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 | u439312138 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n\nfor i in S:\n if i != 1:\n print(i)\n exit()\n \nprint(S[0])', 'S = input()\nK = int(input())\n\nc = "1"\n\nfor i in range(K):\n if S[i] != "1":\n c = S[i]\n break\nprint(c)'] | ['Wrong Answer', 'Accepted'] | ['s238386838', 's188218140'] | [2940.0, 2940.0] | [17.0, 17.0] | [95, 107] |
p03282 | u453068434 | 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? | ['l=list(input())\nk=int(input())\nm=[0]*len(l)\nfor i in range(len(l)):\n var=pow(int(l[i]),5*pow(10,15))\n if(i==0):\n m[i]+=var\n if(i>=1):\n m[i]+=(m[i-1]+var)\n if(k<=m[i]):\n print(l[i])\n exit()', 'l=list(input())\nk=int(input())\nm=[0]*len(l)\nfor i in range(len(l)):\n var=pow(int(l[i]),100)\n if(i==0):\n m[i]+=var\n if(i>=1):\n m[i]+=(m[i-1]+var)\n if(k<=m[i]):\n print(l[i])\n exit()'] | ['Time Limit Exceeded', 'Accepted'] | ['s035021906', 's659011909'] | [197056.0, 3064.0] | [2106.0, 17.0] | [228, 219] |
p03282 | u453642820 | 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())\nk=1\nfor s in S:\n if s!=1:\n print(s)\n exit()\n if k==K:\n print(s)\n exit()\n k+=1', 'S=input()\nk=int(input())\nfor s in S:\n if s!=1:\n print(s)\n exit()', 'S=input()\nK=int(input())\nfor s in S:\n if s!=1:\n print(s)\n break', 'S=input()\nK=int(input())\nfor s in S:\n if s!=1:\n print(s)\n break\nelse:\n print(1)', 'S=input()\nK=int(input())\nk=0\nfor s in S:\n k+=1\n if k==K:\n print(s)\n exit()\n if s!="1":\n print(s)\n break\nelse:\n print(1)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s385506373', 's589470767', 's636784622', 's918069582', 's983026178'] | [2940.0, 2940.0, 2940.0, 2940.0, 3060.0] | [17.0, 17.0, 18.0, 17.0, 18.0] | [139, 81, 80, 99, 159] |
p03282 | u455317716 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n\nfor i in S[K+1]:\n if i != "1":\n print(i)\n break\nelse:\n print(i)', 'S = input()\nK = int(input())\n\ncount = 0\nwhile count < K:\n if S[count] != "1":\n print(S[count])\n break\n count += 1\nelse:\n print("1")'] | ['Runtime Error', 'Accepted'] | ['s884744496', 's318270490'] | [2940.0, 2940.0] | [19.0, 17.0] | [113, 154] |
p03282 | u464205401 | 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\nif S[0]=='1':\n if k==1:\n print(S[0]) \n elif k<(10**14+1):\n print(S[2])\n else:\n print(S[1])\n\nelif S[0]=='2':\n if k<(10**14+1):\n print(S[0])\n else:\n if S[1]!=1:\n print(S[1])\n else:\n print(S[2])\n \nelse:\n print(S[0])", "import re\nS=str(input())\nk=int(input())\n\nif set(S)=={'1'}:\n print(1)\n \nelse:\n x=re.search(r'[2-9]',S).start()\n \n if S[0]=='1':\n if k<=x:\n print(S[0])\n else:\n print(S[x]) \n else:\n print(S[0])\n"] | ['Wrong Answer', 'Accepted'] | ['s227514038', 's221076883'] | [3064.0, 3188.0] | [17.0, 19.0] | [326, 224] |
p03282 | u464912173 | 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? | ['b=list(input())\na=int(input())\nif b[0]=="1":\n print(1)\nelse:\n print(b[1])\n', 's=input()\nk=int(input())\nfor i in range(k):\n if s[i]!="1":\n break\nprint(s[i])'] | ['Runtime Error', 'Accepted'] | ['s851087394', 's509765630'] | [2940.0, 2940.0] | [17.0, 17.0] | [76, 81] |
p03282 | u466331465 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['N = list(input())\nK = int(input())\nfor i in range(len(K)) :\n a = int(N[i])\n if a!=1:\n print(a)\n exit()\nprint(1)', 'N = list(input())\nK = int(input())\nfor i in range(len(K)) :\n a = int(i)\n if a!=1:\n print(a)\n exit()\nprint(1)', 'N = list(input())\nK = int(input())\nfor i in N :\n a = int(i)\n ans += a**(5*10**15)\n if ans>=K:\n print(a)', 'N = list(input())\nK = int(input())\nfor i in range(K) :\n a = int(N[i])\n if a!=1:\n print(a)\n exit()\nprint(1)'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s229976991', 's573338431', 's901535031', 's509315209'] | [2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0] | [119, 116, 109, 114] |
p03282 | u469281291 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["import sys\nimport math\n\ns=input()\nk = int(input())\ncnt = False\nans=0\nfor i in range(len(s),k):\n if s[i] != '1':\n ans = s[i]\n cnt = True\n break\n cnt+=1\nif (not cnt):\n print(1)\nelse:\n print(ans)\n", "import sys\nimport math\n\ns=input()\nk = int(input())\ncnt = False\nans=0\nfor i in range(0,k):\n if s[i] != '1':\n ans = s[i]\n cnt = True\n break\nif (not cnt):\n print(1)\nelse:\n print(ans)\n"] | ['Runtime Error', 'Accepted'] | ['s804650695', 's226292528'] | [3060.0, 3060.0] | [18.0, 18.0] | [226, 210] |
p03282 | u469953228 | 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(s[i])\n exit()\nprint(1)', 's = input()\n\nfor i in range(len(s)):\n if s[i] != 1:\n print(s[i])\n exit()\nprint(1)', 's = 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', 'Wrong Answer', 'Accepted'] | ['s069152613', 's908938265', 's843326409'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [98, 88, 105] |
p03282 | u473633103 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['# coding: utf-8\n# Your code here!\n\ns = input()\nr = int(input())\n\nif r <= len(s):\n print(s[r-1])\nelse:\n for i in s:\n if(i!="1"):\n print(i)\n break', '# coding: utf-8\n# Your code here!\n\ns = list(input())\nk = int(input())\n\nfor i in range(k):\n if s[i] == "1":\n ans = 1\n else:\n ans = s[i]\n break\n \nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s868251775', 's011589218'] | [2940.0, 2940.0] | [23.0, 18.0] | [179, 184] |
p03282 | u474423089 | 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())\ntmp = 0\nfor i in S:\n if i=='1':\n tmp += 1\n else:\n break\nif K >= tmp:\n print(1)\nelse:\n print(S[tmp])", "S=input()\nK=int(input())\ntmp = 0\nfor i in S:\n if i=='1':\n tmp += 1\n else:\n break\nif K <= tmp:\n print(1)\nelse:\n print(S[tmp])"] | ['Runtime Error', 'Accepted'] | ['s110455797', 's032567220'] | [2940.0, 2940.0] | [18.0, 17.0] | [150, 150] |
p03282 | u476435125 | 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? | ['l=list(map(int,input().split()))\nn=l[0]\nm=l[1]\nq=l[2]\nlr=[]\npq=[]\nfor i in range(m):\n lr.append(input())\nfor i in range(q):\n pq.append(input())\nw=0 \nfor i in range(q):\n ppqq=list(map(int,pq[i].split()))\n w=0\n for j in range(m):\n llrr=list(map(int,lr[j].split()))\n if llrr[0]>= ppqq[0] and llrr[1]<=ppqq[1]:\n w+=1\n print(w)\n', 'ss=input()\nk=int(input())\ns=[int(i) for i in ss ]\n#ls=len(s)\nj=0\nq=0\nw=0\nwhile q==0:\n if s[j]==1:\n w+=1\n if k<=w:\n print(1)\n q=1\n j+=1\n else:\n print(s[j])\n q=1'] | ['Runtime Error', 'Accepted'] | ['s967626812', 's754346110'] | [3064.0, 3064.0] | [17.0, 17.0] | [369, 222] |
p03282 | u478266845 | 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\n\nind = 0\nnum = 1\n\n\nfor i in range(len(s)):\n ind = i+1\n if s[i] != 1:\n num = int(s[i])\n break\n \nif k>= ind:\n ans = num\nelse:\n ans = 1\n \nprint(ans)', 's = str(input())\nk = int(input())\n\n\nind = 0\nnum = 1\n\n\nfor i in range(len(s)):\n ind = i+1\n if s[i] != 1:\n num = int(s[i])\n break\n \nif k>= ind:\n ans = num\nelse:\n ans = 1\n \nprint(ans)', "k = int(input())\ns = str(input())\n\n\n\n\nnum = 1\n\n\nfor i in range(len(s)):\n ind = i+1\n if s[i] != '1':\n num = int(s[i])\n break\n \nif k>= ind:\n ans = num\nelse:\n ans = 1\n \nprint(ans)", "s = str(input())\nk = int(input())\n\n\n\nnum = 1\n\n\nfor i in range(len(s)):\n ind = i+1\n if s[i] != '1':\n num = int(s[i])\n break\n \nif k>= ind:\n ans = num\nelse:\n ans = 1\n \nprint(ans)"] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s130215568', 's218752227', 's469249851', 's482955231'] | [3060.0, 3060.0, 3316.0, 3060.0] | [17.0, 17.0, 21.0, 17.0] | [216, 216, 212, 211] |
