text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
from collections import deque
N, M = map(int, input().split())
F_C_of_S = [{} for i in range(N)]
# Companies of Station
C_of_S = {}
# def Union-find
parent = [-1]*M
def root(x):
if (parent[x] < 0): return x
else:
parent[x] = y = root(parent[x])
return y
def unite(x, y):
px = root(x); py = root(y);
if (px == py): return False
if (parent[px] > parent[py]):
px = root(y); py = root(x);
parent[px] += parent[py]
parent[py] = px
return True
# Union-find
pqcs = []
for i in range(M):
p, q, c = map(int, input().split())
p -= 1; q -= 1;
pqcs += [(p, q, c)]
# The first
if not c in F_C_of_S[p]:
F_C_of_S[p][c] = i
# else
else:
unite(F_C_of_S[p][c], i)
C_of_S.setdefault(p, set()).add(c)
# The first
if not (c in F_C_of_S[q]):
F_C_of_S[q][c] = i
# else
else:
unite(F_C_of_S[q][c], i)
C_of_S.setdefault(q, set()).add(c)
# stations of company
S_of_C = {}
for i in range(M):
p, q, c = pqcs[i]
c_set = S_of_C.setdefault(root(i), set())
c_set.add(p); c_set.add(q);
Q = deque([(0, 0, 0)])
# cost to go to the stations from station_0
dist = [float('inf')] * N
dist[0] = 0
gdist = [float('inf')] * M
while Q:
cost, v, flag = Q.popleft()
if not (flag): # (flag == 0) then gdist_update
for l in F_C_of_S[v].values():
l = root(l)
if (cost < gdist[l]):
gdist[l] = cost
Q.appendleft((cost, l, 1))
else: # (flag == 1) then dist_update
for s in S_of_C[v]:
if (cost + 1 < dist[s]):
dist[s] = cost + 1
Q.append((cost + 1, s, 0))
print(dist[N - 1] if dist[N - 1] < float('inf') else -1)
```
Yes
| 93,400 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
from sys import exit, setrecursionlimit, stderr
from functools import reduce
from itertools import *
from collections import *
from bisect import bisect
from heapq import *
def read():
return int(input())
def reads():
return [int(x) for x in input().split()]
N, M = reads()
edges = [[] for _ in range(N)]
edgesc = defaultdict(lambda: [])
for _ in range(M):
p, q, c = reads()
p, q = p-1, q-1
edges[p].append((q, c))
edges[q].append((p, c))
edgesc[p, c].append(q)
edgesc[q, c].append(p)
INF = 1 << 30
def dijkstra(edges, s):
result = defaultdict(lambda: INF)
result[s, -1] = 0
que = deque([(0, s, -1)])
while len(que) > 0:
(d, u, c) = que.popleft()
if result[u, c] < d:
continue
if c < 0:
for (t, c2) in edges[u]:
if d + 1 < result[t, c2]:
result[t, c2] = d + 1
que.append((d + 1, t, c2))
else:
for t in edgesc[u, c]:
if d < result[t, c]:
result[t, c] = d
que.appendleft((d, t, c))
if d < result[u, -1]:
result[u, -1] = d
que.appendleft((d, u, -1))
return result
dist = dijkstra(edges, 0)
try:
print(min(d for (u, _), d in dist.items() if u == N-1))
except ValueError:
print(-1)
```
Yes
| 93,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
def main():
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
from collections import deque, defaultdict
n, m = map(int, readline().split())
graph = [defaultdict(set) for _ in range(n + 1)]
for _ in range(m):
p, q, c = map(int, readline().split())
graph[p][c].add(q)
graph[q][c].add(p)
q = deque([(1, 0, -1)])
vis = set()
while q:
now, v, c = q.popleft()
if now == n:
print(v)
exit()
if (now, c) in vis:
continue
vis.add((now, c))
if c == 0:
for cc in graph[now]:
for next in graph[now][cc]:
if (next, cc) not in vis:
q.append((next, v + 1, cc))
else:
for next in graph[now][c]:
if (next, c) not in vis:
q.appendleft((next, v, c))
if (now, 0) not in vis:
q.appendleft((now, v, 0))
print(-1)
if __name__ == '__main__':
main()
```
Yes
| 93,402 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
import heapq
import sys
input = sys.stdin.buffer.readline
class UnionFind:
"""素集合を木構造として管理する"""
def __init__(self, n):
self.parent = [-1] * n
self.cnt = n
def root(self, x):
"""要素xの根を求める"""
if self.parent[x] < 0:
return x
else:
self.parent[x] = self.root(self.parent[x])
return self.parent[x]
def merge(self, x, y):
"""要素xを含む集合と要素yを含む集合を統合する"""
x = self.root(x)
y = self.root(y)
if x == y:
return
if self.parent[x] > self.parent[y]:
x, y = y, x
self.parent[x] += self.parent[y]
self.parent[y] = x
self.cnt -= 1
def is_same(self, x, y):
"""要素x, yが同じ集合に属するかどうかを求める"""
return self.root(x) == self.root(y)
def dijkstra():
dp[sup_v][0] = 1
q = [(0, sup_v, 0)] # q = [(startからの距離, 現在地)]
while q:
d, v1, v2 = heapq.heappop(q)
if dp[v1][v2] < d:
continue
for nxt_v1, nxt_v2, cost in graph[v1][v2]:
if dp[v1][v2] + cost < dp[nxt_v1][nxt_v2]:
dp[nxt_v1][nxt_v2] = dp[v1][v2] + cost
heapq.heappush(q, (dp[nxt_v1][nxt_v2], nxt_v1, nxt_v2))
n, m = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
INF = 10 ** 18
uf = UnionFind(n)
sup_v = 10 ** 6
graph = {sup_v: {}}
dp = {sup_v: {}}
for a, b, c in info:
a -= 1
b -= 1
if c not in graph:
graph[c] = {}
dp[c] = {}
if a not in graph[c]:
graph[c][a] = []
graph[c][a].append((sup_v, a, 1))
dp[c][a] = INF
if b not in graph[c]:
graph[c][b] = []
graph[c][b].append((sup_v, b, 1))
dp[c][b] = INF
if a not in graph[sup_v]:
graph[sup_v][a] = []
dp[sup_v][a] = INF
if b not in graph[sup_v]:
graph[sup_v][b] = []
dp[sup_v][b] = INF
graph[c][a].append((c, b, 0))
graph[c][b].append((c, a, 0))
graph[sup_v][a].append((c, b, 0))
graph[sup_v][b].append((c, a, 0))
uf.merge(a, b)
if not uf.is_same(0, n - 1):
print(-1)
exit()
dijkstra()
ans = INF
for i in dp:
for j in dp[i]:
if j == n - 1:
ans = min(dp[i][j], ans)
print(ans)
```
No
| 93,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
from collections import deque
N, M = map(int, input().split())
# Companies of Station
C_of_S = {}
# def Union-find
parent = {i:i for i in range(M)}
def root(x):
if x == parent[x]:
return x
parent[x] = y = root(parent[x])
return y
def unite(x, y):
px = root(x); py = root(y);
if px < py:
parent[py] = px
else:
parent[px] = py
# Union-find
pqcs = []
#stations of company
S_of_C = {}
for i in range(M):
p, q, c = map(int, input().split())
p -= 1; q -= 1;
pqcs += [(p, q, c)]
if not 10**6+c in parent:
parent[10**6+c] = i
else:
unite(10**6+c, i)
C_of_S.setdefault(p, set()).add(root(i))
C_of_S.setdefault(q, set()).add(root(i))
c_set = S_of_C.setdefault(root(i), set())
c_set.add(p)
c_set.add(q)
if (M == 0) or not (0 in C_of_S) or not (N-1 in C_of_S):
print(-1)
exit()
Q = deque([(0, 0, 0)])
# cost to go to the station from station_0
dist = [float('inf')]*N
dist[0] = 0
# cost to get on the line
line_dist = [float('inf')]*M
while Q:
cost, v, flag = Q.popleft()
if not (flag): # (flag == 0) then line_dist_update
# v is station
for l in C_of_S[v]:
if (cost < line_dist[l]):
line_dist[l] = cost
Q.appendleft((cost, l, 1))
else: # (flag == 1) then dist_update
# v is company
for s in S_of_C[v]:
if (cost + 1 < dist[s]):
dist[s] = cost + 1
Q.append((cost + 1, s, 0))
print(dist[N - 1] if dist[N - 1] < float('inf') else -1)
```
No
| 93,404 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
def main():
from heapq import heappush,heappop,heapify
inf=2**31-1
n,m,*t=map(int,open(0).read().split())
if not t:
print(-1)
exit()
z=[]
for a,b,c in zip(t[::3],t[1::3],t[2::3]):
z.extend([a,b,a+n*c,b+n*c])
z={i:v for v,i in enumerate(sorted(set(z)))}
edge=[[]for _ in range(len(z))]
d={}
l=[]
for a,b,c in zip(t[::3],t[1::3],t[2::3]):
if b<a:a,b=b,a
i,j,x,y=z[a],z[b],z[a+n*c],z[b+n*c]
if b==n:
l.append(y)
d[y]=inf
edge[i].append((1,x))
edge[j].append((1,y))
edge[x].extend([(0,i),(0,y)])
edge[y].extend([(0,j),(0,x)])
used=[False]+[True]*~-len(z)
edgelist=edge[0]
heapify(edgelist)
while edgelist:
m=heappop(edgelist)
if not used[m[1]]:
continue
d[m[1]]=m[0]
used[m[1]]=False
for e in edge[m[1]]:
if used[e[1]]:
heappush(edgelist,(e[0]+m[0],e[1]))
a=min([d[i]for i in l]or[inf])
print([a,-1][a==inf])
if __name__=='__main__':
main()
```
No
| 93,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.
