282 | u812576525 | 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 i in range(K):\n if S[i] != '1':\n ans = s\n break\n else:\n ans = 1\n continue\nprint(ans)", "S = list(str(input()))\nK = int(input())\n#rint(S)\nans = 1\nfor i in range(K):\n #print(i)\n if S[i] != '1' :\n ans = S[i]\n break\nprint(ans)\n"] | ['Runtime Error', 'Accepted'] | ['s957158830', 's940376998'] | [2940.0, 2940.0] | [17.0, 17.0] | [158, 155] |
p03282 | u813102292 | 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())\nc = 0\nfor i in range(len(s)):\n if s[i] == "1":\n c += 1\n else:\n continue\nif c >= k:\n print(1)\n break\nfor i in range(len(s)):\n if s[i] != "1":\n print(s[i])\n break\n', 's = str(input())\nk = int(input())\nc = 0\nfor i in range(len(s)):\n if s[i] == "1":\n c += 1\n else:\n break\n\nif c >= k:\n print("1")\nelse:\n for i in range(len(s)):\n if s[i] != "1":\n print(s[i])\n break\n'] | ['Runtime Error', 'Accepted'] | ['s649007782', 's990484214'] | [2940.0, 2940.0] | [17.0, 17.0] | [238, 250] |
p03282 | u814608389 | 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\ns=int(input())\nk=int(input())\n\nfor i in range(len(s)):\n if s[i] != '1':\n print(s[i])\n break\n elif i == k - 1:\n break \n ", "import sys\n \ns=input()\nk=int(input())\n \nfor i in range(len(s)):\n if s[i] != '1':\n print(s[i])\n break\n elif i == k - 1:\n print(1)\n break"] | ['Runtime Error', 'Accepted'] | ['s434789719', 's533298601'] | [2940.0, 2940.0] | [17.0, 17.0] | [152, 149] |
p03282 | u814781830 | 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())\nret = "1"\nfor i in K:\n if len(S) <= K:\n break\n if i != "1":\n ret = i\n break\nprint(ret)\n', 'S = input()\nK = int(input())\nret = "1"\nfor i in K:\n if len(S) <= K:\n break\n if S[i] != "1":\n ret = S[i]\n break\nprint(ret)\n', 'S = input()\nK = int(input())\nret = "1"\nfor i in range(K):\n if len(S) < i-1:\n break\n if S[i] != "1":\n ret = S[i]\n break\nprint(ret)\n'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s561765863', 's935785504', 's147555210'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [143, 149, 157] |
p03282 | u816116805 | 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 = int(input())\n \nif a<105:\n print(0)\nelif 105<= a < 135:\n print(1)\nelif 135<= a < 165:\n print(2)\nelse:\n print(3)\n', "s = input()\nk = int(input())\n\nfor i,a in enumerate(s):\n if (a != '1' ) or k==i+1:\n ans = int(a)\n break\n\nprint (ans)\n"] | ['Wrong Answer', 'Accepted'] | ['s552921146', 's667059644'] | [2940.0, 2940.0] | [18.0, 17.0] | [127, 133] |
p03282 | u820047642 | 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==1:\n print(S[0])\nelse:\n ans=1\n for i in range(len(S)):\n if S[i]==1:\n pass\n else:\n ans=S[i]\n break\n print(ans)', 'S=input()\nK=int(input())\n\ni=0\nwhile i<=len(S)-1:\n if S[i]=="1":\n i+=1\n else:\n break\n\nif K<=i:\n print("1")\nelse:\n print(S[i])'] | ['Wrong Answer', 'Accepted'] | ['s708196225', 's555883945'] | [3064.0, 2940.0] | [17.0, 17.0] | [198, 150] |
p03282 | u821262411 | 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\nfor i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()\nelif len(S) > K:\n print(S[K])\nelse:\n print(S[K-1])\n\n", "S=str(input())\nK=int(input())\n\nfor i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()\n \nprint(S[K-1])"] | ['Runtime Error', 'Accepted'] | ['s083979437', 's656968468'] | [2940.0, 2940.0] | [17.0, 17.0] | [163, 127] |
p03282 | u834415466 | 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 pass\n else:\n if i+1<=k:\n print(s[i])\n break\nprint(1)", "s=str(input())\nk=int(input())\na=0\nfor i in range(len(s)):\n if s[i]=='1':\n pass\n else:\n if i+1<=k:\n print(s[i])\n a=1\n break\nif a==0:\n print(1)"] | ['Wrong Answer', 'Accepted'] | ['s805758161', 's188952857'] | [2940.0, 2940.0] | [17.0, 22.0] | [160, 193] |
p03282 | u835732324 | 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(str(input()))\nk=int(input())\nfor i in range(len(S)):\n si=int(S[i])\n ti=si**(5*10**15)\n if k<=ti:\n print(S[i])\n break\n else:\n k-=ti', 'S=list(str(input()))\nk=int(input())\nA=[]\nfor i in range(len(S)):\n si=int(S[i])\n ti=si**(k-1)\n while ti > 0:\n A.append(S[i])\n ti=ti-1\nprint("".join(A))', "s = input()\nk = int(input())\n \nans = '1'\nfor i in range(k):\n if s[i] != '1':\n ans = s[i]\n break\n \nprint(ans)"] | ['Time Limit Exceeded', 'Wrong Answer', 'Accepted'] | ['s042719413', 's105966803', 's158905895'] | [197056.0, 194704.0, 2940.0] | [2106.0, 2108.0, 18.0] | [170, 173, 115] |