p03282 | u488934106 | 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 To_Infinity(s , k):\n\n ans = 0\n\n if s[0] == 1:\n if k == 1:\n ans = 1\n else:\n ans = s[1]\n else:\n ans = s[0]\n\n return ans\ndef main():\n s = list(int(input()))\n k = int(input())\n print(To_Infinity(s , k))\n\nif __name__ == '__main__':\n main()", 'def To_Infinity(s , k):\n\n ans = ""\n\n s1 = s.lstrip("1")\n\n if k - (len(s) - len(s1)) > 0:\n ans = s1[0]\n else:\n ans = "1"\n\n return ans\ndef main():\n s = str(input())\n k = int(input())\n print(To_Infinity(s , k))\n\nif __name__ == \'__main__\':\n main()'] | ['Runtime Error', 'Accepted'] | ['s777306993', 's518717447'] | [3060.0, 3060.0] | [17.0, 17.0] | [305, 284] |
p03282 | u492030100 | 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,K=map(int,input().split())\nprint(1 if N%K != 0 else 0)', 'S=list(input())\nK=int(input())\n#print(S)\nfor i in range(K):\n n=int(S[i])\n if n==1:\n if K-1==i:\n print(1)\n exit(0)\n else:\n print(n)\n exit(0)'] | ['Runtime Error', 'Accepted'] | ['s446781673', 's792158983'] | [2940.0, 2940.0] | [18.0, 18.0] | [56, 191] |
p03282 | u494058663 | 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 i>=len(S):\n print('1')\n exit()\n if S[i]!='1':\n print(S[i])\n exit()\n else:\n print('1')\n exit()\n \n ", "S = list(input())\nK = int(input())\n\nfor i in range(min(len(S),K)):\n if S[i]!='1':\n print(S[i])\n exit()\nprint('1')\n \n \n \n"] | ['Wrong Answer', 'Accepted'] | ['s825987113', 's005434293'] | [2940.0, 2940.0] | [17.0, 17.0] | [213, 146] |
p03282 | u497625442 | 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 K == 1:\n\tprint(S[0])\n\texit()\ni = 0\nwhile S[i] == "1":\n\ti += 1\nprint(S[i])\n', 'S = input()\nK = int(input())\n\nif K == 1:\n\tprint(S[0])\nelse:\n\ti = 0\n\tK -= 1\n\twhile S[i] == "1" and K > 0:\n\t\ti += 1\n\t\tK -= 1\n\tprint((i,S[i]))\n', 'S = input()\nK = int(input())\nif K == 1:\n\tprint(S[0])\n\texit()\nelse:\n\ti = 0\n\tK -= 1\n\twhile S[i] == "1" and K > 0:\n\t\ti += 1\n\t\tK -= 1\n\tprint(S[i])\n\n'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s160139579', 's726523333', 's589968682'] | [2940.0, 3064.0, 2940.0] | [22.0, 18.0, 18.0] | [111, 140, 144] |
p03282 | u498486375 | 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())\na=[]\nwhile S !=0:\n s=S%10\n a.append(s)\n S //=10\nm=0\nfor i in range(1,len(a)+1):\n m += a[-i]*5000000000000000\n if m>=K:\n ans=i\n break\nprint(a[-ans])', 'S=int(input())\nK=int(input())\n\na=[]\nwhile S !=0:\n s=S%10\n a.append(s)\n S //=10\nm=0\n\nfor i in range(1,len(a)+1):\n if a[-i]!=1:\n if i>K:\n ans=1\n else:\n ans=i\n break\n else:\n ans=1\n\nprint(a[-ans])'] | ['Runtime Error', 'Accepted'] | ['s596625052', 's810865409'] | [3064.0, 3064.0] | [17.0, 17.0] | [188, 219] |
p03282 | u498620941 | 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())\nprint(S)\nsize = len(S)\nK = int(input())\nfor i in range(size):\n if S[i] == "1" and K == 1 :\n print("1")\n exit()\n else :\n if S[i] == "1":\n K -= 1\n else : \n print(S[i])\n exit()', 'S = list(input())\nsize = len(S)\nK = int(input())\nfor i in range(size):\n if S[i] == "1" and K == 1 :\n print("1")\n exit()\n else :\n if S[i] == "1":\n K -= 1\n else : \n print(S[i])\n exit()'] | ['Wrong Answer', 'Accepted'] | ['s146462289', 's593321048'] | [3060.0, 3060.0] | [17.0, 17.0] | [258, 249] |
p03282 | u500990280 | 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,K = map(int,input().split())\nlistn = list(str(N))\na = []\nif len(str(N)) < K:\n for i in range(len(str(N))):\n if listn[i] != str(1):\n a.append(str(listn[i]))\nelse:\n for i in range(K):\n if listn[i] != str(1):\n a.append(str(listn[i]))\nif a == []:\n print(1)\nelse:\n print(a[0])\n', 'N,K = map(int,input().split())\nlistn = list(str(N))\nfor i in range(K):\n if listn[i] != str(1):\n a.append(str(listn[i]))\nif a == []:\n print(1)\nelse:\n print(a[0])', 'N, K = [int(input()) for _ in range(2)]\nlistn = list(str(N))\na = []\nif len(str(N)) < K:\n for i in range(len(str(N))):\n if listn[i] != str(1):\n a.append(str(listn[i]))\nelse:\n for i in range(K):\n if listn[i] != str(1):\n a.append(str(listn[i]))\nif a == []:\n print(1)\nelse:\n print(a[0])'] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s275916720', 's806419919', 's935113799'] | [3060.0, 3060.0, 3064.0] | [18.0, 18.0, 17.0] | [322, 176, 330] |
p03282 | u503111914 | 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())\nresult = [0,0]\nfor i in len(s):\n if S[i] != 1:\n result[0] = i+1\n result[1] = S[i]\nprint(1 if K<result[0] else result[1] )\n', 'import sys\nS = input()\nK = int(input())\nfor i in S:\n if i != 1:\n print(i)\n sys.exit()\nprint(1)', 'import sys\nS = input()\nK = int(input())\nresult = [0,""]\nfor i in S:\n if i != "1":\n result[1] = i\n break\n else:\n result[0] += 1\nprint(1 if K<=result[0] else result[1] )\n'] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s033053787', 's314553687', 's660263417'] | [2940.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0] | [169, 101, 179] |
p03282 | u505420467 | 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()\nli=''.join(S)\nS1=len(li)\nm=[]\n\nK=int(input())\nans=str(K)\nk1=''.join(ans)\nk2=len(k1)\nprint(k2)\nfor i in range(0,S1):\n if li[i]!='1' and (i+1)<K:\n if li[i]==2 and k2>=1505000000000001+i:\n pass\n else:\n print(li[i])\n sys.exit()\n\nprint('1')\n", "import sys\nS=input()\nli=''.join(S)\nS1=len(li)\nm=[]\n\nK=int(input())\nans=str(K)\nk1=''.join(ans)\nk2=len(k1)\n\nfor i in range(0,S1):\n if li[i]!='1' and i<K:\n print(li[i])\n sys.exit()\n\nprint('1')\n"] | ['Wrong Answer', 'Accepted'] | ['s633286208', 's467833238'] | [3064.0, 3060.0] | [17.0, 17.0] | [307, 207] |
p03282 | u506910932 | 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()\n\nif s[0] == 1:\n print(1)\nelse:\n print(s[0])', '# coding: utf-8\ns = input()\nk = int(input())\n\n\nif s[0] == 1:\n for i in range(k):\n if s[i] != 1:\n print(s[i])\n exit()\n \n print(1)\nelse:\n print(s[0])', "# 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()\n\nprint(1)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s011454297', 's784264690', 's385816868'] | [2940.0, 3060.0, 2940.0] | [18.0, 17.0, 17.0] | [77, 192, 129] |
p03282 | u514401521 | 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 = input()\n\nfor i in range(len(s)):\n if s[i] != "1" and i + 1 < k:\n print(s[i])\n sys.exit(0)\nprint(1)\n', 'import sys\ns = input()\nk = int(input())\n\nfor i in range(k):\n if s[i] != "1" :\n print(s[i])\n sys.exit(0)\nprint(1)\n'] | ['Runtime Error', 'Accepted'] | ['s909222040', 's727447888'] | [2940.0, 2940.0] | [17.0, 17.0] | [143, 130] |
p03282 | u515052479 | 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\n\nprint(math.log10(2**5000000000000000))', "s = list(str(input()))\nk = int(input())\nans = '1'\n\nfor i in range(k):\n if s[i] != '1':\n ans = s[i]\n break\n if len(s)-1 == i:\n break\n \nprint(int(ans))"] | ['Time Limit Exceeded', 'Accepted'] | ['s549534606', 's340930027'] | [196996.0, 2940.0] | [2106.0, 17.0] | [51, 163] |