The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.
You can change trains at a station where multiple lines are available.
The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.
Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)
Input
The input is given from Standard Input in the following format:
N M
p_1 q_1 c_1
:
p_M q_M c_M
Output
Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.
Examples
Input
3 3
1 2 1
2 3 1
3 1 2
Output
1
Input
8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5
Output
2
Input
2 0
Output
-1
Submitted Solution:
```
import sys
from collections import deque
N, M = map(int, input().split())
Edge = [[] for _ in range(N+1)]
comp = set()
mod = 10**9+7
for _ in range(M):
p, q, c = map(int, input().split())
Edge[p].append(mod*q+c)
Edge[q].append(mod*p+c)
comp.add(c)
dist = {mod: 0}
Q = deque()
Q.append(mod)
visited = set()
while Q:
vn, cp = divmod(Q.pop(), mod)
if mod*vn+cp in visited:
continue
if vn == N:
break
visited.add(mod*vn+cp)
for t in Edge[vn]:
vf, cf = divmod(t, mod)
if mod*vf+cp in visited:
continue
if cp == cf:
dist[mod*vf+cf] = dist.get(mod*vn+cp, 10**9)
Q.append(mod*vf+cf)
else:
dist[mod*vf+cf] = 1 + dist.get(mod*vn+cp, 10**9)
Q.appendleft(mod*vf+cf)
L = [dist.get(N*mod+c, 10**9) for c in comp]
if not L or min(L) == 10**9:
print(-1)
else:
print(min(L))
```
No
| 93,406 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
INF=1e8
path=[]
dp=[]
while(True):
try:
temp=list(map(int,input().split(',')))
path.append(temp)
dp.append([INF for i in range(len(temp))])
except:
break
dp[0][0]=path[0][0]
for i in range(1,len(dp)):
for j in range(len(dp[i])):
if len(dp[i])>len(dp[i-1]):
start=0 if j-1<0 else j-1
dp[i][j]=max(dp[i-1][start:j+1])+path[i][j]
else:
dp[i][j]=max(dp[i-1][j:j+2])+path[i][j]
print(dp[-1][-1])
```
| 93,407 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
import sys
s=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(s)):
for j in range(len(s[i])):
b=len(s[i])>len(s[i-1])
s[i][j]+=max(s[i-1][(j-b)*(j>0):j+2-b])
print(*s[-1])
```
| 93,408 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
path=[]
while(True):
try:
temp=list(map(int,input().split(',')))
path.append(temp)
except:
break
path[0][0]
for i in range(1,len(path)):
for j in range(len(path[i])):
if len(path[i])>len(path[i-1]):
start=0 if j<1 else j-1
path[i][j]+=max(path[i-1][start:j+1])
else:
path[i][j]+=max(path[i-1][j:j+2])
print(path[-1][-1])
```
| 93,409 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
import sys
s=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(s)):
k=len(s[i])
for j in range(k):
t=j-(k>len(s[i-1]))
s[i][j]+=max(s[i-1][t*(j>0):t+2])
print(*s[-1])
```
| 93,410 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
diamond=[]
path=[]
while True:
try:
s=[int(i) for i in input().split(",")]
if(len(diamond)==0):
diamond.append(s)
path.append(s)
elif(len(diamond[-1])<len(s)):
diamond.append(s)
_path=[path[-1][0]+s[0]]
for i in range(len(s)-2):
_path.append(max(path[-1][i],path[-1][i+1])+s[i+1])
_path.append(path[-1][-1]+s[-1])
path.append(_path)
else:
diamond.append(s)
_path=[]
for i in range(len(s)):
_path.append(max(path[-1][i],path[-1][i+1])+s[i])
path.append(_path)
if len(s)==1:
print(path[-1][0])
diamond=[]
path=[]
except EOFError:
break
```
| 93,411 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
while 1:
try:
count = 0
rom = []
while count != 2:
N = list(map(int, input().split(",")))
if len(N) == 1:
count += 1
rom.append(N)
half = len(rom) // 2
for i in range(1, len(rom)):
for j in range(len(rom[i])):
if i <= half:
if j != 0 and j != len(rom[i]) - 1:
rom[i][j] = max(rom[i - 1][j], rom[i - 1][j - 1]) + rom[i][j]
elif j == 0:
rom[i][j] = rom[i - 1][j] + rom[i][j]
else:
rom[i][j] = rom[i - 1][j - 1] + rom[i][j]
else:
rom[i][j] = max(rom[i - 1][j], rom[i - 1][j + 1]) + rom[i][j]
print(rom[-1][0])
except:
break
```
| 93,412 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
import sys
path=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(path)):
for j in range(len(path[i])):
if len(path[i])>len(path[i-1]):
start=0 if j<1 else j-1
path[i][j]+=max(path[i-1][start:j+1])
else:
path[i][j]+=max(path[i-1][j:j+2])
print(path[-1][-1])
```
| 93,413 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
"Correct Solution:
```
import sys
s=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(s)):
k=len(s[i]);b=k>len(s[i-1])
for j in range(k):
t=j-b;s[i][j]+=max(s[i-1][t*(j>0):t+2])
print(*s[-1])
```
| 93,414 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
import sys
s=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(s)):
k=len(s[i]);b=k>len(s[i-1])
for j in range(k):s[i][j]+=max(s[i-1][(j-b)*(j>0):j-b+2])
print(*s[-1])
```
Yes
| 93,415 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
table = [[0 for i in range(100)] for j in range(100)]
L = list(get_input())
N = len(L) // 2 + 1
for l in range(N):
points = [int(i) for i in L[l].split(",")]
for i in range(len(points)):
table[len(points)-i-1][i] = points[i]
for l in range(N, 2*N-1):
#print(l)
points = [int(i) for i in L[l].split(",")]
for i in range(len(points)):
#print(len(points)-i-1+(l-N+1),i+(l-N+1))
table[len(points)-i-1+(l-N+1)][i+(l-N+1)] = points[i]
for c in range(1,2*N):
for i in range(c):
if c-i-2 < 0:
table[c-i-1][i] += table[c-i-1][i-1]
elif i-1 < 0:
table[c-i-1][i] += table[c-i-2][i]
else:
table[c-i-1][i] += max(table[c-i-2][i], table[c-i-1][i-1])
#for i in range(N):
# for j in range(N):
# print(table[i][j], end=" ")
# print("")
print(table[N-1][N-1])
```
Yes
| 93,416 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
ans = [int(input())]
while True:
try:
inp = list(map(int, input().split(",")))
l = len(inp)
if l > len(ans):
tmp = [-float("inf")] + ans + [-float("inf")]
else:
tmp = ans
ans = []
for i in range(l):
ans.append(max([tmp[i] + inp[i], tmp[i + 1] + inp[i]]))
except:
break
print(*ans)
```
Yes
| 93,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
ans = [int(input())]
import sys
for e in sys.stdin:
inp = list(map(int, e.split(",")))
l = len(inp)
if l > len(ans):
tmp = [-float("inf")] + ans + [-float("inf")]
else:
tmp = ans
ans = []
for i in range(l):
ans += [max([tmp[i] + inp[i], tmp[i + 1] + inp[i]])]
print(*ans)
```
Yes
| 93,418 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
count = 0
a = []
while(1):
try:
count += 1
a.append([int(i) for i in input().split(",")])
except EOFError:
break
dim = int(count/2)
b = [[0 for i in range(dim+1)] for j in range(dim+1)]
for i in range(dim):
for j in range(i+1):
b[i-j][j] = a[i][j]
for i in range(dim-1):
for j in range(dim-i-1):
b[dim-1-j][i+1+j] = a[i+dim][j]
dp_ = [[True if i < dim else 0 for i in range(dim+1)] if j < dim else [0 for i in range(dim+1)] for j in range(dim+1)]
def dp(i,j):
if dp_[i][j]:
dp_[i][j] = b[i][j] + max(dp(i+1,j),dp(i,j+1))
return dp_[i][j]
else:
return dp_[i][j]
print(dp(0,0))
```
No
| 93,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
def dmax(l):
for k in range(len(l)-1):
for (i,j) in enumerate(l[k]):
l[k+1].insert(2*i+1,j)
l[k+1]=[l[k+1][i+1]+j for (i,j) in enumerate(l[k+1][:-1])]
o=l[k+1][1:-1]
l[k+1]=[l[k+1][0]]+[o[i] if o[i]>=o[i+1] else o[i+1] for i in range(0,len(o),2)]+[l[k+1][-1]]
return max(l[-1])
cnt=0
l=[]
while 1:
t=list(map(int,input().split(",")))
l.extend([t])
if len(t)==1 and cnt>0:break
cnt+=1
b=len(l)//2+1
m=dmax(l[:b])+dmax(l[b:][::-1])
```
No
| 93,420 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
import sys
s=[list(map(int,e.split(',')))for e in sys.stdin]
for i in range(1,len(s)):
for j in range(len(s[i])):
t=j-len(s[i])>len(s[i-1])
s[i][j]+=max(s[i-1][t*(j>0):t+2])
print(*s[-1])
```
No
| 93,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Arrange integers (0 or more and 99 or less) in a rhombus as illustrated in Fig. 1. Create a program that reads the data representing the rhombus and outputs the maximum value of the sum of the integers that pass when starting from the top and proceeding to the bottom according to the following rules.
* At each step, you can proceed to the lower left diagonal or the lower right diagonal.
For example, in the example of Fig. 1, as shown in Fig. 2, when 7,3,8,7,5,7,8,3,7 is selected and passed, the sum is the maximum 55 (7 + 3 + 8). + 7 + 5 + 7 + 8 + 3 + 7 = 55).
Input
As shown in the input example, a comma-separated sequence of integers is given to the diamond. Each line does not contain whitespace. The input example corresponds to Fig. 1. There are less than 100 rows of data.