p03282 | u839224193 | 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()\nif len(K) > 100:\n K = 110\nelse:\n K = int(K)\n\n\ndef main():\n if S[0] != '1':\n print(S[0])\n return\n\n i = 0\n while i < len(S) - 1 and S[i] == '1':\n i += 1\n\n if K-1 <= i:\n print('1')\n return\n\n if i == len(S) - 1:\n print(S[i])\n return\n\n print(S[i+1])\n\n\nif __name__ == '__main__':\n main()\n", "S = input()\nK = int(input()) - 1\n\n\ndef main():\n if S[0] != '1':\n print(S[0])\n return\n\n i = 0\n while i < len(S) - 1 and S[i] == '1':\n i += 1\n\n if K <= i:\n print('1')\n return\n\n print(S[i+1])\n\n\nif __name__ == '__main__':\n main()\n", "S = input()\nK = input()\nif len(K) > 100:\n K = 110\nelse:\n K = int(K)\n\n\ndef main():\n if S[0] != '1':\n print(S[0])\n return\n\n for i in range(len(S)):\n if S[i] != '1':\n break\n\n if i == len(S) - 1:\n print('1')\n return\n\n if K <= i:\n print('1')\n return\n\n print(S[i])\n\n\nif __name__ == '__main__':\n main()\n"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s368889256', 's443779376', 's965944687'] | [3064.0, 2940.0, 3060.0] | [17.0, 17.0, 17.0] | [382, 279, 380] |
p03282 | u839675792 | 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])\n", "s = input()\nk = int(input())\n\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n break\nelse:\n print('1')\n"] | ['Runtime Error', 'Accepted'] | ['s506348494', 's340319640'] | [2940.0, 2940.0] | [17.0, 17.0] | [62, 124] |
p03282 | u840958781 | 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:\n ans=s[i]\n break\n else:\n bns+=1\nif k<=bns:\n print(1)\nelse:\n print(ans)', 's=input()\nk=int(input())\nans=0\nfor i in range(len(s)):\n if s[i]!=1:\n print(s[i])\n break\n else:\n ans+=1\n if k<ans:\n print(1)\n break', 's=input()\nk=int(input())\nfor i in range(len(s)):\n if s[i]!=1:\n print(s[i])\n break', 's=input()\nk=int(input())\nans,bns=0,0\nfor i in range(len(s)):\n if s[i]!="1":\n ans=s[i]\n break\n else:\n bns+=1 \nif k<=bns: \n print(1)\nelse:\n print(ans)\n \n'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s013600940', 's292101053', 's347843733', 's536925888'] | [3064.0, 2940.0, 3064.0, 2940.0] | [17.0, 17.0, 17.0, 17.0] | [171, 186, 98, 210] |
p03282 | u842964692 | 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())\nkazu=5*10**5\n\nfor i in range(len(S)):\n print(K)\n if K<=int(S[i])**kazu:\n print(int(S[i]))\n break\n else:\n K-=int(S[i])**kazu\n', 'S=input()\nK=int(input())\nkazu=5*10**5\n\nfor i in range(len(S)):\n print(K)\n if K<=int(S[i])**kazu:\n print(int(S[i]))\n break\n else:\n K=K-int(S[i])**kazu\n', 'S=input()\nK=int(input())\nkazu=5*10**5\n\nfor i in range(len(S)):\n if K<=int(S[i])**kazu:\n print(int(S[i]))\n break\n else:\n K=K-int(S[i])**kazu\n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s552044018', 's970298055', 's412762655'] | [3756.0, 3756.0, 3756.0] | [92.0, 92.0, 92.0] | [179, 180, 167] |
p03282 | u844646164 | 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\nS = list(map(int, input()))\nK = int(input())\nafter_S = 0\ndays = 5000000000000000-1\nfor n in S:\n if n == 1:\n after_S += 1\n if math.log(after_S) >= math.log(n):\n print(n)\n break\n\n after_S += days * math.log(n)\n if after_S >= math.log(K):\n print(n)\n break', 'S = list(map(int, input()))\nK = int(input())\nif len(S) == 1:\n print(S[0])\n exit()\n\ncnt = 0\nfor s in S:\n if s == 1:\n cnt += 1\nif K <= cnt:\n print(1)\nelse:\n print(S[cnt])\n', 'S = list(map(int, input()))\nK = int(input())\nif len(S) == 1:\n print(S[0])\n exit()\n\ncnt = 0\nfor s in S:\n if s == 1:\n cnt += 1\n else:\n break\n\nif K <= cnt:\n print(1)\nelse:\n print(S[cnt])\n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s165352068', 's426289187', 's720491096'] | [3060.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [323, 191, 216] |
p03282 | u844789719 | 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(len(S), K)):\n if S[i] > 1:\n print(S[i])\n exit()\nprint(1)', 'S = input()\nK = int(input())\nfor i in range(min(len(S), K)):\n if int(S[i]) > 1:\n print(S[i])\n exit()\nprint(1)'] | ['Runtime Error', 'Accepted'] | ['s346136158', 's619016254'] | [2940.0, 2940.0] | [18.0, 18.0] | [121, 126] |
p03282 | u853900545 | 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())\nn=0\nfor i in s:\n if int(i)==1:\n n+=1\n else:\n break\nif n < k:\n pritn(s[n])\nelse:\n print(1)', "s=input()\nk=int(input())\nn=0\nfor i in s:\n if i =='1':\n n+=1\n else:\n break\nif n < k:\n print(s[n])\nelse:\n print(1)"] | ['Runtime Error', 'Accepted'] | ['s422413111', 's212707488'] | [3060.0, 2940.0] | [17.0, 17.0] | [140, 138] |