p03282 | u518042385 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['n=list(input())\nm=int(input())\ni=0\nfor t in range(len(n)):\n if n[t]!="1":\n i=t+1\n break\nif m<i:\n print(1)\nelse:\n print(int(n[i]))\n', 'n=list(input())\nm=int(input())\ni=0\nfor t in range(100):\n if n[t]=="1":\n pass\n else:\n i=t+1\nif m=>i:\n print(str(n[i]))\nelse:\n print(1)\n', 'n=list(input())\nm=int(input())\ni=0\nfor t in range(len(n)):\n if n[t]=="1":\n pass\n else:\n i=t+1\n break\nif m>=i:\n print(int(n[i]))\nelse:\n print(1)', 'n=list(input())\nm=int(input())\ni=0\nfor t in range(len(n)):\n if n[t]!="1":\n i=t+1\n break\nif m<i:\n print(1)\nelse:\n print(int(n[i-1]))'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s573199261', 's872653450', 's988774204', 's616528999'] | [2940.0, 2940.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0, 17.0] | [139, 144, 156, 140] |
p03282 | u518987899 | 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().strip())\nfor i in range(K):\n if S[i] == 1:\n continue\n else:\n print(S[i])\n exit()\nprint(S[K])', "S = input().strip()\nK = int(input().strip())\nfor i in range(K):\n if S[i] == '1':\n continue\n else:\n print(S[i])\n exit()\nprint(S[K-1])\n"] | ['Wrong Answer', 'Accepted'] | ['s254244183', 's533756831'] | [2940.0, 2940.0] | [17.0, 17.0] | [139, 144] |
p03282 | u519604525 | 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\tif S[i] != 1:\n\t\tbreak\n\t\nif S[i] != 1:\n\tprint(S[i])\nelse:\n\tprint(1)', 'S = list(input())\nK = int(input())\n\nfor i in range(K):\n\tif S[i] != 1:\n\t\tbreak\n\t\nif S[i] != 1:\n\tprint(S[i])\nelse:\n\tprint(1)', "S = list(input())\nK = int(input())\n\nfor i in range(K):\n\tif S[i] != '1':\n\t\tbreak\n\t\nif S[i] != 1:\n\tprint(S[i])\nelse:\n\tprint(1)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s483570772', 's485735677', 's357481590'] | [2940.0, 2940.0, 2940.0] | [18.0, 17.0, 18.0] | [122, 122, 124] |
p03282 | u534998883 | 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\nn = -1\nfor i, ki = enumerate(s):\n if ki!='1':\n\tn = i+1\n break\n \nif n<0 or k<n:\n print(1)\nelse:\n print(s[n-1])\n", "s = input()\nk = int(input())\n\nn = -1\nfor i, ki in enumerate(s):\n if ki!='1':\n n = i+1\n break\n \nif n<0 or k<n:\n print(1)\nelse:\n print(s[n-1])\n"] | ['Runtime Error', 'Accepted'] | ['s955693918', 's958403591'] | [2940.0, 2940.0] | [17.0, 18.0] | [149, 153] |
p03282 | u535827181 | 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]) \nelse:\n print(S[K-1])', 'S = str(input())\nK = int(input())\nfor i in range(K):\n if S[i] != "1":\n print(S[i])\n break\nelse:\n print(1)'] | ['Runtime Error', 'Accepted'] | ['s126828053', 's221633454'] | [2940.0, 2940.0] | [17.0, 17.0] | [109, 113] |
p03282 | u536113865 | 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 = 1\nfor i in range(min(k,s)):\n if s[i]!='1':\n flag = 0\n break\nif flag:\n print(1)\nelse:\n print(s[1])\n", "s = input()\nk = int(input())\nflag = 1\nfor i in range(min(k,len(s))):\n if s[i]!='1':\n flag = 0\n a = s[i]\n break\nif flag:\n print(1)\nelse:\n print(a)\n"] | ['Runtime Error', 'Accepted'] | ['s877688299', 's066364949'] | [2940.0, 2940.0] | [17.0, 17.0] | [157, 176] |
p03282 | u539517139 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=input()\nk=int(input())\ni=0\nwhile i<k:\n if s[i]!=1:\n x=s[i]\n break\n i+=1\nif i==k:\n x=1\nprint(x)', "s=input()\nk=int(input())\nf=0\nfor i in range(k):\n if s[i]!='1':\n x=s[i]\n f=1\n break\nprint(x) if f==1 else print(1)"] | ['Wrong Answer', 'Accepted'] | ['s268368114', 's145788513'] | [2940.0, 3064.0] | [17.0, 18.0] | [105, 123] |
p03282 | u543954314 | 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 and s[0] == '1':\n print(s[0])\nelif s[0] == 1:\n print(s[1])\nelse:\n print(s[0])", "s = input()\nk = int(input())\ni = 0\nl = len(s)\nwhile s[i] == '1':\n if i >= l-1:\n break\n i += 1\nif i == 0 or k <= i:\n print(s[0])\nelse:\n print(s[i])"] | ['Wrong Answer', 'Accepted'] | ['s052802399', 's987632543'] | [3060.0, 3060.0] | [18.0, 17.0] | [119, 153] |
p03282 | u549497563 | 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 s[:k]:\n if i!= "1":\n ans = i\n exit()\nprint(ans)', 's = input()\nk = int(input())\nans = "1"\nfor i in s[:k]:\n if a!= "1":\n ans = a\n exit()\nprint(ans)', 's = input()\nk = int(input())\nans = "1"\nfor i in s[:k]:\n if i != "1":\n ans = i\n break\nprint(ans)'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s686008661', 's767903333', 's799121472'] | [3064.0, 2940.0, 2940.0] | [18.0, 18.0, 18.0] | [112, 112, 112] |
p03282 | u550943777 | 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? | ['inf = 5*10**15\nS = input()\nK = int(input())\n\nif S[0] == 1:\n print(S[1])\nelse:\n print(S[0])', 'S = input()\nK = int(input())\n\nfor i in S:\n if K == 1:\n print(i)\n break\n elif i == "1":\n K -= 1\n else:\n print(i)\n break'] | ['Wrong Answer', 'Accepted'] | ['s732734150', 's021157700'] | [2940.0, 2940.0] | [17.0, 18.0] | [96, 162] |
p03282 | u551373727 | 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 = input()\n\nif k == '1':\n print(s[0])\nelse:\n n = min(int(k), len(s))\n f = True\n for i in range(n):\n if s[i] != '1':\n print(s[i])\n f = False\n if f:\n print(1)", "s = list(input())\nk = input()\n\nif k == '1':\n print(s[0])\nelse:\n n = min(int(k), len(s))\n f = True\n for i in range(n):\n if s[i] != '1':\n print(s[i])\n f = False\n break\n if f:\n print(1)"] | ['Wrong Answer', 'Accepted'] | ['s152055484', 's471836855'] | [3060.0, 3060.0] | [17.0, 17.0] | [226, 244] |
p03282 | u560464565 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['#!/usr/bin/env python3\n# -*- coding: utf-8 -*-\n\nS = input()\nK = input()\nS = str(S)\nK = int(K)\n\nfor i in range(K):\n if S[i] != 1:\n break\n else:\n print("1")\n\nprint(S[i])\n', '#!/usr/bin/env python3\n# -*- coding: utf-8 -*-\n\nS = str(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'] | ['s527828774', 's195944028'] | [2940.0, 2940.0] | [17.0, 18.0] | [188, 177] |
p03282 | u560988566 | 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(len(s)):\n if s[i] != 1:\n if i <= k:\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != 1:\n ans = s[i]\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != 1:\n if i < k:\n ans = s[i]\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != 1:\n if i <= k-1:\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nanstr = ""\nfor i in range(len(s)):\n anstr += s[i]*(int(s[i])**5000000000000000)\n if len(anstr) > k:\n break\nprint(anstr[k-1])', 's = input()\nk = int(input())\nanstr = ""\nfor i in range(len(s)):\n anstr += s[i]*(int(s[i])**5000000000000000)\nprint(anstr[k-1])', 's = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != 1:\n if i < k:\n ans = int(s[i])\n break\nprint(ans)', 's = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != "1":\n if i <= k-1:\n ans = int(s[i])\n break\nprint(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Wrong Answer', 'Time Limit Exceeded', 'Time Limit Exceeded', 'Wrong Answer', 'Accepted'] | ['s006173896', 's046710338', 's112791338', 's512355153', 's657799310', 's898576627', 's987875567', 's642493046'] | [2940.0, 2940.0, 2940.0, 2940.0, 197056.0, 197056.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0, 17.0, 2105.0, 2108.0, 17.0, 17.0] | [136, 112, 130, 138, 158, 127, 135, 140] |
p03282 | u561992253 | 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()) - 1\n\ncnt1 = 0\nfor i in range(len(s)):\n if s[i] == 1:\n cnt1 += 1\n\nif cnt1 >= k:\n print(1)\nelse:\n print(s[cnt1])', "s = input()\nk = int(input()) - 1\n \ncnt1 = 0\nfor i in range(len(s)):\n if s[i] == '1':\n cnt1 += 1\n else:\n break\n \nprint(cnt1, k)\nif cnt1 > k:\n print(1)\nelse:\n print(s[cnt1])", "s = input()\nk = int(input()) - 1\n \ncnt1 = 0\nfor i in range(len(s)):\n if s[i] == '1':\n cnt1 += 1\n else:\n break\n \n#print(cnt1, k)\nif cnt1 > k:\n print(1)\nelse:\n print(s[cnt1])"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s501324806', 's635113706', 's732500051'] | [2940.0, 2940.0, 2940.0] | [17.0, 18.0, 17.0] | [145, 181, 182] |