Output
Outputs the maximum value of the sum of integers that pass according to the rules on one line.
Example
Input
7
3,8
8,1,0
2,7,4,4
4,5,2,6,5
2,7,4,4
8,1,0
3,8
7
Output
55
Submitted Solution:
```
path=[]
while(True):
try:
temp=list(map(int,input().split(',')))
path.append(temp)
except:
break
```
No
| 93,422 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0221
"""
import sys
from sys import stdin
from collections import deque
input = stdin.readline
def fb_gen(count=1):
while True:
if count % 15 == 0:
ans = 'FizzBuzz'
elif count % 5 == 0:
ans = 'Buzz'
elif count % 3 == 0:
ans = 'Fizz'
else:
ans = str(count)
yield ans
count += 1
def main(args):
while True:
m, n = map(int, input().split())
if m == 0 and n == 0:
break
players = deque(range(1, m+1))
fb = fb_gen()
for _ in range(n):
p = input().strip()
if p != fb.__next__():
if len(players) > 1:
players.popleft()
else:
players.rotate(-1)
result = list(players)
result.sort()
print(*result)
if __name__ == '__main__':
main(sys.argv[1:])
```
| 93,423 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
while True:
m, n = map(int, input().split())
if m == 0:
break
plst = [i for i in range(1, m + 1)]
length = m
ind = 0
for i in range(1, n + 1):
s = input()
if length == 1:
continue
if i % 15 == 0:
if s != "FizzBuzz":
plst.pop(ind)
length -= 1
ind %= length
else:
ind += 1
ind %= length
elif i % 5 == 0:
if s != "Buzz":
plst.pop(ind)
length -= 1
ind %= length
else:
ind += 1
ind %= length
elif i % 3 == 0:
if s != "Fizz":
plst.pop(ind)
length -= 1
ind %= length
else:
ind += 1
ind %= length
else:
if s != str(i):
plst.pop(ind)
length -= 1
ind %= length
else:
ind += 1
ind %= length
print(*plst)
```
| 93,424 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
def fizz_buzz():
c = 1
while True:
res = ''
if c % 3 == 0:
res = res + 'Fizz'
if c % 5 == 0:
res = res + 'Buzz'
if res == '':
yield str(c)
else:
yield res
c += 1
while True:
m,n = map(int,input().split())
if m == 0: break
player = list(range(m))
p = 0
fb = fizz_buzz()
for i in range(n):
inp = input()
if len(player) > 1:
if inp != next(fb):
del player[p]
p = p % len(player)
else:
p = (p+1) % len(player)
result = str(player[0]+1)
if len(player) > 1:
for pi in player[1:]:
result += ' ' + str(pi+1)
print(result)
```
| 93,425 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
def solve():
from itertools import cycle
from sys import stdin
f_i = stdin
m_i = map(lambda x: x.rstrip(), f_i)
while True:
m, n = map(int, next(m_i).split())
if m == 0:
break
player = list(range(1, m + 1))
ans = ('FizzBuzz' if i % 15 == 0 else 'Fizz' if i % 3 == 0 else
'Buzz' if i % 5 == 0 else str(i) for i in range(1, n + 1))
while len(player) > 1:
for p, a in zip(cycle(player), ans):
n -= 1
if next(m_i) != a:
break
else:
break
idx = player.index(p)
player = player[idx+1:] + player[:idx]
while n:
next(m_i)
n -= 1
player.sort()
print(*player)
solve()
```
| 93,426 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
while(True):
m,n = map(int,input().split())
if not m: break
a = [True]*m
j = -1
for i in range(1,n+1):
if sum(a) < 2: input(); continue
j = (j+1+((a+a)[j+1:]).index(True))%m
s = input()
if not i%15:
if s == "FizzBuzz": continue
else: a[j] = False; continue
if not i%5:
if s == "Buzz": continue
else: a[j] = False; continue
if not i%3:
if s == "Fizz": continue
else: a[j] = False; continue
if s != str(i): a[j] = False;continue
print(" ".join(str(e+1) for e in range(m) if a[e]))
```
| 93,427 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
while(True):
m,n = map(int,input().split())
if not m: break
a = list(range(1,m+1))
b = [input() for _ in range(n)]
t = [str(i+1) for i in range(n)]
t[2::3] = ["Fizz"]*len(t[2::3])
t[4::5] = ["Buzz"]*len(t[4::5])
t[14::15] = ["FizzBuzz"]*len(t[14::15])
i=0
for j in range(n):
if len(a) <2: break
if b[j] != t[j]: del a[i]; i = 0 if i >= len(a) else i; continue
i += 1
i = i%len(a)
print(" ".join(str(e) for e in a))
```
| 93,428 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0221
"""
import sys
from sys import stdin
from collections import deque
input = stdin.readline
def fb_gen(count=1):
while True:
if count % 15 == 0:
ans = 'FizzBuzz'
elif count % 5 == 0:
ans = 'Buzz'
elif count % 3 == 0:
ans = 'Fizz'
else:
ans = str(count)
yield ans
count += 1
def main(args):
while True:
m, n = map(int, input().split())
if m == 0 and n == 0:
break
players = deque(range(1, m+1))
fb = fb_gen()
ok_count = 0
for _ in range(n):
p = input().strip()
if p != fb.__next__():
if len(players) > 1:
players.rotate(ok_count)
players.popleft()
ok_count = 0
else:
ok_count -= 1
result = list(players)
result.sort()
print(*result)
if __name__ == '__main__':
main(sys.argv[1:])
```
| 93,429 |
Provide a correct Python 3 solution for this coding contest problem.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
"Correct Solution:
```
while 1:
m,n=map(int,input().split())
if m==0:break
a,i,c=list(range(1,m+1)),0,0
while i<n:
i+=1
b,f=input(),0
if m<2:continue
if i%15==0:
if b!='FizzBuzz':
del a[c]
f=1
elif i%5==0:
if b!='Buzz':
del a[c]
f=1
elif i%3==0:
if b!='Fizz':
del a[c]
f=1
elif b!=str(i):
del a[c]
f=1
if f:m-=1
else:c+=1
c%=m
print(*a)
```
| 93,430 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
while(True):
m,n = map(int,input().split())
if not m: break
a = list(range(1,m+1))
b = [input() for _ in range(n)]
t = [str(i+1) for i in range(n)]
t[2::3] = ["Fizz"]*len(t[2::3])
t[4::5] = ["Buzz"]*len(t[4::5])
t[14::15] = ["FizzBuzz"]*len(t[14::15])
i=0
for j in range(n):
if len(a) <2: break
if b[j] != t[j]: del a[i]; i = i%len(a); continue
i += 1
i = i%len(a)
print(" ".join(str(e) for e in a))
```
Yes
| 93,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
hoge = ["FizzBuzz" if i % 15 == 0 else("Fizz" if i % 3 == 0 else("Buzz" if i % 5 == 0 else str(i))) for i in range(10001)]
while True:
m, n = map(int, input().split())
if m == 0:
break
man = list(range(1, m+1))
game = [input() for _ in range(n)]
idx = 0
for i in range(n):
if game[i] != hoge[i+1]:
del man[idx]
if len(man) == 1:
break
else:
idx += 1
idx %= len(man)
print(*man)
```
Yes
| 93,432 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
def fizzbuzz(i):
if i % 15 == 0:
return 'FizzBuzz'
elif i % 5 == 0:
return 'Buzz'
elif i % 3 == 0:
return 'Fizz'
else:
return str(i)
import sys
f = sys.stdin
while True:
m, n = map(int, f.readline().split())
if m == n == 0:
break
member = list(range(1, m + 1))
s = [f.readline().strip() for _ in range(n)]
pos = 0
for i in range(n):
if s[i] != fizzbuzz(i + 1):
del member[pos]
m = len(member)
if m == 1:
break
else:
pos += 1
pos %= m
print(*member)
```
Yes
| 93,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
def FB(n):
if n%3==0 and n%5==0: return 'FizzBuzz'
if n%3==0: return 'Fizz'
if n%5==0: return 'Buzz'
return str(n)
while True:
M,N=map(int,input().split())
if M==0 and N==0: break
TF=[1]*M
S=[input() for _ in range(N)]
k=0
for i,s in enumerate(S):
if sum(TF)==1: break
while True:
if not TF[k]:
k=(k+1)%M
else:
if FB(i+1)==s:
k=(k+1)%M
break
else:
TF[k]=0
k=(k+1)%M
break
ans=[]
for j,p in enumerate(TF):
if p:
ans.append(str(j+1))
print(' '.join(ans))
```
Yes
| 93,434 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
# Aizu Problem 0221: Fizz Buzz
import sys, math, os, struct
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def valid(k, a):
if k % 3 == 0 and k % 5 == 0:
return a == "FizzBuzz"
elif k % 3 == 0:
return a == "Fizz"
elif k % 5 == 0:
return a == "Buzz"
else:
return int(a) == k
def fizz_buzz(m, n, A):
players = list(range(1, m + 1))
p = 0
k = 0
while len(A) > 0:
k += 1
a = A.pop(0)
if valid(k, a):
p = (p + 1) % m
else:
if len(players) == 0:
print()
return
players.pop(p)
if len(players) == 1:
break
m -= 1
if p == m:
p = 0
print(' '.join([str(p) for p in players]))
while True:
m, n = [int(_) for _ in input().split()]
if m == 0:
break
A = [input().strip() for _ in range(n)]
fizz_buzz(m, n, A)
```
No
| 93,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
while(True):
m,n = map(int,input().split())
if not m: break
a = [True]*m
j = -1
for i in range(1,n+1):
if not sum(a): input(); continue
j = (j+1+((a+a)[j+1:]).index(True))%m
s = input()
if not i%15:
if s == "FizzBuzz": continue
else: a[j] = False; continue
if not i%5:
if s == "Buzz": continue
else: a[j] = False; continue
if not i%3:
if s == "Fizz": continue
else: a[j] = False; continue
if s != str(i): a[j] = False;continue
print(" ".join(str(e+1) for e in range(m) if a[e]))
```
No
| 93,436 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0221
"""
import sys
from sys import stdin
from collections import deque
input = stdin.readline
def FizzBuzz(count=1):
while True:
if count % 15 == 0:
ans = 'FizzBuzz'
elif count % 5 == 0:
ans = 'Buzz'
elif count % 3 == 0:
ans = 'Fizz'
else:
ans = str(count)
yield ans
count += 1
def main(args):
while True:
m, n = map(int, input().split())
if m == 0 and n == 0:
break
players = deque(range(1, m+1))
fb = FizzBuzz()
for i in range(1, n+1):
p = input().strip('\n')
if p != fb.__next__():
players.popleft()
else:
players.rotate(-1)
if len(players) == 1:
break
result = list(players)
result.sort()
print(*result)
if __name__ == '__main__':
main(sys.argv[1:])
```
No
| 93,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a game that uses numbers called "Fizz Buzz". In this game, multiple players count the numbers one by one, starting with 1, and each player says only one number after the previous player. At that time, you must say "Fizz" if it is divisible by 3, "Buzz" if it is divisible by 5, and "FizzBuzz" if it is divisible by both. For example, the first 16 statements are as follows:
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16, ・ ・ ・
Taro decided to play "Fizz Buzz" with his friends. Taro and his colleagues decided the rules as follows.