p03282 | u853952087 | 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=int(input())\nx=5*pow(10,15)\ni=0\nwhile k>0:\n k-=pow(int(a[i]),x)\n i+=1\nprint(a[i-1])', "a=input()\nk=int(input())\nif int(a)<10:\n a=str(int(a)*11)\ni=0\nx=0\nwhile k>0:\n if a[i]=='1':\n k-=1\n i+=1\n x=1\n else: k=-1 \nprint(a[i-x])", 'a=input()\nk=int(input())\nx=5*pow(10,15)\ni=0\nwhile k>0:\n k-=int(str(a)[i])*x\n i+=1\nprint(str(a)[i-1])', 'a=input()\nk=int(input())\nx=5*math.pow(10,15)\ni=0\nwhile k>0:\n k-=int(str(a)[i])*x\n i+=1\nprint(str(a)[i-1])', "\na=input()\nk=int(input())\nif int(a)<10:\n a=str(int(a)*11)\ni=0\nx=0\n\nwhile k>=0:\n if a[i]=='1':\n k-=1\n i+=1\n if k==0:\n x=1\n k=-1\n else: \n k=-1\n\n\nprint(a[i-x])"] | ['Time Limit Exceeded', 'Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s013967817', 's186709791', 's477298318', 's875280088', 's815498783'] | [197056.0, 3060.0, 3060.0, 2940.0, 3064.0] | [2105.0, 17.0, 17.0, 17.0, 17.0] | [101, 166, 106, 111, 636] |
p03282 | u854144714 | 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())\nwhile True:\n if S[0]!="1":\n print(S[0])\n break\n else:\n continue', 'S=input()\nK=int(input())\nfor i in range(len(S)):\n if S[i]!=1:\n print(S[i])\n break\n else:\n continue', 'S=input()\nK=int(input())\nfor i in range(len(S)):\n if S[i]=="1":\n if i==K-1:\n print(1)\n break\n else:\n continue\n else:\n print(S[i])\n break'] | ['Time Limit Exceeded', 'Wrong Answer', 'Accepted'] | ['s621417749', 's884056866', 's300417455'] | [2940.0, 2940.0, 3060.0] | [2104.0, 18.0, 22.0] | [115, 125, 203] |
p03282 | u855985627 | 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\ncount_1=0\nwhile s[0]==1:\n count_1+=1\n s.remove[0]\nif count_1>=k:\n print('1')\nelse:\n print(s[0])", "s=input()\nS=''\nk=int(input())\nfor num in s:\n S+=num*(int(num)**5000000000000000)\nprint(S[k])", "s=input()\nk=int(input())\n\ncount_1=0\nwhile s!='':\n if s[0]=='1':\n count_1+=1\n s=s.lstrip('1')\nif count_1>=k:\n print('1')\nelse:\n print(s[0])\n", "s=input()\nk=int(input())\n\ncount_1=0\nwhile s!='':\n if s[0]=='1':\n count_1+=1\n s=s[1:]\n else:\n break\nif count_1>=k:\n print('1')\nelse:\n print(s[0])\n"] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s031787558', 's455783658', 's631340794', 's544034829'] | [2940.0, 197056.0, 2940.0, 2940.0] | [17.0, 2106.0, 2104.0, 17.0] | [125, 93, 148, 158] |
p03282 | u859897687 | 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(i) for i in input()]\nk=int(input())\na=5000000000000000\nfor i in s:\n k-=i**a\n if k<0:\n print(i)', 's=[int(i) for i in input()]\nk=int(input())\n#2^10^15=\na=0\nfor i in s:\n if i==1:\n a+=1\n else:\n break\nif k<=a:\n print(1)\nelse:\n print(s[a])'] | ['Wrong Answer', 'Accepted'] | ['s399726299', 's891213314'] | [197056.0, 2940.0] | [2106.0, 17.0] | [106, 146] |
p03282 | u860657719 | 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())\nn=1\nwhile n <= 5000e12:\n if len(S) > K:\n S = S[:K-1]\n str_new_number = ""\n for str_number in S:\n number = int(str_number)\n str_new_number = str_new_number + str_number*number\n S = str_new_number\n n += 1\nprint(S[K-1])\n', 'S = input()\nK = int(input())\n\nk = 1\nwhile k <= K:\n if S[k-1] != "1":\n print(S[k-1])\n break\n k += 1\nelse:\n print(1)\n'] | ['Time Limit Exceeded', 'Accepted'] | ['s797892095', 's434714130'] | [44232.0, 2940.0] | [2106.0, 18.0] | [282, 138] |
p03282 | u861141787 | 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\nans = "1"\nfor i in range(K):\n if S[i] != "1":\n ans = S[i]\n break\n\nprint(ans)\n '] | ['Wrong Answer', 'Accepted'] | ['s120845245', 's027823940'] | [2940.0, 2940.0] | [17.0, 17.0] | [81, 128] |
p03282 | u863397945 | 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 print(S[i])\n break\nprint("1")', 'S = input()\nK = int(input())\njudge = 1\n\nfor i in range(K):\n if S[i] != "1":\n print(S[i])\n judge = 0\n break\n\nif judge != 0:\n print("1")'] | ['Wrong Answer', 'Accepted'] | ['s748587623', 's268590801'] | [9160.0, 9152.0] | [27.0, 29.0] | [113, 161] |
p03282 | u864197622 | 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\na=0\nb=0\nfor i in range(len(S)):\n if int(S[i]) > 1:\n a=i\n b=S[i]\n break\n# print(a)\n# print(b)\nif K< a:\n print (S[a])\n \nelse:\n print (b)\n \nelse:\n print (b)', 'S = input()\nK = int(input())\n\nfor i in range(K):\n if S[i] != "1":\n print(S[i])\n break\nelse:\n print("1")'] | ['Runtime Error', 'Accepted'] | ['s182866505', 's044659492'] | [2940.0, 2940.0] | [17.0, 17.0] | [223, 123] |
p03282 | u867848444 | 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 i in s:\n count+=int(i)**(5*(10**10))\n if count>=k:\n print(i)\n break', "s=input()\nk=int(input())\ncount=0\nans=''\nfor i in range(k):\n if s[i]=='1':\n count+=1\n else:\n ans=s[i]\n break\nprint(1 if count==k else ans)"] | ['Time Limit Exceeded', 'Accepted'] | ['s808322867', 's047397885'] | [168344.0, 2940.0] | [2105.0, 18.0] | [124, 164] |
p03282 | u870518235 | 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\nref = 0\nfor i in range(len(S)):\n if S[i] == "1":\n ref += 1\n else:\n break\n\nans = ""\nif ref == 0:\n ans = S[0]\nelse:\n if ref >= K:\n ans = "1"\n else:\n ans = S[ref-1]\n\nprint(ans)', 'S = list(input())\nK = int(input())\n\nref = 0\nfor i in range(len(S)):\n if S[i] == "1":\n ref += 1\n else:\n break\n\nans = ""\nif ref == 0:\n ans = S[0]\nelse:\n if ref >= K:\n ans = "1"\n else:\n ans = S[ref]\n\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s940030447', 's202188378'] | [9184.0, 9172.0] | [26.0, 28.0] | [252, 250] |