p03282 | u566428756 | 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=''\nfor v in s:\n if int(v)*(5**12)<=k:\n break\nprint(v)", "s=input()\nlen_s=len(s)\ncnt=0\nk=int(input())\nfor v in s:\n if v!='1':\n print(int(v))\n break\n else:\n cnt+=1\n if k==cnt:\n print(1)\n break"] | ['Wrong Answer', 'Accepted'] | ['s453151756', 's244284273'] | [2940.0, 2940.0] | [17.0, 17.0] | [90, 189] |
p03282 | u569272329 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = str(input())\nK = int(input())\nif S.count("1") == S(len):\n print(1)\n exit(0)\nfor s in S:\n if s != "1":\n print(s)\n exit()\n', 'S = str(input())\nK = int(input())\nfor i in range(0, K):\n if S[i] != "1":\n print(S[i])\n exit()\nprint(1)\n'] | ['Runtime Error', 'Accepted'] | ['s709061193', 's912579744'] | [2940.0, 2940.0] | [17.0, 17.0] | [147, 120] |
p03282 | u569776981 | 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 exit()\n \n if i == k - 1:\n print('1')\n exit()\n\n\n", "S = input()\nK = int(input())\n\nfor i in range(K):\n if S[i] != '1':\n print(S[i])\n exit()\n \n if i == K - 1:\n print('1')\n exit()\n\n\n"] | ['Runtime Error', 'Accepted'] | ['s047736871', 's623235176'] | [9020.0, 9156.0] | [24.0, 32.0] | [164, 164] |
p03282 | u570232767 | 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(0, len(S)):\n if int(S[i])*5000000000000000 > K:\n print(S[i])\n break\n\n', "S = input()\nK = int(input())\n\nfor i in range(0, len(S)):\n if S[i] == '1':\n if K == i+1:\n print(S[i])\n break\n elif S[i+1] != '1':\n print(S[i+1])\n break\n else:\n print(S[0])\n break\n\n"] | ['Wrong Answer', 'Accepted'] | ['s713007498', 's407436659'] | [2940.0, 2940.0] | [17.0, 17.0] | [131, 257] |
p03282 | u571444155 | 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 \nflag = True\n \nfor i in range(len(k)):\n if s[i] != 1:\n print(s[i])\n flag = False\n break\n \nif flag:\n print("1")', 's = input()\nk = input()\n\nflag = True\n\nfor i in range(len(k)):\n if s[i] != 1:\n prtin(s[i])\n flag = false\n break\n \nif flag:\n print("1")\n', 's = list(input())\nk = int(input())\n \nflag = True\n\nfor i in range(k):\n if s[i] != "1":\n print(s[i])\n flag = False\n break\n\nif flag:\n print("1")'] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s199119459', 's613455366', 's505338075'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [149, 148, 152] |
p03282 | u574326823 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['# coding: utf-8\n\ns,k = [ input() for i in range(2) ]\n\ncount = 0\nwhile count != len(s) and s[count]=="1":\n count += 1\n\nprint(count)\n\nif int(k) <= count:\n print(1)\nelse:\n print(s[count])\n', '# coding: utf-8\n\ns,k = [ input() for i in range(2) ]\n\ncount = 0\nwhile count != len(s) and s[count]=="1":\n count += 1\n\nif int(k) <= count:\n print(1)\nelse:\n print(s[count])'] | ['Wrong Answer', 'Accepted'] | ['s548187256', 's330902439'] | [2940.0, 3060.0] | [17.0, 17.0] | [194, 179] |
p03282 | u580697892 | 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\nN = input()\nK = int(input())\nif N[0] == 1:\n for i in range(1, K+1):\n if N[i] != 1:\n print(N[i])\n break\n else:\n print(1)\n break\nelse:\n print(N[0])', '#coding: utf-8\nN = input()\nK = int(input())\nif N[0] == 1:\n for i in range(1, len(N)):\n if N[i] != 1:\n print(N[i])\n break\nelse:\n print(N[0])', '#coding: utf-8\nN = input()\nK = int(input())\nans = 0\nfor i in range(min(K, len(N))):\n if int(N[i]) != 1:\n ans = N[i]\n #print(N[i])\n break\nprint(ans if ans != 0 else 1)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s187956749', 's438394432', 's982216463'] | [2940.0, 3060.0, 2940.0] | [18.0, 18.0, 17.0] | [224, 174, 190] |
p03282 | u580904613 | 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())\nif len(n) > 1:\n if k > 1:\n if n[0] is not '1':\n print(n[0])\n else:\n print(1)\n else:\n print(n[0])\nelse:\n print(n[0])", "n = input()\nk = int(input())\nif len(n) > 1:\n for i in range(k):\n if n[i] is not '1':\n print(n[i])\n exit()\nprint(n[0])"] | ['Wrong Answer', 'Accepted'] | ['s871373294', 's040278985'] | [3068.0, 2940.0] | [18.0, 17.0] | [166, 131] |
p03282 | u583276018 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["n = input()\nk = int(input())\n\nif(k == 1):\n print(n[0])\nelif(len(n)==1):\n print(n)\nelif(n[0]== '1'):\n i = 1\n while(n[i] == '1' or i + 1 < len(n)):\n i += 1\n if(k <= i):\n print(1)\n else:\n print(n[i])\nelse:\n print(n[0])", 's = input()\nif(s[0] != 1):\n print(s[0])\nelse:\n i = 1\n while(s[i] == 1):\n i += 1\n print(s[i])', "n = input()\nk = int(input())\n\nif(k == 1):\n print(n[0])\nelif(len(n)==1):\n print(n)\nelif(n[0]== '1'):\n i = 1\n while(n[i] == '1'):\n i += 1\n if(i == len(n)):\n break\n if(k <= i):\n print(1)\n else:\n print(n[i])\nelse:\n print(n[0])"] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s098933247', 's226331157', 's448438501'] | [3064.0, 2940.0, 3064.0] | [17.0, 17.0, 17.0] | [257, 101, 282] |
p03282 | u589726284 | 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" or K != 1 or len(S) == 1:\n print(S[0])\nelse :\n print(S[1])', 'S = input()\nK = int(input())\nresult = "1"\nfor i in range(K):\n if S[i] != "1":\n result = S[i]\n break\nprint(result)'] | ['Wrong Answer', 'Accepted'] | ['s902589361', 's026339063'] | [2940.0, 2940.0] | [17.0, 17.0] | [104, 120] |
p03282 | u595289165 | 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 i in s:\n if i == 1:\n cnt += 1\n if cnt == 4:\n print(1)\n else:\n print(i)\n break\n', 's = input()\nk = int(input())\ncnt = 0\nans = 1\nfor i in s:\n if i == 1:\n cnt += 1\n else:\n ans = i\n break\n\nif k <= cnt:\n print(1)\nelse:\n print(ans)', 's = input()\nk = int(input())\ncnt = 0\nans = 1\nfor i in s:\n if i == "1":\n cnt += 1\n else:\n ans = i\n break\n\nif k <= cnt:\n print(1)\nelse:\n print(ans)'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s116916187', 's896252739', 's521401199'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [164, 176, 178] |
p03282 | u597374218 | 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 len(s)==1 or s[0]!=1 or k==1:print(s[0])\nelse:print(s[1])', 's=input()\nk=int(input())-1\ni=0\nwhile s[i]=="1" and k:k,i=k-1,i+1\nprint(s[i])'] | ['Wrong Answer', 'Accepted'] | ['s595100207', 's171810826'] | [2940.0, 2940.0] | [18.0, 18.0] | [85, 76] |
p03282 | u598812605 | 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\ni = 0\nwhile i < len(s):\n if int(s[i]) ** (500 * 10 **12) >= k:\n# print(i, s[i], int(s[i]) ** (500 * 10 *12))\n break\n i += 1\nprint(s[i])', 'import math\ns = input()\nk = int(input())\n\ni = 0\nwhile i < len(s):\n if int(math.log(s[i])) * (500 * 10 **12) >= math.log(k):\n# print(i, s[i], int(s[i]) ** (500 * 10 *12))\n break\n i += 1\nprint(s[i])', "s = input()\nk = int(input())\n\ni = 0\nwhile i < k - 1 and i < len(s):\n if s[i] != '1':\n break\n i += 1\nprint(s[i])\n"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s064985880', 's247438872', 's224102285'] | [158228.0, 2940.0, 2940.0] | [2106.0, 17.0, 17.0] | [186, 217, 125] |