"The wrong person will drop out. The next person will start with the next number after the wrong number. That is, if you say 1, 2, 3, you made a mistake with 3, so you will start with 4."
I play the game according to this rule, but because I am not used to the game, I sometimes do not notice that I made a mistake, and I cannot make a fair decision. So you decided to create a program that outputs the remaining people at the end of the set number of remarks so that Taro and his friends can enjoy this game.
Create a program that inputs the number of players, the number of times spoken during the game, and each statement, and outputs the number of the remaining players at the end of the input in ascending order. However, the players are numbered from 1, and the order of speaking is also from the first player, and when the speaking is completed, the first player will speak again. If the player in turn has already dropped out, the next player will speak. Also, the program must ignore subsequent statements when the player is alone.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
m n
s1
s2
::
sn
The first line gives the number of players m (2 ≤ m ≤ 1000) and the number of remarks n (1 ≤ n ≤ 10000).
The next n lines are given the i-th statement s1. si is an integer, Fizz, Buzz, or a string (up to 8 characters) that indicates FizzBuzz.
The number of datasets does not exceed 50.
Output
For each input dataset, the number of the player remaining when the specified number of remarks is input is output in ascending order.
Example
Input
5 7
1
2
Fizz
4
Buzz
6
7
3 5
1
2
3
4
5
0 0
Output
2 3 4 5
1
Submitted Solution:
```
while 1:
m,n=map(int,input().split())
if m==0:break
a,i,c=[i+1 for i in range(m)],0,0
while i<n:
i+=1
b,f=input(),0
if len(a)==1:continue
if i%15==0:
if b!='FizzBuzz':
del a[c]
f=1
elif i%5==0:
if b!='Buzz':
del a[c]
f=1
elif i%3==0:
if b!='Fizz':
del a[c]
f=1
else:
if b!=str(i):
del a[c]
f=1
if f:m-=1
else:c=(c+1)%m
print(*a)
```
No
| 93,438 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
def ms_number(n):
count = 1
msnumber_list = [6]
while msnumber_list[-1] < n:
msnumber_list.append(7 * count + 1)
msnumber_list.append(7 * count + 6)
count += 1
return msnumber_list[:-1]
def ms_prime(n, msnumber_list):
msprime_list = []
msprime_flag = [0 for i in range(n + 1)]
for i in msnumber_list:
if msprime_flag[i] == 0:
msprime_list.append(i)
for j in msnumber_list:
if i * j > n:
break
msprime_flag[i * j] = 1
return msprime_list
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5) + 1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n // i)
divisors.sort()
return divisors
msnumber_list = ms_number(300000)
msprime_list = ms_prime(300000, msnumber_list)
answer = []
input_list = []
while True:
n = int(input())
if n == 1:
break
input_list.append(n)
divisors = make_divisors(n)
answer.append([d for d in divisors if d in msprime_list])
for i in range(len(input_list)):
print(str(input_list[i]) + ": ", end="")
print(*answer[i])
```
| 93,439 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
def modifiedSieve():
L = 300001
msPrimes = []
isMSPrime = [ True for _ in range(L) ]
for i in range(6):
isMSPrime[i] = False
for i in range(6, L):
if isMSPrime[i]:
if i % 7 == 1 or i % 7 == 6:
msPrimes.append(i)
for j in range(i * i, L, i):
isMSPrime[j] = False
else:
isMSPrime[i] = False
return msPrimes
def getMSFactors(num, msPrimes):
factors = set()
for msp in msPrimes:
if msp > num:
break
if num % msp == 0:
factors.add(msp)
if num in msPrimes:
factors.add(num)
return sorted(list(factors))
if __name__ == '__main__':
msPrimes = modifiedSieve()
while True:
num = int(input())
if num == 1:
break
factors = getMSFactors(num, msPrimes)
print("{}: {}".format(num, " ".join(list(map(str, factors)))))
```
| 93,440 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
a=[0]*6+[1,0,1,0,0,0,0]*42857
p=[]
for i in range(6,300000,7):
if a[i]:
p+=[i]
for j in range(i**2,300000,i):
a[j]=0
if a[i+2]:
p+=[i+2]
for j in range((i+2)**2,300000,i+2):
a[j]=0
while 1:
n=int(input())
if n==1:break
b=[x for x in p if n%x==0]
print('{}: {}'.format(n, ' '.join(map(str, b))))
```
| 93,441 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
def main():
while True:
data = int(input())
if data == 1: break
i = 1
factlist = []
while True:
item = 7 * i - 1
if item > int(data / 6) + 1: break
if data % item == 0:
quo = int(data / item)
if isms(quo) and ismsp(quo):
factlist.append(quo)
item = 7 * i + 1
if item > int(data / 6) + 1:break
if data % item == 0:
quo = int(data / item)
if isms(quo) and ismsp(quo):
factlist.append(quo)
i += 1
factlist.sort()
if factlist == []: factlist.append(data)
print(str(data) + ":", end = "")
for item in factlist:
print(" " + str(item), end = "")
print("")
def isms(num):
if num % 7 == 1 or num % 7 == 6: return True
else: return False
def ismsp(num):
i = 1
while True:
item = 7 * i - 1
if item > int(num / 6) + 1: break
if num % item == 0: return False
item = 7 * i + 1
if item > int(num / 6) + 1: break
if num % item == 0: return False
i += 1
return True
main()
```
| 93,442 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Prime():
def __init__(self, n):
self.M = m = int(math.sqrt(n)) + 10
self.A = a = [True] * m
a[0] = a[1] = False
self.T = t = []
for i in range(2, m):
if not a[i]:
continue
if not (i % 7 == 1 or i % 7 == 6):
continue
t.append(i)
for j in range(i*i,m,i):
a[j] = False
def is_prime(self, n):
return self.A[n]
def division(self, n):
d = []
for c in self.T:
if n < c:
break
if n % c == 0:
d.append(c)
return d
def main():
rr = []
pr = Prime(300000**2)
while True:
n = I()
if n == 1:
break
d = pr.division(n)
rr.append('{}: {}'.format(n,' '.join(map(str,d))))
return '\n'.join(map(str, rr))
print(main())
```
| 93,443 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
N = 300000
a = [((i % 7 ==1)or(i % 7 ==6)) for i in range(N)]
#print(a[11])
a[1]=0
p =[]
for i in range(6,N):
if a[i]:
p+=[i]
for j in range(i * i ,N,i):
a[j] = 0
while True:
n = int(input())
if n == 1:
break
print(n,end = ":")
for x in p:
if n % x ==0:
print(" "+str(x),end = "")
print()
```
| 93,444 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
from math import floor
# 約数全列挙(ソート済み)
# n = 24 -> [1,2,3,4,6,8,12,24]
def divisor(n):
left = [1]
right = [n]
sup = floor(pow(n,1/2))
for p in range(2,sup+1):
if n % p == 0:
if p == n//p:
left.append(p)
continue
left.append(p)
right.append(n//p)
res = left + right[::-1]
return res
ans_list = []
def main():
while True:
ans = solve()
if ans == "end":
break
ans_list.append(ans)
for ans in ans_list:
print(ans)
def solve():
n = int(input())
if n == 1:
return "end"
cand = [c for c in divisor(n) if c % 7 in (1,6)]
ans = "{}:".format(n)
for c in cand:
flag = True
for p in [p for p in divisor(c) if p % 7 in (1,6)]:
if p not in (1,c):
flag = False
break
if c > 1 and flag is True:
ans += " {}".format(c)
return ans
main()
```
| 93,445 |
Provide a correct Python 3 solution for this coding contest problem.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
"Correct Solution:
```
p = [1]*300010
p[0]=p[1]=0
for i in range(2, 300010):
if not(i%7==1 or i%7==6):
p[i] = 0
for i in range(2, 300010):
if p[i]:
for j in range(i*2, 300010, i):
p[j] = 0
doyososu = []
for i in range(2, 300010):
if p[i]:
doyososu.append(i)
while True:
n = int(input())
if n==1:
break
ans = []
for e in doyososu:
if n%e==0:
ans.append(e)
print(n, ":", sep="", end=" ")
print(*ans)
```
| 93,446 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def monsat(num):
L = [False for i in range(num+1)]
for i in range(0,num,7):
if i == 1 or i == 0:
continue
for d in [-1,1]:
if L[i+d] == False:
L[i+d] = True
for j in range((i+d)*2,num,i+d):
if L[j] is False:
L[j] = [i+d]
else:
L[j].append(i+d)
return L
L = monsat(3*10**5+1)
while True:
n = int(input())
if n == 1:
break
print(str(n) + ':',n if L[n] is True else ' '.join(map(str,L[n])))
```
Yes
| 93,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
sieve = [0] * 300001
for i in range(6, 300000, 7):
sieve[i] = sieve[i+2] = 1
MSprimes = []
for i in range(6, 300000, 7):
if sieve[i] == 1:
MSprimes.append(i)
for j in range(2*i, 300000, i):
sieve[j] = 0
if sieve[i+2] == 1:
MSprimes.append(i+2)
for j in range(2*(i+2), 300000, i+2):
sieve[j] = 0
while True:
N = int(input())
if N == 1: break
ansp = [x for x in MSprimes if N % x == 0]
print('{}: {}'.format(N, ' '.join(map(str, ansp))))
```
Yes
| 93,448 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