p03282 | u870793476 | 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\na = 0\nb = 0\n\nfor i in range(len(S)):\n if S[i] = 1:\n a = i+1\n else:\n b = S[i]\n break\n\nif K <= a:\n print(1)\nelse:\n print(b)\n', 'S = input()\nK = int(input())\n\na = 0\nb = 0\n\nfor i in range(len(S)):\n if S[i] == "1":\n a = i+1\n else:\n b = S[i]\n break\n\nif K <= a:\n print(1)\nelse:\n print(b)\n'] | ['Runtime Error', 'Accepted'] | ['s900044291', 's602844502'] | [2940.0, 2940.0] | [17.0, 18.0] | [185, 188] |
p03282 | u883048396 | 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? | ['sS = input().rstrip()\n#iK = int(input())\nif sS[0] == "1":\n if iK !=1:\n print(sS[1])\n else:\n print(sS[0])\nelse:\n print(sS[0])\n', '\nsS = input().rstrip()\niK = int(input())\n\nfor i in range(iK):\n if sS[i] != "1" :\n print(sS[i])\n exit()\nprint(sS[i])\n'] | ['Runtime Error', 'Accepted'] | ['s432930718', 's603651264'] | [2940.0, 2940.0] | [24.0, 17.0] | [148, 133] |
p03282 | u890807039 | 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 c in range(len(s)):\n if s[c] == 1:\n k -= 1\n if k >= 0:\n print(1)\n break\n else:\n print(s[c])\n break', 's = input()\nk = int(input())\nfor c in range(len(s)):\n if s[c] == 1:\n k -= 1\n if k < 0:\n print(1)\n break\n else:\n print(s[c])\n break', 's = input()\nk = int(input())\nfor c in range(len(s)):\n if s[c] == 1:\n k -= 1\n if k > 0:\n print(1)\n break\n else:\n print(s[c])\n break', 's = input()\nk,counter,n = int(input()),0,0\nwhile counter < len(s):\n if s[counter] == "1":\n n += 1\n counter += 1\nif n == len(s):\n print(1)\nelse:\n counter = 0\n while counter < k:\n if s[counter] != "1":\n print(s[counter])\n break', 's = input()\nk = int(input())\nfor c in range(min(len(s),k)):\n if s[c] != "1":\n print(s[c])\n exit()\nprint(1)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Time Limit Exceeded', 'Accepted'] | ['s603162603', 's744079041', 's958416950', 's970068079', 's463602012'] | [3060.0, 2940.0, 2940.0, 3060.0, 2940.0] | [17.0, 17.0, 18.0, 2104.0, 17.0] | [187, 186, 186, 276, 123] |
p03282 | u891635666 | 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().strip()\nk = int(input())\n\ncount = 0\nfor i in range(k):\n c = s[i]\n if c != '1':\n print(c)\n else:\n count += 1\nelse:\n print('1')", "s = input().strip()\nk = int(input())\n\ncount = 0\nfor i in range(min(len(s), k)):\n c = s[i]\n if c != '1':\n print(c)\n break\n else:\n count += 1\nelse:\n print('1')"] | ['Runtime Error', 'Accepted'] | ['s129709864', 's816844173'] | [3060.0, 3060.0] | [17.0, 17.0] | [163, 190] |
p03282 | u896726004 | 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())\nflag = True\n\nif len(s)==1:\n print(int(s))\n exit()\n\nnew = ''\nfor i in len(k):\n if s[i]!='1':\n flag = False\n new += i\n\nif flag:\n print(1)\nelse:\n print(new[0])", "s = input()\nk = int(input())\nflag = False\n\nfor i in range(min(len(s), k)):\n if s[i] != '1':\n ans = int(s[i])\n flag = True\n break\n\nif flag:\n print(ans)\nelse:\n print(1)"] | ['Runtime Error', 'Accepted'] | ['s774758984', 's231945881'] | [3060.0, 3064.0] | [17.0, 17.0] | [214, 196] |
p03282 | u898421873 | 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? | ['"""\nC - To Infinity\n"""\ns_list = input()\nK = int(input())\n\nprint(s_list[0])', '"""\nC - To Infinity\n"""\ns_list = input()\nK = int(input())\n\n\nfor s in range(min(len(s_list), K)):\n if s_list[s] != "1":\n print(s_list[s])\n break\nelse:\n print(s_list[0])'] | ['Wrong Answer', 'Accepted'] | ['s649444744', 's504503229'] | [2940.0, 3060.0] | [17.0, 17.0] | [75, 187] |
p03282 | u906428167 | 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 <= len(s):\n print(s[k-1])\nelif s[:k] == '1'*k:\n print(1)\nelse:\n print(s.translate(str.maketrans('','','1'))[0])\n ", "s = input()\nk = int(input())\n\ni = 0\nwhile s[i] == '1' or i+1 < k:\n\ti += 1 \nprint(int(s[i])) ", "s = input()\nk = int(input())\n\na = 0\nfor i in range(k+1):\n if s[i] != '1':\n a = 1\n print(int(s[i]))\n break\nif a = 0:\n print(1)", "s = list(map(int,list(input())))\nk = int(input())\n\nfor j in range(1,5*(10**15)):\n a = ''\n for i in range(len(s)):\n a += str(s[i])*s[i]**j\n if len(a) >= k:\n print(a[k-1])\n break\n", "s = input()\nk = int(input())\n\ni = 0\n\nwhile s[i] == '1' and i < k-1:\n i += 1\n\nprint(int(s[i]))"] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s462085711', 's650898907', 's955241386', 's983093470', 's119629619'] | [3060.0, 2940.0, 2940.0, 3572.0, 2940.0] | [17.0, 17.0, 17.0, 2231.0, 17.0] | [152, 94, 136, 189, 94] |