p03282 | u599547273 | 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[k] != "1":\n print(s[k])\n exit()\nprint(1)', 's = input()\nk = int(input())\n\nfor k_i in range(k):\n if s[k_i] != "1":\n print(s[k_i])\n exit()\nprint(1)'] | ['Runtime Error', 'Accepted'] | ['s043048581', 's651922844'] | [2940.0, 2940.0] | [17.0, 18.0] | [112, 118] |
p03282 | u600261652 | 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 resolve():\n S = input()\n K = int(input())\n ans = []\n for i in range(len(S)):\n if S[i] == "1":\n ans.append("1")\n else:\n if len(ans) < K:\n print(S[i])\n else:\n print("1")\nresolve()', 'def resolve():\n S = input()\n K = int(input())\n for i in range(K):\n if S[i] != "1":\n print(S[i])\n exit()\n print("1")\nresolve()'] | ['Wrong Answer', 'Accepted'] | ['s804466926', 's739882114'] | [9084.0, 9076.0] | [30.0, 28.0] | [270, 166] |
p03282 | u602252807 | 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())\nkey = 0\nfor i in range(0,len(s)):\n if s[i] != 1:\n key = i\n break\nif k < key:\n print(1)\nelse:\n print(s[key])', 's = input()\nk = int(input())\nkey = 0\nfor i in range(len(s)):\n if s[i] != str(1):\n key = i\n break \nif k < key + 1:\n print(1)\nelse:\n print(s[key])'] | ['Wrong Answer', 'Accepted'] | ['s584197130', 's147431728'] | [2940.0, 2940.0] | [18.0, 17.0] | [159, 167] |
p03282 | u602677143 | 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 s[0] == "1":\n print(1)\nelif len(s) == 1:\n print(s)\nelse:\n print(s[0])', 'import sys\ns = list(input())\nk = int(input())\nif len(s) == 1:\n print(s[0])\nelif k == 1:\n print(s[0])\nelse:\n for i in range(k):\n if s[i] != "1":\n print(s[i])\n sys.exit()\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s578189306', 's118454811'] | [2940.0, 3064.0] | [17.0, 17.0] | [116, 219] |
p03282 | u606523772 | 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 print(S[1])', 'S = input()\nK = int(input())\nans = 0\nfor i in range(K):\n if S[i]!="1":\n ans = S[i]\n break\nprint(1) if ans==0 else print(ans)'] | ['Runtime Error', 'Accepted'] | ['s008848975', 's305720441'] | [2940.0, 2940.0] | [18.0, 17.0] | [81, 141] |
p03282 | u608088992 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S = input()\nK = int(input())\n \nif K == 1:\n print(int(S[0]))\nelif len(S) == 1:\n print(int(S))\nelif S[:K] == "1"*K:\n print(1)\nelse:\n for i in range(K+1):\n if S[i] != "!":\n print(int(S[i]))\n break', 'S = input()\nK = int(input())\n \nif K == 1:\n print(int(S[0]))\nelif len(S) == 1:\n print(int(S))\nelif S[0] == "1":\n print(int(S[0]))\nelse:\n print(int(S[1]))', 'S = input()\nK = int(input())\n \none = True\nfor i in range(K):\n if S[i] != "1":\n print(int(S[i]))\n one = False\n break\nif one:\n print(1)\n '] | ['Runtime Error', 'Wrong Answer', 'Accepted'] | ['s015066888', 's039728345', 's120879747'] | [3060.0, 3060.0, 2940.0] | [17.0, 17.0, 17.0] | [210, 156, 147] |
p03282 | u612721349 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['#!/usr/bin/env python\n# -*- coding: utf-8 -*-\n\n\ndef solve():\n s = input()\n n = int(input())\n idx = 0\n slen = len(s)\n while idx < slen and s[idx] == "1":\n idx += 1\n if idx < n or idx == slen:\n print(1)\n else:\n print(s[idx])\n\n\nif __name__ == "__main__":\n solve()\n', 's, _ = [input() for _ in range(2)]\ns = s.rstrip("1")\nprint(s[0] if s else "1")\n', '#!/usr/bin/env python\n# -*- coding: utf-8 -*-\n\n\ndef solve():\n s = input()\n n = int(input())\n idx = 0\n slen = len(s)\n while idx < slen and s[idx] == "1":\n idx += 1\n if idx == n:\n print(1)\n return\n if idx == slen:\n print(1)\n else:\n print(s[idx])\n\n\nif __name__ == "__main__":\n solve()\n'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s204535491', 's934604544', 's996610549'] | [3060.0, 2940.0, 3060.0] | [17.0, 18.0, 17.0] | [306, 79, 356] |
p03282 | u616489782 | 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=input();\nK=input();\ni = 0;\nfor i in range(len(S)) :\n if int(S[i]) > 1 :\n print(S[i]);\n break;\n elif i >= K :\n print(1);\n', 'S=input();\nK=input();\ni = 0;\nfor i in range(len(S)) :\n if int(S[i]) > 1 :\n print(S[i]);\n break;\n elif i + 1 >= K :\n print(1);\n', 'S=input();\nK=input();\ni = 0;\nfor i in range(len(S)) :\n if int(S[i]) > 1 :\n print(S[i]);\n break;\n elif i + 1 >= K :\n print(1);\n', 'S=input();\nK=input();\ni = 0;\nfor i in range(len(S)) :\n if int(S[i]) > 1 :\n print(S[i]);\n break;\n elif i + 1 >= int(K) :\n print(1);\n break;\n'] | ['Runtime Error', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s205989084', 's265291601', 's332481098', 's509433413'] | [9064.0, 9092.0, 9108.0, 9164.0] | [22.0, 25.0, 27.0, 28.0] | [162, 153, 153, 173] |
p03282 | u620755587 | 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 if s[:k] == "1"*k:\n print(s[0])\n else:\n print(s[k-1])\nelse:\n print(s[0])', 's = input()\nk = int(input())\nonecnt = 0\nfor i in range(len(s)):\n if s[i] == "1":\n onecnt += 1\n else:\n break\n \nif onecnt == 0:\n print(s[0])\n exit()\nif onecnt >= k:\n print(1)\n exit()\nelse:\n print(s[onecnt])'] | ['Runtime Error', 'Accepted'] | ['s026005510', 's867856856'] | [3060.0, 3060.0] | [17.0, 17.0] | [127, 220] |
p03282 | u623814058 | 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(int(s[i]))\n break\nelse:\n print(1)", "S=input()\nK=int(input())\n\nfor i in range(K):\n if S[i]!='1':\n print(int(S[i]))\n break\nelse:\n print(1)"] | ['Runtime Error', 'Accepted'] | ['s241961462', 's665421707'] | [9120.0, 9168.0] | [22.0, 31.0] | [120, 120] |
p03282 | u623819879 | 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())\nfl=0\nt=0\nans='1'\nfor i in range(len(s)):\n if s[i]=='1' and fl==0:\n t+=1\n else:\n if fl==0:\n ans=s[i]\n fl=1\nif t<=k:\n ans='1'\nprint(ans)", "s=input()\nk=int(input())\nfl=0\nt=0\nfor i in range(len(s)):\n if s[i]=='1' and fl=0:\n t+=1\n else:\n if fl==0:\n ans=s[i]\n fl=1\nif t<=k:\n ans='1'\nprint(ans)", "s=input()\nk=int(input())\nfl=0\nt=0\nans='1'\nfor i in range(len(s)):\n if s[i]=='1' and fl==0:\n t+=1\n else:\n if fl==0:\n ans=s[i]\n fl=1\nif t>=k:\n ans='1'\nprint(ans)"] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s343191177', 's687461844', 's303876545'] | [3064.0, 3188.0, 3064.0] | [18.0, 18.0, 17.0] | [178, 169, 178] |
p03282 | u626228246 | 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()\nn = int(input())\nif n == 1:\n\tprint(s[0])\n\tsys.exit()\nif \nfor i in s:\n\tif i != '1':\n\t\tprint(i)\n\t\tsys.exit()\nprint(1)", "import sys\ns = input()\nn = int(input())\ncnt = 0\nif n == 1:\n\tprint(s[0])\n\tsys.exit()\nfor i in s:\n\tif i == '1':\n\t\tcnt += 1\n\telse:\n\t\tbreak\nif n <= cnt:\n\tprint(1)\n\tsys.exit()\n\t\nfor i in s:\n\tif i != '1':\n\t\tprint(i)\n\t\tsys.exit()\nprint(1)"] | ['Runtime Error', 'Accepted'] | ['s445074419', 's998220366'] | [8884.0, 9216.0] | [25.0, 27.0] | [138, 231] |
p03282 | u629287683 | 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\n\nprint(s[0])\n\nif s[0] == 1:\n if k == 1:\n print(1)\n else:\n print(s[1])\nelse:\n print(s[0])', 's = str(input())\nk = int(input())\n\nans = 1\nfor i in range(k):\n #print("hoge")\n if s[i] != "1":\n ans = s[i]\n break\nprint(ans)'] | ['Wrong Answer', 'Accepted'] | ['s607085999', 's480049245'] | [2940.0, 2940.0] | [17.0, 17.0] | [146, 144] |