# coding: utf-8
isgdprime=[1 for i in range(300001)]
gdprimes=[]
for i in range(6):
isgdprime[i]=0
for i in range(300001):
if isgdprime[i]==1 and (i%7==1 or i%7==6):
gdprimes.append(i)
j=i*2
while j<300001:
isgdprime[j]=0
j+=i
else:
isgdprime[i]=0
while 1:
n=int(input())
if n==1:
break
print(n,end=':')
for i in range(len(gdprimes)):
if n%gdprimes[i]==0:
print('',gdprimes[i],end='')
print()
```
Yes
| 93,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
from array import array
size = 300000
s = array('i', range(size + 1))
for i in range(size+1):
if i % 7 != 1 and i % 7 != 6:
s[i] = 0
s[1] = 0
sqrtn = int(round(size ** 0.5))
for i in range(6, sqrtn + 1):
if s[i]:
s[i * i: size + 1: i] = array('i', [0] * len(range(i*i, size+1, i)))
primes = list(filter(None, s))
while True:
n = int(input())
if n == 1: break
i = 0
ls = []
for p in primes:
if p > n: break
if n % p == 0:
ls.append(p)
print(n, end="")
print(':', end=" ")
print(*ls)
```
Yes
| 93,450 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
this_prime_num = [0 for i in range(300001)]
for i in range(300001):
if i % 7 == 1 or i % 7 == 6:
this_prime_num[i] = 1
for i in range(2, 300001):
j = i * 2
while this_prime_num[i] == 1 and j < len(this_prime_num):
#print(j)
this_prime_num[j] = 0
j += i
this_primenum_list = []
for i in range(2, len(this_prime_num)):
if this_prime_num[i]:
this_primenum_list.append(i)
print(this_primenum_list[:30])
ans_list = []
while True:
n = int(input())
if n == 1:
break
i = 0
ans = [n]
while this_primenum_list[i] <= n:
#print(i)
if n % this_primenum_list[i] == 0:
ans.append(this_primenum_list[i])
i += 1
if i >= len(this_primenum_list):
break
ans_list.append(ans)
for i in ans_list:
print(str(i[0]) + ":", end = "")
for j in i[1:]:
print("", end = " ")
print(j, end = "")
print()
```
No
| 93,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
def factors(n):
sieve = [1]*(n+10)
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] == 0: continue
for pp in range(2*p,n+1,p):
sieve[pp] = 0
m += 1
ret = []
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] and n % p == 0:
ret.append(p)
m += 1
return(ret)
while True:
N = int(input())
if N == 1: break
ans = ' '.join(list(map(str,factors(N))))
print('{0}: {1}'.format(N,ans))
```
No
| 93,452 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
def factors(n):
sieve = [1]*(n+10)
ret = []
m = 1
while m*7-1 <= n:
for p in [m*7-1,m*7+1]:
if sieve[p] == 0: continue
if n%p == 0:
ret.append(p)
for pp in range(2*p,n+1,p):
sieve[pp] = 0
m += 1
return(ret)
while True:
N = int(input())
if N == 1: break
ans = ' '.join(list(map(str,factors(N))))
print('{0}: {1}'.format(N,ans))
```
No
| 93,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chief Judge's log, stardate 48642.5. We have decided to make a problem from elementary number theory. The problem looks like finding all prime factors of a positive integer, but it is not.
A positive integer whose remainder divided by 7 is either 1 or 6 is called a 7N+{1,6} number. But as it is hard to pronounce, we shall call it a Monday-Saturday number.
For Monday-Saturday numbers a and b, we say a is a Monday-Saturday divisor of b if there exists a Monday-Saturday number x such that ax = b. It is easy to show that for any Monday-Saturday numbers a and b, it holds that a is a Monday-Saturday divisor of b if and only if a is a divisor of b in the usual sense.
We call a Monday-Saturday number a Monday-Saturday prime if it is greater than 1 and has no Monday-Saturday divisors other than itself and 1. A Monday-Saturday number which is a prime in the usual sense is a Monday-Saturday prime but the converse does not always hold. For example, 27 is a Monday-Saturday prime although it is not a prime in the usual sense. We call a Monday-Saturday prime which is a Monday-Saturday divisor of a Monday-Saturday number a a Monday-Saturday prime factor of a. For example, 27 is one of the Monday-Saturday prime factors of 216, since 27 is a Monday-Saturday prime and 216 = 27 × 8 holds.
Any Monday-Saturday number greater than 1 can be expressed as a product of one or more Monday-Saturday primes. The expression is not always unique even if differences in order are ignored. For example, 216 = 6 × 6 × 6 = 8 × 27 holds.
Our contestants should write a program that outputs all Monday-Saturday prime factors of each input Monday-Saturday number.
Input
The input is a sequence of lines each of which contains a single Monday-Saturday number. Each Monday-Saturday number is greater than 1 and less than 300000 (three hundred thousand). The end of the input is indicated by a line containing a single digit 1.
Output
For each input Monday-Saturday number, it should be printed, followed by a colon `:' and the list of its Monday-Saturday prime factors on a single line. Monday-Saturday prime factors should be listed in ascending order and each should be preceded by a space. All the Monday-Saturday prime factors should be printed only once even if they divide the input Monday-Saturday number more than once.
Sample Input
205920
262144
262200
279936
299998
1
Output for the Sample Input
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Example
Input
205920
262144
262200
279936
299998
1
Output
205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998
Submitted Solution:
```
S = [0 for i in range(300000)]
A = [-1, 1]
for i in range(1, int(300000 / 7) + 1):
d = 7 * i
for a in A:
da = d + a
if da < 300000 and S[da] == 0:
S[da] = 1
for j in range(2, int(300000 / 2)):
if da * j >= 300000:break
S[da * j] = -1
while 1:
n = int(input())
if n == 1:break
res = [str(n) + ':']
for i in range(2, n+1):
if S[i] == 1 and n % i == 0:
res.append(str(i))
print(' '.join(res))
```
No
| 93,454 |
Provide a correct Python 3 solution for this coding contest problem.
Do you know "sed," a tool provided with Unix? Its most popular use is to substitute every occurrence of a string contained in the input string (actually each input line) with another string β. More precisely, it proceeds as follows.
1. Within the input string, every non-overlapping (but possibly adjacent) occurrences of α are marked. If there is more than one possibility for non-overlapping matching, the leftmost one is chosen.
2. Each of the marked occurrences is substituted with β to obtain the output string; other parts of the input string remain intact.
For example, when α is "aa" and β is "bca", an input string "aaxaaa" will produce "bcaxbcaa", but not "aaxbcaa" nor "bcaxabca". Further application of the same substitution to the string "bcaxbcaa" will result in "bcaxbcbca", but this is another substitution, which is counted as the second one.
In this problem, a set of substitution pairs (αi, βi) (i = 1, 2, ... , n), an initial string γ, and a final string δ are given, and you must investigate how to produce δ from γ with a minimum number of substitutions. A single substitution (αi, βi) here means simultaneously substituting all the non-overlapping occurrences of αi, in the sense described above, with βi.
You may use a specific substitution (αi, βi ) multiple times, including zero times.
Input
The input consists of multiple datasets, each in the following format.
n
α1 β1
α2 β2
.
.
.
αn βn
γ δ
n is a positive integer indicating the number of pairs. αi and βi are separated by a single space. You may assume that 1 ≤ |αi| < |βi| ≤ 10 for any i (|s| means the length of the string s), αi ≠ αj for any i ≠ j, n ≤ 10 and 1 ≤ |γ| < |δ| ≤ 10. All the strings consist solely of lowercase letters. The end of the input is indicated by a line containing a single zero.
Output
For each dataset, output the minimum number of substitutions to obtain δ from γ. If δ cannot be produced from γ with the given set of substitutions, output -1.
Example
Input
2
a bb
b aa
a
bbbbbbbb
1
a aa
a
aaaaa
3
ab aab
abc aadc
ad dee
abc
deeeeeeeec
10
a abc
b bai
c acf
d bed
e abh
f fag
g abe
h bag
i aaj
j bbb
a
abacfaabe
0
Output
3
-1
7
4
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [LS() for _ in range(n)]
s = S()
k = S()
l = len(k)
if s == k:
rr.append(0)
continue
u = set([s])
c = set([s])
t = 0
r = -1
while c:
t += 1
ns = set()
for cs in c:
for d,e in a:
nt = cs.replace(d,e)
if len(nt) > l:
continue
ns.add(nt)
if k in ns:
r = t
break
c = ns - u
u |= ns
rr.append(r)
return '\n'.join(map(str,rr))
print(main())
```
| 93,455 |
Provide a correct Python 3 solution for this coding contest problem.