p03282 | u910352296 | 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? | ['str = map(int,list(input()))\nfor i in str :\n if i != 1 :\n print (i)\n break\nprint(1)', 'str = map(int,list(input()))\nk = int (input())\ncounter = 1\nfor i in str :\n if i != 1 :\n ans = i\n break\n ans = 1\n if counter >= k:\n break\n counter += 1\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s013354331', 's169414172'] | [2940.0, 2940.0] | [17.0, 17.0] | [90, 176] |
p03282 | u920299620 | 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())\n\nfor i in range(len(s)):\n if(s[i]!=1):\n print(s[i])\n sys.exit(0)\n if(i+1==k):\n print(1)\n sys.exit()\nprint(1)', 'import sys\ns=input()\nk=input()\n\nfor i in range(len(s)):\n if(s[i]!=1):\n print(s[i])\n sys.exit(0)\nprint(1)', 'import sys\ns=input()\nk=int(input())\n\nfor i in range(len(s)):\n if(s[i]!="1"):\n print(s[i])\n sys.exit(0)\n if(i+1==k):\n print(1)\n sys.exit(0)\nprint(1)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s479537300', 's771927426', 's841766232'] | [2940.0, 2940.0, 3064.0] | [17.0, 17.0, 17.0] | [178, 121, 181] |
p03282 | u922416423 | 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\nans = 1\nindex = 0\nfor j in range(len(S)):\n if S[j] != '1':\n ans = int(s[j])\n index = j\n break\n\nif ans == 1:\n print(ans)\nelif index < K-1:\n print(ans)\nelse:\n print(1)\n ", "S = str(input())\nK = int(input())\n\nans = 1\nindex = 0\nfor j in range(len(S)):\n if S[j] != '1':\n ans = int(S[j])\n index = j\n break\n\nif ans == 1:\n print(ans)\nelif index > K-1:\n print(1)\nelse:\n print(ans)\n "] | ['Runtime Error', 'Accepted'] | ['s549876783', 's190894024'] | [2940.0, 3060.0] | [19.0, 18.0] | [242, 242] |
p03282 | u923077302 | 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? | ['main():\n S = input()\n K = int(input())\n output = solve(S, K)\n return output\n\ndef solve(S, K):\n for s in S:\n if s != "1":\n return s\n return "1"', 'def main():\n S = input()\n K = int(input())\n output = solve(S, K)\n return output\n\ndef solve(S, K):\n for s in S:\n if s != "1":\n return s\n return "1"\n\nif __name__ == "__main__":\n main()', 'def main():\n S = input()\n K = int(input())\n output = solve(S, K)\n print(output)\n return\n\ndef solve(S, K):\n if K <= len(S) and S[K-1] == "1":\n \n \n \n return "1"\n if K == 1:\n for s in S:\n if not s == "1":\n return s \n return S[0]\n\nif __name__ == "__main__":\n main()', 'main():\n S = input()\n K = int(input())\n output = solve(S, K)\n return output\n\ndef solve(S, K):\n for s in S:\n if s != "1":\n return s\n return "1"\n\nif __name__ == "__main__":\n main()', 'def main():\n S = input()\n K = int(input())\n output = solve(S, K)\n print(output)\n return\n\ndef solve(S, K):\n if K <= len(S) and S[K-1] == "1":\n \n \n \n return "1"\n if K == 1:\n for s in S:\n if not s == "1":\n return s \n return S[0]\n\nif __name__ == "__main__":\n main()', 'def main():\n S = input()\n K = int(input())\n output = solve(S, K)\n print(output)\n return\n\ndef solve(S, K):\n \n for i in range(K):\n if S[i] != "1":\n return S[i]\n return "1"\n\nif __name__ == "__main__":\n main()'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Accepted'] | ['s008868289', 's301084631', 's680767023', 's718582741', 's903754213', 's649798891'] | [2940.0, 2940.0, 3060.0, 2940.0, 3060.0, 2940.0] | [17.0, 17.0, 18.0, 17.0, 17.0, 17.0] | [178, 221, 591, 217, 591, 284] |
p03282 | u923279197 | 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 i!='1':\n print(i)\n exit()\nprint(s[k-1])", "s=input()\nk=int(input())\nfor i in range(max(k,len(s))):\n if i!='1':\n print(i)\n exit()\nprint(s[k-1])", 'S = [int(i) for i in input()]\nK = int(input())\nfor s in S:\n if s == 1:\n K -= 1\n if K <= 0:\n print(1)\n break\n else:\n print(s)\n break\n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s954066602', 's954756699', 's889346411'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [116, 116, 188] |
p03282 | u923712635 | 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())\nif(S[0]!='1'):\n print(S[0])\nelse:\n i=0\n while(S[i]=='1'):\n i += 1\n if(K<=i):\n print('1')\n else:\n print(S[i])", "S = input()\nK = int(input())\nif(S[0]!='1'):\n print(S[0])\nelse:\n i=0\n while(S[i]=='1'):\n i += 1\n if(i==len(S)):\n break\n if(K<=i):\n print('1')\n else:\n print(S[i])"] | ['Runtime Error', 'Accepted'] | ['s519440624', 's423054525'] | [3060.0, 3060.0] | [18.0, 17.0] | [156, 182] |
p03282 | u924406834 | 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())\nnumbers = (int(x) for x in str(S))\nfor i in range(5*(10**15)):\n times = 0\n temporary = []\n while times <= K:\n for a in numbers:\n times += a\n for j in renge(a):\n temporary.append(a)\n numbers = temporary\nprint(numbers[K-1])', "s = input()\nk = int(input())\nif s[0] != '1':\n print(s[0])\nelse:\n ans = 1\n for i in range(k):\n if s[i] != '1':\n ans = s[i]\n break\n print(ans)"] | ['Runtime Error', 'Accepted'] | ['s409823578', 's834658699'] | [3060.0, 2940.0] | [17.0, 17.0] | [311, 181] |