p03282 | u631914718 | 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, string, fractions, heapq, re, array, bisect, sys, random, time, copy\nimport itertools as it\nimport operator as op\nimport collections as cl\nimport functools as ft\n\ndef LI(): return list(map(int, input().split()))\ndef LI_(): return list(map(lambda x: int(x) - 1, input().split()))\ndef LF(): return list(map(float, input().split()))\ndef LS(): return input().split()\ndef I(): return int(input())\ndef F(): return float(input())\ndef S(): return input()\n\n\nwait_time = 5 * 10**15\n\nif __name__ == "__main__":\n s = S()\n k = input()\n\n result = \'\'\n for letter in s:\n if letter != \'1\':\n result = letter\n break\n\n if not result:\n result = \'1\'\n', 'import math, string, fractions, heapq, re, array, bisect, sys, random, time, copy\nimport itertools as it\nimport operator as op\nimport collections as cl\nimport functools as ft\n\ndef LI(): return list(map(int, input().split()))\ndef LI_(): return list(map(lambda x: int(x) - 1, input().split()))\ndef LF(): return list(map(float, input().split()))\ndef LS(): return input().split()\ndef I(): return int(input())\ndef F(): return float(input())\ndef S(): return input()\n\n\ndef count_first_1(s):\n count = 0\n first_letter = \'\'\n\n for letter in s:\n if letter == \'1\':\n count += 1\n else:\n first_letter = letter\n break\n\n return count, first_letter\n\n\nif __name__ == "__main__":\n s = S()\n k = I()\n\n count, first_letter = count_first_1(s)\n\n if not first_letter:\n result = \'1\'\n\n elif count >= k:\n result = \'1\'\n\n else:\n result = first_letter\n\n print(result)\n'] | ['Wrong Answer', 'Accepted'] | ['s748916860', 's554073807'] | [7880.0, 6864.0] | [356.0, 225.0] | [692, 936] |
p03282 | u634046173 | 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\nind = -1\nfor i in range(len(S)):\n if S[i] == 1\n ind = i\n else:\n break\n\nif ind == -1:\n print(S[0])\nelif ind + 1 >= K:\n print(1)\nelse:\n print(S[ind+1])\n', "S = input()\nK = int(input())\n\nind = -1\nfor i in range(len(S)):\n if S[i] == '1':\n ind = i\n else:\n break\n\nif ind == -1:\n print(S[0])\nelif ind + 1 >= K:\n print(1)\nelse:\n print(S[ind+1])\n"] | ['Runtime Error', 'Accepted'] | ['s469943025', 's174879026'] | [9032.0, 9180.0] | [20.0, 27.0] | [191, 194] |
p03282 | u634159866 | 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[0])\n \nelse:\n lst = []\n for i in range(len(S)):\n if S[i]!='1':\n lst.append(i)\n \n if lst==[]:\n print('1')\n else \n if K<lst[0]:\n print('1')\n else:\n print(S[lst[0]])", "S = input()\nK = int(input())\n\nif len(S)==1:\n print(S[0])\n \nelse:\n lst = []\n for i in range(len(S)):\n if S[i]!='1':\n lst.append(i)\n \n if lst==[]:\n print('1')\n else:\n if K<lst[0]+1:\n print('1')\n else:\n print(S[lst[0]])"] | ['Runtime Error', 'Accepted'] | ['s028052213', 's162015137'] | [2940.0, 3060.0] | [18.0, 17.0] | [303, 302] |
p03282 | u634461820 | 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())\nans = 10\n\nlist = [i for i in str(S)]\nlist.sort()\n\nif K <= len(list):\n if int(list[K-1]) == 1:\n ans = 1\nelse:\n for i in list:\n if int(i) > 1:\n ans = min(ans,int(i))\n\nprint(ans)', 'S = int(input())\nK = int(input())\nf = 0\nans = 10\n\nlist = [i for i in str(S)]\n \nfor i in range(K):\n if int(list[i]) > 1:\n ans = min(ans,int(list[i]))\n f = 1\n break\n\nif f == 0:\n print(1)\nelse:\n print(ans)'] | ['Wrong Answer', 'Accepted'] | ['s536763942', 's799639932'] | [3064.0, 3064.0] | [18.0, 18.0] | [240, 232] |
p03282 | u636162168 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s,k=input(),int(input())\nfor i in s[:k]:\n if i!="1":\n print(i)\n break\n print(1)\n break', 's,k=input(),int(input())\na=True\nfor i in s[:k]:\n if i!="1":\n print(i)\n a=True\n break\n a=False\nif not a:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s248853449', 's233372892'] | [2940.0, 2940.0] | [18.0, 17.0] | [109, 143] |
p03282 | u636311816 | 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 c in s:\n if c!=1:\n res=c\n break\n\nprint(res)', 's = input()\nk = int(input())\n\ni=0\nfor c in s:\n if c!="1":\n res=c\n #print(i,res)\n break\n i+=1\n\nif k>i:\n print(res)\nelse:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s290610260', 's216090722'] | [2940.0, 2940.0] | [18.0, 17.0] | [94, 162] |
p03282 | u636775911 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['n=int(input())\nk=int(input())\nj=0\nfor i in range(100,-1,-1):\n if(int(n%pow(10,i))==0):\n else:\n s.append(int(n%pow(10,i)))\n n-=s[j]\n j+=1\nif(s[0]=1):\n print(s[1])\nelse:\n print(s[0])', "#coding:utf-8\nn=input()\ncount=0\nl=int(input())\nfor i in range(len(n)):\n if(n[i]=='1'):\n count+=1\n if(l==count):\n print(n[i])\n break;\n continue\n else:\n print(n[i])\n break;"] | ['Runtime Error', 'Accepted'] | ['s736350559', 's547426630'] | [2940.0, 2940.0] | [19.0, 17.0] | [193, 197] |
p03282 | u640922335 | 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 range(len(S)):\n if S[i]==1:\n count+=1\n else:\n break\nif count>=K:\n print(1)\nelse:\n print(S[count])', 'S=input()\nK=int(input())\ncount=0\nfor i in range(len(S)):\n if int(S[i])==1:\n count+=1\n else:\n break\n\nif count>=K:\n print(1)\nelse:\n print(S[count])'] | ['Wrong Answer', 'Accepted'] | ['s939187209', 's366778392'] | [9048.0, 9100.0] | [28.0, 32.0] | [149, 155] |
p03282 | u641406334 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['s=list(input())\nk=int(input())\nif k==1 or len(s)==1:\n print(s[0])\nelse:\n for i in range(len(s)):\n num = int(s[i])**500000000\n if k-num<=0:\n print(s[i])\n break\n ', "s = input()\nk = int(input())\nfor i in range(k):\n if s[i]!='1':\n break\nprint(s[i])"] | ['Wrong Answer', 'Accepted'] | ['s488250018', 's112361574'] | [165752.0, 2940.0] | [2105.0, 17.0] | [183, 91] |
p03282 | u642528832 | 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())\nans = 1\n\nfor i in range(len(S)):\n if S[i] != 1 :\n ans = S[i]\n break\nprint(ans)', "S = list(input())\nK = int(input())\nans = 1\n\nfor i in range(K):\n if S[i] != '1' :\n ans = S[i]\n break\nprint(ans)"] | ['Wrong Answer', 'Accepted'] | ['s144547532', 's750270087'] | [9112.0, 9096.0] | [31.0, 29.0] | [130, 127] |
p03282 | u646130340 | 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_ones = 0\ninf_n = 0\nfor c in S:\n n = int(c)\n if n == 1:\n count_ones += 1\n if n > 2:\n inf_n = n\n break\nif K <= count_ones:\n print(1)\nelse:\n print(inf_n)\n ', 'S = input()\nK = int(input())\ncount_ones = 0\ninf_n = 0\nfor c in S:\n n = int(c)\n if n == 1:\n count_ones += 1\n else:\n inf_n = n\n break\nif K <= count_ones:\n print(1)\nelse:\n print(inf_n)'] | ['Wrong Answer', 'Accepted'] | ['s327940773', 's860631567'] | [2940.0, 3064.0] | [17.0, 17.0] | [204, 195] |
p03282 | u648315264 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['import math\nfrom math import gcd,pi,sqrt\nINF = float("inf")\n\nimport sys\nsys.setrecursionlimit(10**6)\nimport itertools\nfrom collections import Counter,deque\ndef i_input(): return int(input())\ndef i_map(): return map(int, input().split())\ndef i_list(): return list(i_map())\ndef i_row(N): return [i_input() for _ in range(N)]\ndef i_row_list(N): return [i_list() for _ in range(N)]\ndef s_input(): return input()\ndef s_map(): return input().split()\ndef s_list(): return list(s_map())\ndef s_row(N): return [s_input for _ in range(N)]\ndef s_row_str(N): return [s_list() for _ in range(N)]\ndef s_row_list(N): return [list(s_input()) for _ in range(N)]\n\ndef main():\n s = s_input()\n k = i_input()\n if k == 1:\n print(s[0])\n else:\n for i in range(k-1):\n if s[i] != "1":\n print(s[i+1])\n exit()\n print(s[-1])\n\n\nif __name__=="__main__":\n main()\n', "from collections import defaultdict,deque\nimport sys,heapq,bisect,math,itertools,string,queue,datetime\nsys.setrecursionlimit(10**8)\nINF = float('inf')\nmod = 10**9+7\neps = 10**-7\ndef inpl(): return list(map(int, input().split()))\ndef inpl_s(): return list(input().split())\n\nS = input()\nK = int(input())\n\nfor i in range(K):\n\tif S[i] == '1':\n\t\tcontinue\n\telse:\n\t\tprint(S[i])\n\t\tsys.exit()\n\nprint(S[K-1])\n"] | ['Runtime Error', 'Accepted'] | ['s893325459', 's216408635'] | [9496.0, 10712.0] | [31.0, 40.0] | [907, 399] |