Do you know "sed," a tool provided with Unix? Its most popular use is to substitute every occurrence of a string contained in the input string (actually each input line) with another string β. More precisely, it proceeds as follows.
1. Within the input string, every non-overlapping (but possibly adjacent) occurrences of α are marked. If there is more than one possibility for non-overlapping matching, the leftmost one is chosen.
2. Each of the marked occurrences is substituted with β to obtain the output string; other parts of the input string remain intact.
For example, when α is "aa" and β is "bca", an input string "aaxaaa" will produce "bcaxbcaa", but not "aaxbcaa" nor "bcaxabca". Further application of the same substitution to the string "bcaxbcaa" will result in "bcaxbcbca", but this is another substitution, which is counted as the second one.
In this problem, a set of substitution pairs (αi, βi) (i = 1, 2, ... , n), an initial string γ, and a final string δ are given, and you must investigate how to produce δ from γ with a minimum number of substitutions. A single substitution (αi, βi) here means simultaneously substituting all the non-overlapping occurrences of αi, in the sense described above, with βi.
You may use a specific substitution (αi, βi ) multiple times, including zero times.
Input
The input consists of multiple datasets, each in the following format.
n
α1 β1
α2 β2
.
.
.
αn βn
γ δ
n is a positive integer indicating the number of pairs. αi and βi are separated by a single space. You may assume that 1 ≤ |αi| < |βi| ≤ 10 for any i (|s| means the length of the string s), αi ≠ αj for any i ≠ j, n ≤ 10 and 1 ≤ |γ| < |δ| ≤ 10. All the strings consist solely of lowercase letters. The end of the input is indicated by a line containing a single zero.
Output
For each dataset, output the minimum number of substitutions to obtain δ from γ. If δ cannot be produced from γ with the given set of substitutions, output -1.
Example
Input
2
a bb
b aa
a
bbbbbbbb
1
a aa
a
aaaaa
3
ab aab
abc aadc
ad dee
abc
deeeeeeeec
10
a abc
b bai
c acf
d bed
e abh
f fag
g abe
h bag
i aaj
j bbb
a
abacfaabe
0
Output
3
-1
7
4
"Correct Solution:
```
from collections import deque
def testcase_ends():
n = int(input())
if n == 0:
return 1
sub = [input().split() for i in range(n)]
gamma = input()
delta = input()
q = deque()
q.append((gamma, 0))
res = 10000
while q:
s, i = q.popleft()
for alpha, beta in sub:
if alpha not in s: continue
t = s.replace(alpha, beta)
if t == delta:
res = min(res, i+1)
continue
elif len(t) >= len(delta):
continue
q.append((t, i+1))
if res == 10000: res = -1
print(res)
return 0
def main():
while not testcase_ends():
pass
if __name__ == '__main__':
main()
```
| 93,456 |
Provide a correct Python 3 solution for this coding contest problem.
Do you know "sed," a tool provided with Unix? Its most popular use is to substitute every occurrence of a string contained in the input string (actually each input line) with another string β. More precisely, it proceeds as follows.
1. Within the input string, every non-overlapping (but possibly adjacent) occurrences of α are marked. If there is more than one possibility for non-overlapping matching, the leftmost one is chosen.
2. Each of the marked occurrences is substituted with β to obtain the output string; other parts of the input string remain intact.
For example, when α is "aa" and β is "bca", an input string "aaxaaa" will produce "bcaxbcaa", but not "aaxbcaa" nor "bcaxabca". Further application of the same substitution to the string "bcaxbcaa" will result in "bcaxbcbca", but this is another substitution, which is counted as the second one.
In this problem, a set of substitution pairs (αi, βi) (i = 1, 2, ... , n), an initial string γ, and a final string δ are given, and you must investigate how to produce δ from γ with a minimum number of substitutions. A single substitution (αi, βi) here means simultaneously substituting all the non-overlapping occurrences of αi, in the sense described above, with βi.
You may use a specific substitution (αi, βi ) multiple times, including zero times.
Input
The input consists of multiple datasets, each in the following format.
n
α1 β1
α2 β2
.
.
.
αn βn
γ δ
n is a positive integer indicating the number of pairs. αi and βi are separated by a single space. You may assume that 1 ≤ |αi| < |βi| ≤ 10 for any i (|s| means the length of the string s), αi ≠ αj for any i ≠ j, n ≤ 10 and 1 ≤ |γ| < |δ| ≤ 10. All the strings consist solely of lowercase letters. The end of the input is indicated by a line containing a single zero.
Output
For each dataset, output the minimum number of substitutions to obtain δ from γ. If δ cannot be produced from γ with the given set of substitutions, output -1.
Example
Input
2
a bb
b aa
a
bbbbbbbb
1
a aa
a
aaaaa
3
ab aab
abc aadc
ad dee
abc
deeeeeeeec
10
a abc
b bai
c acf
d bed
e abh
f fag
g abe
h bag
i aaj
j bbb
a
abacfaabe
0
Output
3
-1
7
4
"Correct Solution:
```
def substitute(text, a, b):
pos = 0
newText = text
L1 = len(a)
L2 = len(b)
while True:
idx = newText.find(a, pos)
if idx < 0:
return newText
newText = newText[:idx] + b + newText[idx + L1:]
pos = idx + L2
return newText
def transform(orig, goal, count):
global subs
global minCount
if len(orig) > len(goal):
return
if orig == goal:
minCount = min(minCount, count)
return
for key in subs:
newStr = substitute(orig, key, subs[key])
if newStr != orig:
transform(newStr, goal, count + 1)
if __name__ == '__main__':
while True:
N = int(input())
if N == 0:
break
subs = {}
for _ in range(N):
a, b = input().strip().split()
subs[a] = b
orig = input().strip()
goal = input().strip()
minCount = 999999999
transform(orig, goal, 0)
if minCount == 999999999:
print(-1)
else:
print(minCount)
```
| 93,457 |
Provide a correct Python 3 solution for this coding contest problem.
Do you know "sed," a tool provided with Unix? Its most popular use is to substitute every occurrence of a string contained in the input string (actually each input line) with another string β. More precisely, it proceeds as follows.
1. Within the input string, every non-overlapping (but possibly adjacent) occurrences of α are marked. If there is more than one possibility for non-overlapping matching, the leftmost one is chosen.
2. Each of the marked occurrences is substituted with β to obtain the output string; other parts of the input string remain intact.
For example, when α is "aa" and β is "bca", an input string "aaxaaa" will produce "bcaxbcaa", but not "aaxbcaa" nor "bcaxabca". Further application of the same substitution to the string "bcaxbcaa" will result in "bcaxbcbca", but this is another substitution, which is counted as the second one.
In this problem, a set of substitution pairs (αi, βi) (i = 1, 2, ... , n), an initial string γ, and a final string δ are given, and you must investigate how to produce δ from γ with a minimum number of substitutions. A single substitution (αi, βi) here means simultaneously substituting all the non-overlapping occurrences of αi, in the sense described above, with βi.
You may use a specific substitution (αi, βi ) multiple times, including zero times.
Input
The input consists of multiple datasets, each in the following format.
n
α1 β1
α2 β2
.
.
.
αn βn
γ δ
n is a positive integer indicating the number of pairs. αi and βi are separated by a single space. You may assume that 1 ≤ |αi| < |βi| ≤ 10 for any i (|s| means the length of the string s), αi ≠ αj for any i ≠ j, n ≤ 10 and 1 ≤ |γ| < |δ| ≤ 10. All the strings consist solely of lowercase letters. The end of the input is indicated by a line containing a single zero.
Output
For each dataset, output the minimum number of substitutions to obtain δ from γ. If δ cannot be produced from γ with the given set of substitutions, output -1.
Example
Input
2
a bb
b aa
a
bbbbbbbb
1
a aa
a
aaaaa
3
ab aab
abc aadc
ad dee
abc
deeeeeeeec
10
a abc
b bai
c acf
d bed
e abh
f fag
g abe
h bag
i aaj
j bbb
a
abacfaabe
0
Output
3
-1
7
4
"Correct Solution:
```
# coding: utf-8
import queue
while 1:
n=int(input())
if n==0:
break
dic={}
for i in range(n):
k,s=input().split()
dic[k]=s
s=input()
ans=input()
q=queue.Queue()
q.put((s,0))
find=False
while not q.empty():
s=q.get()
if s[0]==ans:
print(s[1])
q=queue.Queue()
find=True
if not find:
for k in dic.keys():
t=s[0].replace(k,dic[k])
if len(t)<=len(ans) and s[0]!=t:
q.put((t,s[1]+1))
if not find:
print(-1)
```
| 93,458 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you know "sed," a tool provided with Unix? Its most popular use is to substitute every occurrence of a string contained in the input string (actually each input line) with another string β. More precisely, it proceeds as follows.
1. Within the input string, every non-overlapping (but possibly adjacent) occurrences of α are marked. If there is more than one possibility for non-overlapping matching, the leftmost one is chosen.
2. Each of the marked occurrences is substituted with β to obtain the output string; other parts of the input string remain intact.
For example, when α is "aa" and β is "bca", an input string "aaxaaa" will produce "bcaxbcaa", but not "aaxbcaa" nor "bcaxabca". Further application of the same substitution to the string "bcaxbcaa" will result in "bcaxbcbca", but this is another substitution, which is counted as the second one.
In this problem, a set of substitution pairs (αi, βi) (i = 1, 2, ... , n), an initial string γ, and a final string δ are given, and you must investigate how to produce δ from γ with a minimum number of substitutions. A single substitution (αi, βi) here means simultaneously substituting all the non-overlapping occurrences of αi, in the sense described above, with βi.
You may use a specific substitution (αi, βi ) multiple times, including zero times.
Input
The input consists of multiple datasets, each in the following format.
n
α1 β1
α2 β2
.
.
.