p03282 | u928784113 | 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())\nL = [int(i) for i in S]\ncount = 0\nfor i in L:\n if i != 1:\n if count >= K:\n print(i)\n else:\n print(1)\n exit()\n else:\n count += 1\nprint(1)', 'S = input()\nK = int(input())\nL = [int(i) for i in S]\ncount = 1\nfor i in L:\n if i != 1:\n if count <= K:\n print(i)\n else:\n print(1)\n exit()\n else:\n count += 1\nprint(1)'] | ['Wrong Answer', 'Accepted'] | ['s437591543', 's854766266'] | [2940.0, 2940.0] | [18.0, 18.0] | [221, 221] |
p03282 | u931462344 | 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())\ncnt = 0\nans = False\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n ans = True\nif ans:\n print(1)", "s = input()\nk = int(input())\nans = False\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n ans = True\n break\nif ans:\n print(1)", "s = input()\nk = int(input())\nans = True\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n ans = False\n break\nif ans:\n print(1)"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s782802445', 's847209565', 's845015179'] | [3060.0, 2940.0, 2940.0] | [19.0, 17.0, 18.0] | [149, 154, 154] |
p03282 | u932465688 | 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())\nt = 0\nif S[0] != 1:\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i] != 1:\n t = i\n break\n if t <= K:\n print(1)\n else:\n print(S[t])', 'S = input()\nK = int(input())\nt = 0\nif S[0] != 1:\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i] != 1:\n t = i\n break\n if t+1 > K:\n print(1)\n else:\n print(S[t])', 'S = input()\nK = int(input())\nt = 0\nif S[0] != 1:\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i] != 1:\n t = i\n break\n if t < K:\n print(1)\n else:\n print(S[t])', "S = input()\nK = int(input())\nt = 0\nif S[0] != '1':\n print(S[0])\nelse:\n for i in range(len(S)):\n if S[i] != '1':\n t = i\n break\n if t+1 > K:\n print(1)\n else:\n print(S[t])"] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s399560919', 's515108925', 's866745995', 's603814722'] | [3060.0, 3060.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0, 18.0] | [186, 187, 185, 191] |
p03282 | u932868243 | 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())\ncnt=0\nfor ss in s:\n if ss==1:\n cnt+=1\n else:\n break\nif cnt==0:\n print(s[0])\n exit()\nif k<=cnt:\n print(1)\nelse:\n print(s[cnt])', 's=input()\nk=int(input())\nif s[0]==1:\n print(s[1])\nelse:\n print(s[0])', "s=input()\nk=int(input())\ncnt=0\nfor ss in s:\n if ss=='1':\n cnt+=1\n else:\n break\nif cnt==0:\n print(s[0])\n exit()\nif k<=cnt:\n print(1)\nelse:\n print(s[cnt])"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s173813943', 's979396387', 's772411147'] | [3060.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [162, 70, 164] |
p03282 | u933622697 | 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()\nS_list = list(map(int, S.split()))\nK = int(input())\n\nS_set_under_K = set(S[:K])\n\nif len(S_set_under_K) == 1 and S_list[0] == 1:\n print(1)\nelse:\n print(S_list[1])\n', 'S = input()\nS_list = list(map(int, list(S)))\nK = int(input())\n\nS_set_under_K = set(S[:K])\n\nif len(S) == 1:\n print(S)\nelif len(S_set_under_K) == 1 and S_list[0] == 1:\n print(1)\nelse:\n for c in S:\n if c != "1":\n print(c)\n exit()\n'] | ['Runtime Error', 'Accepted'] | ['s272308200', 's786913201'] | [2940.0, 3060.0] | [17.0, 17.0] | [180, 265] |
p03282 | u934788990 | 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(k):\n if s[i] !=1:\n print(int(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 break\n else:\n print(1)\n break', 'S = input()\nK = int(input())\n\nfor i in range(K):\n if int(S[i]) != 1:\n print(int(S[i]))\n exit()\nprint(1)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s546896144', 's694818332', 's301150884'] | [2940.0, 2940.0, 2940.0] | [18.0, 18.0, 17.0] | [113, 139, 120] |
p03282 | u940102677 | 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()\na = "1"\nfor i in range(int(input())):\n if s[i] != 1:\n a = s[i]\n break\nprint(a)', 's = input()\na = "1"\nfor i in range(int(input())):\n if s[i] != "1":\n a = s[i]\n break\nprint(a)'] | ['Wrong Answer', 'Accepted'] | ['s914385459', 's589851885'] | [2940.0, 2940.0] | [17.0, 17.0] | [97, 99] |
p03282 | u941438707 | 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 ans+=int(s[i])**500000000000\n if ans>=k:\n print(s[i])\n break', 's=input()\nk=int(input())\nfor i in range(k):\n if s[i]!="1":\n print(s[i])\n break\n if i==k-1:\n print(1)'] | ['Time Limit Exceeded', 'Accepted'] | ['s458787674', 's962573039'] | [161972.0, 2940.0] | [2108.0, 17.0] | [136, 127] |