p03282 | u652081898 | 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 = input()\n\nfor i in n:\n if i != "1":\n print(int(i))', 'N = int(input())\nif N < 105:\n print("0")\nelif N < 135:\n print("1")\nelif N < 165:\n print("2")\nelif N < 189:\n print("3")\nelif N < 195:\n print("4")\nelse:\n print("5")\n', 'n = input()\nk = int(input())\n\nfor i in range(len(n)):\n if n[i] != "1":\n x = i\n break\n\nif k <= i:\n print(1)\nelse:\n print(n[i])'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s156387619', 's492424166', 's174627369'] | [2940.0, 2940.0, 3060.0] | [17.0, 17.0, 19.0] | [75, 181, 148] |
p03282 | u652569315 | 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(s[i])\n exit()\nprint(1)', "s=input()\nk=int(input())\nfor i in range(k):\n if s[i]!='1':\n print(s[i])\n exit()\nprint(1)"] | ['Wrong Answer', 'Accepted'] | ['s993939655', 's113096187'] | [2940.0, 2940.0] | [17.0, 17.0] | [93, 95] |
p03282 | u655975843 | 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(1, len(s) + 1):\n if s[i - 1] == '1':\n continue\n else:\n break\nif i == 1:\n print(s[0])\nelif i >= k:\n print(1)\nelse:\n print(s[i])", "s = input()\nk = int(input())\n\nfor i in range(len(s)):\n if s[i] == '1':\n continue\n else:\n break\nif i == 0:\n print(s[0])\nelif i < k:\n print(s[i + 1])\nelse:\n print(1)", "s = input()\nk = int(input())\n\nfor i in range(1, len(s) + 1):\n if s[i - 1] == '1':\n continue\n else:\n break\nif i == 1:\n print(s[0])\nelif i >= k:\n print(1)\nelse:\n print(s[k])", "s = input()\nk = int(input())\n\nfor i in range(len(s)):\n if s[i] == '1':\n continue\n else:\n break\nif i == 0:\n print(s[0])\nelif i < k:\n print(s[i])\nelse:\n print(1)"] | ['Wrong Answer', 'Runtime Error', 'Runtime Error', 'Accepted'] | ['s579277079', 's738242611', 's813228030', 's524388840'] | [3060.0, 3060.0, 2940.0, 3060.0] | [19.0, 18.0, 17.0, 19.0] | [200, 192, 200, 188] |
p03282 | u667024514 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['n = list(str(input()))\nk = int(input())\nfor i in range(n):\n if n[i] != "1":\n print(n[i])\n exit()\nprint("1")', 'n = list(str(input()))\nk = int(input())\nfor i in range(min(k,len(n))):\n if n[i] != "1":\n print(n[i])\n exit()\nprint("1")'] | ['Runtime Error', 'Accepted'] | ['s639326958', 's747892742'] | [2940.0, 3060.0] | [17.0, 17.0] | [114, 126] |
p03282 | u667458133 | 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\nfor i in range(K):\n if S[i] != 1:\n print(S[i])\n flag = False\n break\n\nif flag:\n print(1)\n', 'S = input()\nK = int(input())\nflag = True\n\nfor i in range(K):\n if S[i] != 1:\n print(S[i])\n flag = False\n\nif flag:\n print(1)\n', "S = input()\nK = int(input())\nflag = True\n\nfor i in range(K):\n if S[i] != '1':\n print(S[i])\n flag = False\n break\n\nif flag:\n print(1)\n"] | ['Wrong Answer', 'Runtime Error', 'Accepted'] | ['s228573028', 's300497551', 's782557645'] | [2940.0, 3060.0, 2940.0] | [18.0, 18.0, 17.0] | [157, 143, 159] |
p03282 | u669696235 | 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,list(input())))\nK=int(input())\n\nsum=0\nfor i in range(0,len(S)):\n if(S[i]==1):\n sum+=1\n else:\n tmp=S[i]\n for j in range(2,5000000000000000):\n tmp*=S[i]\n print(tmp)\n if(sum+tmp>=K):\n print(S[i])\n exit()\n else:\n sum+=tmp\n\n', 'S=list(map(int,list(input())))\nK=int(input())\n\nsum=0\nfor i in range(0,len(S)):\n if(S[i]==1):\n sum+=1\n if(sum>=K):\n print(S[i])\n exit()\n else:\n tmp=S[i]\n for j in range(2,5000000000000000):\n tmp*=S[i]\n if(sum+tmp>=K):\n print(S[i])\n exit()\n else:\n sum+=tmp\n\n'] | ['Wrong Answer', 'Accepted'] | ['s738434190', 's575240070'] | [3060.0, 3064.0] | [18.0, 18.0] | [350, 398] |
p03282 | u672475305 | 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 len(s) == 1:\n print(s)\nelse:\n print(s[0])', 's = input()\nk = int(input())\nif len(s) == 1:\n print(s)\nelse:\n if s[0] != 1:\n print(s[0])\n else:\n print(s[1])', "s = input()\nk = int(input())\n\nif len(s) == 1:\n print(s)\nelif len(set(s)) == 1:\n print(s[0])\nelse:\n t = s.replace('1','')\n if k < len(s) - len(t):\n print(1)\n else:\n print(t[0])"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s351524944', 's811563568', 's620354026'] | [2940.0, 2940.0, 3064.0] | [17.0, 17.0, 19.0] | [79, 131, 204] |
p03282 | u672898046 | 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 resolve():\n s = input()\n k = int(input())\n c = 0\n for i in s:\n if i == "1":\n c += 1\n else:\n c += int(i)**((5*10**16)-1)\n if k <= c:\n print(i)\n break\nresolve()', 'def resolve():\n s = input()\n k = int(input())\n c = 0\n for i in range(k):\n if s[i] != "1":\n print(s[i])\n break\n if i == k-1:\n print(1)\nresolve()'] | ['Time Limit Exceeded', 'Accepted'] | ['s062984516', 's721994075'] | [251488.0, 9164.0] | [2211.0, 29.0] | [239, 202] |
p03282 | u683406607 | 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\ntmp = 0\nfor each in s:\n if each > 1:\n print(each)\n break\n tmp += 1\n if tmp == k:\n print(1)\n break', 's = input()\nk = int(input())\n\ntmp = 0\nfor each in s:\n if int(each) > 1:\n print(each)\n break\n tmp += 1\n if tmp == k:\n print(1)\n break'] | ['Runtime Error', 'Accepted'] | ['s157037572', 's275728588'] | [2940.0, 2940.0] | [17.0, 17.0] | [142, 147] |
p03282 | u686036872 | 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\nif len(S) == 1 or K == 1:\n print(S[0])\n\nelif S[0] == 1:\n print(S[1])\n\nelse:\n print(S[0])', 'S = list(input())\nK = int(input())\n\nif S[0] == 1:\n if K >= 2:\n print(S[1])\n else:\n print(1)\nelse:\n print)S[1]', 'S = list(input())\nK = int(input())\n\nif len(S) == 1:\n print(S[0])\n\nelif S[0] == 1:\n print(S[1])\n\nelse:\n print(S[0])', 'S = list(input())\nK = int(input())\n\nif len(S) == 1 or K == 1:\n print(S[0])\n\nelif S[0] == 1 and K >= 2:\n print(S[1])\n\nelif S[1] != 1 and K >= 2:\n print(S[0])', '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)'] | ['Wrong Answer', 'Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s156235303', 's724684105', 's759081896', 's852683152', 's013540624'] | [3060.0, 2940.0, 3060.0, 3060.0, 2940.0] | [17.0, 18.0, 17.0, 17.0, 17.0] | [133, 132, 123, 165, 121] |
p03282 | u686461495 | 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(k):\n if not list[i] == 1:\n ans=list[i]\n break\nprint(ans)\n', '# coding:utf-8\ns=input()\nk=int(input())\nlist=list(s)\nans=1\nfor i in range(k):\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'] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s727156881', 's849129212', 's072006718'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [148, 148, 169] |