αn βn
γ δ
n is a positive integer indicating the number of pairs. αi and βi are separated by a single space. You may assume that 1 ≤ |αi| < |βi| ≤ 10 for any i (|s| means the length of the string s), αi ≠ αj for any i ≠ j, n ≤ 10 and 1 ≤ |γ| < |δ| ≤ 10. All the strings consist solely of lowercase letters. The end of the input is indicated by a line containing a single zero.
Output
For each dataset, output the minimum number of substitutions to obtain δ from γ. If δ cannot be produced from γ with the given set of substitutions, output -1.
Example
Input
2
a bb
b aa
a
bbbbbbbb
1
a aa
a
aaaaa
3
ab aab
abc aadc
ad dee
abc
deeeeeeeec
10
a abc
b bai
c acf
d bed
e abh
f fag
g abe
h bag
i aaj
j bbb
a
abacfaabe
0
Output
3
-1
7
4
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
inline long long GetTSC() {
long long lo, hi;
asm volatile("rdtsc" : "=a"(lo), "=d"(hi));
return lo + (hi << 32);
}
inline double GetSeconds() { return GetTSC() / 2.8e9; }
const long inf = pow(10, 15);
int di[] = {-1, 0, 1, 0};
int dj[] = {0, 1, 0, -1};
tuple<int, int> q[50 * 50 * 50 * 50];
void solve() {
double starttime = GetSeconds();
while (1) {
int w, h;
cin >> w >> h;
if (w == 0 && h == 0) break;
int lp, ls, rp, rs;
bool lb[2500] = {};
bool rb[2500] = {};
bool f[2500][2500] = {};
for (int i = 0; i < h; i++) {
string s;
cin >> s;
for (int j = 0; j < w; j++) {
if (s[j] == '#') lb[i * w + j] = 1;
if (s[j] == '%') lp = i * w + j;
if (s[j] == 'L') ls = i * w + j;
}
cin >> s;
for (int j = 0; j < w; j++) {
if (s[j] == '#') rb[i * w + j] = 1;
if (s[j] == '%') rp = i * w + j;
if (s[j] == 'R') rs = i * w + j;
}
}
f[ls][rs] = 1;
int qi = 0;
int qe = 1;
q[0] = make_tuple(ls, rs);
bool ff = 0;
while (qi < qe) {
int lt = get<0>(q[qi]);
int rt = get<1>(q[qi]);
int li = lt / w;
int lj = lt % w;
int ri = rt / w;
int rj = rt % w;
qi++;
for (int i = 0; i < 4; i++) {
int nli = li + di[i];
int nlj = lj + dj[i];
int nri = ri + di[i];
int nrj = rj - dj[i];
if (nli < 0 || nli >= h || nlj < 0 || nlj >= w || lb[nli * w + nlj]) {
nli = li;
nlj = lj;
}
if (nri < 0 || nri >= h || nrj < 0 || nrj >= w || rb[nri * w + nrj]) {
nri = ri;
nrj = rj;
}
int nlt = nli * w + nlj;
int nrt = nri * w + nrj;
if (nlt == lp && nrt == rp) {
ff = 1;
break;
}
if (nlt == lp || nrt == rp) continue;
if (f[nlt][nrt]) continue;
f[nlt][nrt] = 1;
q[qe++] = make_tuple(nlt, nrt);
}
if (ff) break;
}
if (ff) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
}
return;
}
int main() {
solve();
return 0;
}
```
No
| 93,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Taro hides important books in the school locker, so he manages them more strictly than other people, and in addition to the keys provided by the school, he has installed the following button authentication type keys. ..
<image>
However, Taro, who easily forgets his password, has a habit of making it possible to write his password in one stroke. Even more sweet Taro wrote the candidates on paper when he thought about his password. Jiro, who happened to pick up the paper, tried to open the locker without throwing it away because he had a bad personality.
However, there are as many as 1000 password proposals written on the paper, and it seems that the sun will go down if you try all the inputs. Therefore, Jiro, who knows Taro's habits, decided to pick up only the ones that can be written with one stroke from the 1000 passwords in order to reduce the number of examinations, and asked the programmer to do the work.
There is always one or more candidates for each dataset. The rules for one-stroke writing are shown below.
1. The same characters are not consecutive. For example, AAA is not allowed.
2. One-stroke writing is possible in four directions, up, down, left and right. Do not move diagonally.
3. It does not go out of the frame of the button array and continue. Movement such as ABCA is not allowed.
4. You can pass over the characters that you have passed once.
5. Input of only one character of the button is regarded as one-stroke writing.
Input
The input consists of a group of 1000 character strings as shown below, and each line is an uppercase alphabetic string of 1 to 10 characters from A to I. It is assumed that there is no duplication of character string data.
string1
string2
...
string1000
Output
Extract only the candidate passwords (candidatePassword) from the input and list them as follows. In the following cases, N candidates are listed.
candidatePassword1
candidatePassword2
...
candidatePasswordN
The output is the same as the order of the input strings and the order must not be changed.
Example
Input
ABCFI
ABCABCABC
AEI
EFC
（中略）
DEHED
EEEEE
（以上でちょうど1000個）
Output
ABCFI
EFC
（中略）
DEHED
Submitted Solution:
```
dic = {}
dic["A"] = "BD"
dic["B"] = "ACE"
dic["C"] = "BF"
dic["D"] = "AEG"
dic["E"] = "BDFH"
dic["F"] = "CEI"
dic["G"] = "DH"
dic["H"] = "EGI"
dic["I"] = "FH"
results = []
for i in range(2):
pwd = input()
frag = 1
for j in range(len(pwd) - 1):
if pwd[j] in dic[pwd[j + 1]]:
a = 1
else:
frag = 0
break
if frag:
results.append(pwd)
for i in range(len(results)):
print(results[i])
```
No
| 93,460 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Taro hides important books in the school locker, so he manages them more strictly than other people, and in addition to the keys provided by the school, he has installed the following button authentication type keys. ..
<image>
However, Taro, who easily forgets his password, has a habit of making it possible to write his password in one stroke. Even more sweet Taro wrote the candidates on paper when he thought about his password. Jiro, who happened to pick up the paper, tried to open the locker without throwing it away because he had a bad personality.
However, there are as many as 1000 password proposals written on the paper, and it seems that the sun will go down if you try all the inputs. Therefore, Jiro, who knows Taro's habits, decided to pick up only the ones that can be written with one stroke from the 1000 passwords in order to reduce the number of examinations, and asked the programmer to do the work.
There is always one or more candidates for each dataset. The rules for one-stroke writing are shown below.
1. The same characters are not consecutive. For example, AAA is not allowed.
2. One-stroke writing is possible in four directions, up, down, left and right. Do not move diagonally.
3. It does not go out of the frame of the button array and continue. Movement such as ABCA is not allowed.
4. You can pass over the characters that you have passed once.
5. Input of only one character of the button is regarded as one-stroke writing.
Input
The input consists of a group of 1000 character strings as shown below, and each line is an uppercase alphabetic string of 1 to 10 characters from A to I. It is assumed that there is no duplication of character string data.
string1
string2
...
string1000
Output
Extract only the candidate passwords (candidatePassword) from the input and list them as follows. In the following cases, N candidates are listed.
candidatePassword1
candidatePassword2
...
candidatePasswordN
The output is the same as the order of the input strings and the order must not be changed.
Example
Input
ABCFI
ABCABCABC
AEI
EFC
（中略）
DEHED
EEEEE
（以上でちょうど1000個）
Output
ABCFI
EFC
（中略）
DEHED
Submitted Solution:
```
movable = {"A":("B","D","\n"),"B":("A","C","E","\n"),"C":("B","F","\n"),"D":("A","E","G","\n"),"E":("D","B","F","H","\n"),"F":("C","E","I","\n"),"G":("D","H","I","\n"),"H":("G","E","I","\n"),"I":("H","F","\n")}
A = [""]*1000
for i in range(1000):
A[i] = input()
B = []
def judge(string):
counter = 0
for i in range(len(string)-1):
if string[i+1] in movable[string[i]]:
counter += 1
else :
break
return len(string) == counter
for po in A:
if judge(po):
B.append(po)
else:
continue
for pu in B:
print(pu)
```
No
| 93,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
Taro hides important books in the school locker, so he manages them more strictly than other people, and in addition to the keys provided by the school, he has installed the following button authentication type keys. ..
<image>
However, Taro, who easily forgets his password, has a habit of making it possible to write his password in one stroke. Even more sweet Taro wrote the candidates on paper when he thought about his password. Jiro, who happened to pick up the paper, tried to open the locker without throwing it away because he had a bad personality.
However, there are as many as 1000 password proposals written on the paper, and it seems that the sun will go down if you try all the inputs. Therefore, Jiro, who knows Taro's habits, decided to pick up only the ones that can be written with one stroke from the 1000 passwords in order to reduce the number of examinations, and asked the programmer to do the work.
There is always one or more candidates for each dataset. The rules for one-stroke writing are shown below.
1. The same characters are not consecutive. For example, AAA is not allowed.
2. One-stroke writing is possible in four directions, up, down, left and right. Do not move diagonally.
3. It does not go out of the frame of the button array and continue. Movement such as ABCA is not allowed.
4. You can pass over the characters that you have passed once.
5. Input of only one character of the button is regarded as one-stroke writing.
Input
The input consists of a group of 1000 character strings as shown below, and each line is an uppercase alphabetic string of 1 to 10 characters from A to I. It is assumed that there is no duplication of character string data.
string1
string2
...
string1000
Output
Extract only the candidate passwords (candidatePassword) from the input and list them as follows. In the following cases, N candidates are listed.
candidatePassword1
candidatePassword2
...
candidatePasswordN
The output is the same as the order of the input strings and the order must not be changed.