p03282 | u944209426 | 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())\nt=0\na=1\nfor i in range(len(s)):\n if s[i]!='1':\n t=i+1\n a=s[i]\nif k<t:\n print(s[k-1])\nelse:\n print(a)", "s=input()\nk=int(input())\nt=0\na=1\nfor i in range(len(s)):\n if s[i]!='1':\n t=i+1\n a=s[i]\n break\nif k<t:\n print(s[k-1])\nelse:\n print(a)"] | ['Wrong Answer', 'Accepted'] | ['s851002223', 's143545054'] | [3060.0, 3060.0] | [17.0, 22.0] | [148, 162] |
p03282 | u944325914 | 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,tuple(input())))\nk=int(input())\ni=0\nif 1 in s:\n i=s.index()\n if i+1>=k:\n print(1)\n else:\n print(s[i+1])\nelse:\n print(s[0])', 's=list(map(int,tuple(input())))\nk=int(input())\ni=0\nif 1 in s:\n for i in range(len(s)):\n if s[i]!=1:\n break\n if i>=k:\n print(1)\n else:\n print(s[i])\nelse:\n print(s[0])'] | ['Runtime Error', 'Accepted'] | ['s260156714', 's320620120'] | [9092.0, 9144.0] | [28.0, 33.0] | [151, 187] |
p03282 | u952022797 | 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 -*-\nimport sys\nimport copy\nimport collections\nfrom bisect import bisect_left\nfrom bisect import bisect_right\nfrom collections import defaultdict\nfrom heapq import heappop, heappush\nimport numpy as np\n\t\ndef main():\n\tS = input()\n\tS = list(S)\n\tK = int(input())\n\t\n\tif len(S) == 1:\n\t\tprint(S[0])\n\t\n\t\n\ttmp = []\n\tfor i in S:\n\t\tflg = False\n\t\tif int(i) == 1:\n\t\t\ttmp.append(1)\n\t\telse:\t\n\t\t\ttmp.append(int(i))\n\t\t\tif flg = True:\n\t\t\t\tbreak\n\t\t\tflg = True\n\t\n\ta = np.random.randint(100)\n\t\n\tif len(tmp) < K:\n\t\tif a == 0\n\t\t\tprint(tmp[-1])\n\t\telse:\n\t\t\tprint(tmp[-2])\n\telse:\n\t\ttmp[K-1]\n\t\nif __name__ == "__main__":\n\tmain()\n', '# -*- coding: utf-8 -*-\nimport sys\nimport copy\nimport collections\nfrom bisect import bisect_left\nfrom bisect import bisect_right\nfrom collections import defaultdict\nfrom heapq import heappop, heappush\nimport numpy as np\n\t\ndef main():\n\tS = input()\n\tS = list(S)\n\tK = int(input())\n\t\n\tif len(S) == 1:\n\t\tprint(S[0])\n\t\tsys.exit()\n\t\t\n\ttmp = -1\n\tfor i, val in enumerate(S):\n\t\tif int(val) != 1:\n\t\t\ttmp = int(i)+1\n\t\t\tbreak\n\t\n\tif K >= tmp:\n\t\tprint(S[tmp-1])\n\telse:\n\t\tprint(1)\n\t\nif __name__ == "__main__":\n\tmain()\n'] | ['Runtime Error', 'Accepted'] | ['s544313469', 's697464049'] | [3060.0, 12176.0] | [17.0, 153.0] | [1141, 502] |
p03282 | u957872856 | 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, s in enumerate(S):\n if s == "1" and i+1==k:\n print(s)\n break\n else:\n print(s)\n break\n', 'S = input()\nk = int(input())\nfor s in S:\n if s == "1" and k <= 5000000000000000:\n print(1)\n break\n else:\n print(s)\n break', 'S = input()\nk = int(input())\nd = {}\ncnt = 0\ncnt1 = 0\nfor i, s in enumerate(S):\n if s == "1" and k <= 5000000000000000:\n print(1)\n break\n elif s != "1":\n print(s)\n break\n else:\n print(S[i+1])', 'S = input()\nk = int(input())\nfor i, s in enumerate(S):\n if s == "1" and i+1==k:\n print(s)\n break()\n else:\n print(s)\n break\n', 'S = input()\nk = int(input())\ncnt = 0\nfor i, s in enumerate(S):\n if s == "1":\n if i+1 == k:\n print(s)\n exit()\n else:\n print(s)\n exit()\n'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Runtime Error', 'Accepted'] | ['s291691747', 's719258333', 's838394338', 's925174266', 's805093488'] | [2940.0, 2940.0, 3060.0, 2940.0, 2940.0] | [17.0, 19.0, 17.0, 17.0, 17.0] | [135, 135, 208, 137, 155] |
p03282 | u961674365 | 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 x in s:\n if x!=1:\n print(x)\n exit()', 's=input()\nk=int(input())\nfor x in s:\n if x==1:\n k-=1\n if k==0:\n print(1)\n else:\n print(x)\n exit()', "s=input()\nk=int(input())\nfor x in s:\n if x=='1':\n k-=1\n if k==0:\n print(1)\n exit()\n else:\n print(x)\n exit()"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s812200197', 's942730042', 's729070051'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [71, 116, 131] |
p03282 | u962197874 | 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())\nl = 0\nfor i in S:\n if i == '1':\n l+=1\n \tif l >= K:\n print(1)\n break\n else:\n print(int(i))\n break\n ", "S = input()\nK = int(input())\nl = 0\nfor i in S:\n if i == '1':\n l+=1\n \tif l >= K:\n print(1)\n break\n else:\n print(i)\n break\n ", "S = input()\nK = int(input())\nl = 0\nfor i in S:\n if i == '1':\n l+=1\n if l >= K:\n break\n else:\n break\nprint(i)"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s809310120', 's951392811', 's608905854'] | [2940.0, 2940.0, 2940.0] | [17.0, 18.0, 18.0] | [152, 147, 124] |