p03282 | u690442716 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['N = int(input())\nK = int(input())\nn = N \nans = 0\nc = 0\nco\nwhile n > 10: \n c = c + 1 \n n = n // 10\nwhile c > 0: \n m = N // (10 * c)\n d = m % 10\n if d == 1:\n co = co + 1\n else:\n ans = d\n break\n c = c - i\nif K < co:\n ans = n\nprint(ans)', 'N = int(input())\nK = int(input())\nn = N \nans = 0\nc = 0\nco = 0\nwhile n > 10: \n c = c + 1 \n n = n // 10\nwhile c > 0: \n m = N // (10 * c)\n d = m % 10\n if d == 1: \n co = co + 1\n else:\n ans = d \n break\n c = c - 1\nif K < co: \n ans = 1\nprint(ans)', 'N = int(input())\nans = 1\nd = 1\nwhile N>10:\n N = N //10\nans = N % 10\nprint(ans)', 'N = int(input())\nK = int(input())\nn = N \nans = n\nc = 0\nco = 0\n\nwhile n > 10: \n c = c + 1 \n n = n // 10\n#print(c)\nwhile c > -1: \n m = N // (10 ** c)\n d = m % 10\n #print(m)\n #print(d)\n if d == 1: \n co = co + 1\n else: \n ans = d \n break\n c = c - 1\nif K <= co: \n ans = 1\n\nprint(ans)'] | ['Runtime Error', 'Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s333654318', 's338072192', 's665730590', 's296209260'] | [3064.0, 3064.0, 2940.0, 3064.0] | [17.0, 17.0, 17.0, 17.0] | [296, 516, 81, 569] |
p03282 | u690781906 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ["s = input()\nk = int(input())\nfor i in len(s):\n if s[i] != '1' and i < k:\n print(s[i])\n exit()\nprint('1')", "s = input()\nk = int(input())\nfor i in range(len(s)):\n if s[i] != '1' and i < k:\n print(s[i])\n exit()\nprint('1')\n"] | ['Runtime Error', 'Accepted'] | ['s442432565', 's062099854'] | [2940.0, 2940.0] | [18.0, 17.0] | [121, 129] |
p03282 | u691896522 | 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(len(s)):\n if s[i] != '1' and k <= i:\n print(s[i])\n exit()\nprint(ans)", "s = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != '1' and k < i:\n print(s[i])\n exit()\nprint(ans)", "s = input()\nk = int(input())\nans = 1\nfor i in range(len(s)):\n if s[i] != '1' and k > i:\n print(s[i])\n exit()\nprint(ans)"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s138170992', 's677274612', 's269200828'] | [3060.0, 2940.0, 2940.0] | [20.0, 17.0, 18.0] | [137, 136, 136] |
p03282 | u692453235 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['S, K = map(int, [input() for i in range(2)])\ncount = 0\n\nfor i in str(S):\n if i == 1:\n count += 1\n continue\n else:\n a = i\n break\n\nif count >= K:\n print(1)\nelse:\n print(a)', 'S, K = map(int, [input() for i in range(2)])\ncount = 0\na = 0\n\nfor i in str(S):\n if int(i) == 1:\n count += 1\n continue\n else:\n a = int(i)\n break\n\nif count >= K:\n print(1)\nelse:\n print(a)'] | ['Wrong Answer', 'Accepted'] | ['s281030852', 's102518435'] | [2940.0, 3060.0] | [17.0, 17.0] | [209, 225] |
p03282 | u692632484 | 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 c in S:\n\tK-=1\n if K==0:\n print(c)\n break\n if c!='1':\n print(c)\n break", "S=input()\nK=int(input())\n\nfor c in S:\n K-=1\n if K==0:\n print(c)\n break\n if c!='1':\n print(c)\n break\n"] | ['Runtime Error', 'Accepted'] | ['s838908839', 's161479252'] | [2940.0, 2940.0] | [17.0, 18.0] | [125, 115] |
p03282 | u693007703 | 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\noutput = ''\n\nif len(S) <= K:\n for letter in S:\n if letter != '1':\n output = l\n break\nelse :\n for i in range(K):\n if S[i] != '1':\n output = S[i]\n break\n \nif output:\n print(output)\nelse :\n print('1')", "S = input()\nK = int(input())\n\noutput = ''\n\nif len(S) <= K:\n for letter in S:\n if letter != '1':\n output = letter\n break\nelse :\n for i in range(K):\n if S[i] != '1':\n output = S[i]\n break\n\nif output:\n print(output)\nelse :\n print('1')"] | ['Runtime Error', 'Accepted'] | ['s504961698', 's573771480'] | [3060.0, 3060.0] | [17.0, 18.0] | [300, 301] |
p03282 | u693953100 | 2,000 | 1,024,000 | Mr. Infinity has a string S consisting of digits from `1` to `9`. Each time the date changes, this string changes as follows: * Each occurrence of `2` in S is replaced with `22`. Similarly, each `3` becomes `333`, `4` becomes `4444`, `5` becomes `55555`, `6` becomes `666666`, `7` becomes `7777777`, `8` becomes `88888888` and `9` becomes `999999999`. `1` remains as `1`. For example, if S is `1324`, it becomes `1333224444` the next day, and it becomes `133333333322224444444444444444` the day after next. You are interested in what the string looks like after 5 \times 10^{15} days. What is the K-th character from the left in the string after 5 \times 10^{15} days? | ['#include<stdio.h>\nint main(void)\n{\n char s[101]; \n unsigned long k;\n scanf("%s",&s);\n scanf("%u,&k);\n int len=0;\n int i;\n char ans[2]="1";\n while(s[len++])\n int c=len<k?k:len;\n for(i=0;i<c;i++)\n {\n if(s[i]!=\'1\')\n {ans[0]=s[i]; break;}\n }\n printf("%s",ans);\nreturn 0;\n}', 's=input()\nk=input()\na=int(s[0])\n\nfor num in range(0,max(len(s),k)):\n if int(s[num])==1:\n print(1)\n else:\n print(s[num])', "s = input()\nt =int(input())\nfor i in range(t):\n if s[i]!='1':\n print(s[i])\n exit()\nprint('1')"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s306269824', 's730613857', 's874275385'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [326, 139, 100] |
p03282 | u694810977 | 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 s:\n if i != "1":\n print(str(i))\n break\n else:\n print("1")\n', 's = str(input())\nk = int(input())\nfor i in s:\n if i != "1":\n print(str(i))\n break\n else:\n print("1")\n break\n', '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', 'Wrong Answer', 'Accepted'] | ['s112143238', 's389103647', 's157786043'] | [2940.0, 2940.0, 2940.0] | [17.0, 17.0, 17.0] | [128, 142, 118] |
p03282 | u697658632 | 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(c)\n break\nelse:\n print(s[k - 1])\n", "s = input()\nk = int(input())\nfor i in range(k):\n if s[i] != '1':\n print(c)\n break\nelse:\n print(s[k - 1])\n", "s = input()\nk = int(input())\nfor i in range(min(len(s), k)):\n if s[i] != '1':\n print(s[i])\n break\nelse:\n print(s[k - 1])\n"] | ['Runtime Error', 'Runtime Error', 'Accepted'] | ['s001276167', 's565116693', 's428357771'] | [2940.0, 2940.0, 2940.0] | [18.0, 18.0, 17.0] | [126, 113, 129] |
p03282 | u698416089 | 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(s)[0] == 1:\n print(list(s)[1])\nelse:\n print(list(s)[0])', 's = list(input())\nk = int(input())\nfor i in range(len(s)):\n if s[i] != "1":\n break\nif k-1 < i:\n print("1")\nelse:\n print(s[i])'] | ['Wrong Answer', 'Accepted'] | ['s250050180', 's819475547'] | [2940.0, 2940.0] | [17.0, 17.0] | [94, 131] |
p03282 | u698479721 | 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())\nk = min(k,len(s))\ni = 0\nwhile i < k:\n if s[i] != 1:\n print(s[i])\n sys.exit()\n i += 1\nprint(1)', 'import sys\ns = input()\nk = int(input())\nk = min(k,len(s))\ni = 0\nflag = 0\nwhile i < k:\n if int(s[i]) != 1:\n print(s[i])\n flag = 1\n sys.exit()\n i += 1\nif flag == 0:\n print(1)'] | ['Wrong Answer', 'Accepted'] | ['s277077878', 's527133755'] | [2940.0, 3060.0] | [17.0, 17.0] | [141, 184] |
p03282 | u699765883 | 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\ntime = 5 * 10**15\n\nnum = input()\nk = int(input())\n\nchk = 0\ncount = 0\nwhile chk == 1:\n if num[0] == '1':\n a.pop(0)\n count += 1\n else:\n chk = 1\n\nif k <= count:\n print(1)\nelse:\n print(num[0])\n", "import math\ntime = 5 * 10**15\n\nnum = input()\nk = int(input())\n\nchk = 0\nwhile chk == 1:\n if num[0] == '1':\n a.pop(0)\n else:\n chk = 1\n\nprint(num[0])\n", "import math\ntime = 5 * 10**15\n\nnum = list(input())\nk = int(input())\n\nchk = 0\ncount = 0\nwhile chk == 0:\n if num[0] == '1':\n num.pop(0)\n count += 1\n if len(num)==0:\n break\n else:\n chk = 1\n\nif k <= count:\n print(1)\nelse:\n print(num[0])\n"] | ['Wrong Answer', 'Wrong Answer', 'Accepted'] | ['s266265688', 's558563433', 's611845186'] | [3060.0, 3060.0, 3060.0] | [17.0, 17.0, 17.0] | [234, 167, 284] |
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