p03282 | u962330718 | 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()\nS=[0]*len(s)\nfor i in range(len(s)):\n S[i]=int(s[i])\nk=int(input())\nif len(s)>=k:\n if sum(S[:k])==k:\n print(1)\nelse:\n for i in range(len(S)):\n if S[i]!=1:\n print(S[i])\n break', 's=input()\nS=[0]*len(s)\nfor i in range(len(s)):\n S[i]=int(s[i])\nk=int(input())\nans=0\nif sum(S[:k])==k:\n print(1)\nelse:\n for i in range(len(s)):\n if s[i]!=1:\n print(s[i])\n break', 's=input()\nk=int(input())\nans=0\nfor i in range(len(s)):\n if s[i]==1:\n ans+=1\n if ans==k:\n print(1)\n break\n else:\n print(s[i])\n break', 's=input()\nS=[0]*len(s)\nfor i in range(len(s)):\n S[i]=int(s[i])\nk=int(input())\nans=0\nif len(s)>=k:\n if sum(S[:k])==k:\n print(1)\n ans=1\nif ans!=1:\n for i in range(len(S)):\n if S[i]!=1:\n print(S[i])\n break'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s415602792', 's430446552', 's587947862', 's456214434'] | [3060.0, 3060.0, 2940.0, 3064.0] | [17.0, 17.0, 17.0, 18.0] | [203, 191, 155, 224] |
p03282 | u962423738 | 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\tprint(int(s[1])\nelse:\n\tprint(int(s[0]))', 's=input()\nk=int(input())\ncnt=0\nif s[0]!="1":\n\tprint(int(s[0]))\nelse:\n\tfor i in s:\n\t\tcnt+=1\n\t\tif i!="1":\n\t\t\tif cnt<=k:\n\t\t\t\tprint(int(i))\n\t\t\telse:\n\t\t\t\tprint(1)', 's=input()\nk=int(input())\nl=0\nfor i in s:\n if i=="1":\n l+=1\n else:\n break\nprint("1" if k<=l else s[l])'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s422336397', 's572406365', 's913050120'] | [8876.0, 9076.0, 9052.0] | [25.0, 26.0, 30.0] | [79, 157, 121] |
p03282 | u964299793 | 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())\na=list(map(int,input().split()))\nb=list(map(int,input().split()))\nc=list(map(int,input().split()))\na.sort()\nb.sort()\nc.sort()\nn=len(a)\nj=0\nk=0\nfor i in range(n):\n while j<n and b[j]<=a[i]:\n j+=1\n while k<n and c[k]<=b[j]:\n k+=1\n\n', "s=input()\nk=int(input())\nans=-1\nflag=True \nfor i in range(min(len(s),k)):\n if s[i]!='1':\n flag=False\n ans=int(s[i])\n break\nif flag:\n print(1)\nelse:\n print(ans)\n"] | ['Runtime Error', 'Accepted'] | ['s145057683', 's854058584'] | [3064.0, 2940.0] | [18.0, 18.0] | [264, 193] |
p03282 | u968404618 | 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 s in S[:K+1]:\n if s != "1": \n print(s)\n break\n else:\n print(1)\n ', 'S = input()\nK = int(input())\n\nidx = min(K, len(S))\nfor s in S[:idx]:\n if s != "1": \n print(s)\n break\nelse:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s614227575', 's890155609'] | [9080.0, 9092.0] | [28.0, 28.0] | [122, 130] |
p03282 | u969708690 | 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(len(S)):\n if i==K-1:\n print(S[i])\n exit()\n if S[i]!="1":\n print(S[i])\n exit()\nprint(1)'] | ['Wrong Answer', 'Accepted'] | ['s051298884', 's715676175'] | [9100.0, 9108.0] | [30.0, 29.0] | [98, 140] |
p03282 | u970899068 | 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(len(s)):\n if s[i]!='1':\n if k>=i:\n print('1')\n exit()\n else:\n print(s[i])\n exit()\nprint('1')", "s=input()\nk=int(input())\n\nfor i in range(len(s)):\n if s[i]!='1':\n if k<=i:\n print('1')\n exit()\n else:\n print(s[i])\n exit()\nprint('1')"] | ['Wrong Answer', 'Accepted'] | ['s618056425', 's934132697'] | [2940.0, 2940.0] | [17.0, 17.0] | [194, 194] |
p03282 | u977389981 | 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())\none_count = 0\n\nfor i in range(len(S)):\n if S[i] != '1':\n print(S[i])\n break\n else:\n one_count += 1\n\nif one_count >= K:\n print('1')\nelse:\n print('0')", "S = input()\nk = int(input())\n\nans = '1'\nfor i in range(len(S)):\n if i + 1 <= k and S[i] != '1':\n ans = S[i]\n break\n \nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s208254014', 's000611707'] | [2940.0, 2940.0] | [17.0, 18.0] | [210, 151] |
p03282 | u977642052 | 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 main(s, k):\n for s in s:\n if s != \'1\':\n print(s)\n return\n\n\nif __name__ == "__main__":\n s = int(input())\n k = int(input())\n\n main(s, k)\n', 'def main(s: str, k: int):\n for i in range(k):\n if s[i] != "1":\n print(s[i])\n return\n\n print(1)\n\n\nif __name__ == "__main__":\n s = input()\n k = int(input())\n\n main(s, k)\n'] | ['Runtime Error', 'Accepted'] | ['s200154061', 's227995451'] | [2940.0, 2940.0] | [17.0, 17.0] | [180, 212] |
p03282 | u977661421 | 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 -*-\ns = input()\nk = int(input())\n\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n break\nprint(1)\n", "# -*- coding: utf-8 -*-\ns = input()\nk = int(input())\n\nfor i in range(k):\n if s[i] != '1':\n print(s[i])\n exit()\nprint(1)\n"] | ['Wrong Answer', 'Accepted'] | ['s684336501', 's680179163'] | [2940.0, 2940.0] | [17.0, 17.0] | [136, 137] |
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