Example
Input
ABCFI
ABCABCABC
AEI
EFC
（中略）
DEHED
EEEEE
（以上でちょうど1000個）
Output
ABCFI
EFC
（中略）
DEHED
Submitted Solution:
```
N = 1000
A = ["B","D"]
B = ["A","C","E"]
C = ["B","F"]
D = ["A","E","G"]
E = ["B","D","F","H"]
F = ["C","E","I"]
G = ["D","H"]
H = ["E","G","I"]
I = ["F","I"]
def check(string):
if len(string) == 1:
return True
for i in range(len(string)-1):
if string[i] == "A" and not(string[i+1] in A):
return False
if string[i] == "B" and not(string[i+1] in B):
return False
if string[i] == "C" and not(string[i+1] in C):
return False
if string[i] == "D" and not(string[i+1] in D):
return False
if string[i] == "E" and not(string[i+1] in E):
return False
if string[i] == "F" and not(string[i+1] in F):
return False
if string[i] == "G" and not(string[i+1] in G):
return False
if string[i] == "H" and not(string[i+1] in H):
return False
if string[i] == "I" and not(string[i+1] in I):
return False
return True
for i in range(N):
s = str(input())
if check(s):
print(s)
```
No
| 93,462 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
INF=10**5
while True:
n=int(input())
if n==0:break
imos=[0 for i_ in range(INF)]
for _ in range(n):
a,b=input().split()
s1=list(map(int,a.split(':')))
s2=list(map(int,b.split(':')))
s1=s1[0]*3600+s1[1]*60+s1[2]
s2=s2[0]*3600+s2[1]*60+s2[2]
imos[s1]+=1
imos[s2]-=1
for i in range(INF-1):
imos[i+1]+=imos[i]
print(max(imos))
```
| 93,463 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
from itertools import accumulate
def get_second(time_str):
return int(time_str[:2]) * 3600 + int(time_str[3:5]) * 60 + int(time_str[6:])
while True:
n = int(input())
if n == 0:
break
cnt = [0] * 86400
for _ in range(n):
dep_str, arr_str = input().strip().split()
dep, arr = get_second(dep_str), get_second(arr_str)
cnt[dep] += 1
cnt[arr] -= 1
print(max(accumulate(cnt)))
```
| 93,464 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
def solve():
while True:
N = int(input())
if N==0:
return
firsts = [0]*(86401)
lasts = [0]*(86401)
for i in range(N):
a,b = input().split()
h,m,s = a.split(':')
t = 0
for i,n in enumerate([h,m,s]):
if n[0]=='0':
n = n[1]
t += int(n)*60**(2-i)
firsts[t] += 1
h,m,s = b.split(':')
t = 0
for i,n in enumerate([h,m,s]):
if n[0]=='0':
n = n[1]
t += int(n)*60**(2-i)
lasts[t] += 1
now = 0
ans = 0
for f,l in zip(firsts,lasts):
now += f-l
ans = max(ans,now)
print(ans)
solve()
```
| 93,465 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
# 山手線は34.5Km
# 一周一時間かかる
# 駅は29駅ある
import itertools
while True:
n = int(input()) # n <= 10000
if n == 0:
break
timestamps = [0 for _ in range(24 * 60 * 60 + 1)]
for _ in range(n):
start, end = input().split()
start_h, start_m, start_s = map(int, start.split(":"))
start_h = start_h * 60 * 60
start_m = start_m * 60
start_time = start_h + start_m + start_s
end_h, end_m, end_s = map(int, end.split(":"))
end_h = end_h * 60 * 60
end_m = end_m * 60
end_time = end_h + end_m + end_s
timestamps[start_time + 1] += 1
timestamps[end_time + 1] -= 1
timestamps = list(itertools.accumulate(timestamps))
print(max(timestamps))
```
| 93,466 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
import sys
# import math
# import bisect
# import copy
# import heapq
# from collections import deque
# import decimal
# sys.setrecursionlimit(100001)
# INF = sys.maxsize
# MOD = 10 ** 9 + 7
# ===CODE===
def main():
while True:
n = int(input())
if n == 0:
exit(0)
d = [[[0 for i in range(60)] for j in range(60)] for k in range(24)]
for _ in range(n):
a, b = input().split()
a_h, a_m, a_s = a.split(":")
b_h,b_m,b_s = b.split(":")
d[int(a_h)][int(a_m)][int(a_s)] += 1
d[int(b_h)][int(b_m)][int(b_s)] -= 1
ans = 0
tmp_ans = 0
for i in range(24):
for j in range(60):
for k in range(60):
tmp_ans += d[i][j][k]
ans = max(ans, tmp_ans)
print(ans)
if __name__ == '__main__':
main()
```
| 93,467 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
while True:
n=int(input())
if n==0:break
ans=0
k=0
l1=[]
l2=[]
for i in range(n):
a=input()
l1.append(int(3600*int(a[0:2])+60*int(a[3:5])+int(a[6:8])))
l2.append(int(3600*int(a[9:11])+60*int(a[12:14])+int(a[15:17])))
l1.sort()
l2.sort()
for t in l1:
while t>=l2[0]:
k+=1
l2.pop(0)
if k==0:
ans+=1
else:
k-=1
print(ans)
```
| 93,468 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
import sys
import itertools
# import numpy as np
import time
import math
import heapq
sys.setrecursionlimit(10 ** 7)
from collections import defaultdict
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
INF = 10 ** 9 + 7
# map(int, input().split())
# list(map(int, input().split()))
while True:
N = int(input())
if N == 0:
break
times = [0] * (24 * 60 * 60 + 1)
for i in range(N):
line = input().split()
start, end = 0, 0
for j in range(2):
h, m, s = map(int, line[j].split(":"))
seconds = h * 60 * 60 + m * 60 + s
if j == 0:
start = seconds
else:
end = seconds
times[start] += 1
times[end] -= 1
for i in range(1, 60 * 60 * 24 + 1):
times[i] += times[i - 1]
print(max(times))
```
| 93,469 |
Provide a correct Python 3 solution for this coding contest problem.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
"Correct Solution:
```
while(1):
n = int(input())
if n==0: break
table = [0 for i in range(24*60*60)]
for i in range(n):
s,g=input().split()
s=list(map(int,s.split(':')))
x=(s[0]*60+s[1])*60+s[2]
g=list(map(int,g.split(':')))
y=(g[0]*60+g[1])*60+g[2]
table[x]+=1
table[y]-=1
ret=table[0]
for i in range(1,24*60*60):
table[i]+=table[i-1]
ret=max(ret,table[i])
print(ret)
```
| 93,470 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
# AOJ_osaki.py
def calc_imos(one_day):
day = []
cum_sum = 0
for i in one_day:
cum_sum += i
day.append(cum_sum)
return max(day)
def calc_time(time):
h,m,s = map(int,time.split(":"))
return h*60*60 + m*60 + s
while True:
N = int(input())
if N==0:break
one_day = [0]*(24*60*60+1) # 秒数で1日を管理する
for i in range(N):
start, end = input().split()
# 時間を1日の秒数に直す
# one_dayに累積和を登録していく
start = calc_time(start)
end = calc_time(end)
one_day[start+1] += 1
one_day[end+1] -= 1
print(calc_imos(one_day))
# 累積和を計算して本数を計算する
```
Yes
| 93,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
while(1):
num = 100009
time = [0]*num
n = int(input())
if(n == 0): break
for i in range(n):
t = list(input().split())
a_,b_ = t[0].split(':'),t[1].split(':')
a = int(a_[0])*3600+int(a_[1])*60+int(a_[2])
b = int(b_[0])*3600+int(b_[1])*60+int(b_[2]) -1
#print(a)
#print(b)
time[a] += 1
time[b+1] -= 1
ans = 1
for i in range(num):
if(i != 0): time[i] += time[i-1]
ans = max(ans,time[i])
print(ans)
```
Yes
| 93,472 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
while 1:
n=int(input())
if n==0:break
DP=[0 for _ in range(24*10**4+1)]
for i in range(n):
a,b=map(lambda x:int("".join(x.split(":"))),input().split())
DP[a] +=1
DP[b] -=1
for i in range(1,24*10**4+1):
DP[i]=DP[i]+DP[i-1]
print(max(DP))
```
Yes
| 93,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
from itertools import accumulate
def time_to_int(x):
time = ""
for ch in x:
if ch == ":":
continue
time += ch
return int(time)
while True:
n = int(input())
if n == 0:
exit()
data = [0] * (10**6)
for i in range(n):
s,e = input().split()
int_s = time_to_int(s)
int_e = time_to_int(e)
data[int_s] += 1
data[int_e] -= 1
data = [0] + data
cum_data = list(accumulate(data))
print(max(cum_data))
```
Yes
| 93,474 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
def time_to_i(s):
tt = list(map(int, s.split(':')))
return tt[0] * 3600 + tt[1] * 60 + tt[2]
while True:
n = int(input())
if n == 0: break
tmp = []
for i in range(n):
tmp.append(input().split())
table = []
for t in tmp:
table.append((time_to_i(t[0]), 0))
table.append((time_to_i(t[1]), 1))
table.sort(key=lambda x:x[0])
trains = 1
for t in table:
if t[1] == 0:
if trains > 0:
trains -= 1
else:
trains += 1
print(trains)
```
No
| 93,475 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
times = [[0] * 2 for i in range(n)]
for i in range(n):
l, a = input().split()
lh, lm, ls = map(int, l.split(":"))
ah, am, ass = map(int, a.split(":"))
times[i][0] = lh * 3600 + lm * 60 + ls
times[i][1] = ah * 3600 + am * 60 + ass
times.sort(key = lambda x:x[1])
ans = 0
arrive = 60 * 60 * 24
for i in range(n):
if times[i][0] < arrive:
arrive = times[i][1]
ans += 1
print(ans)
```
No
| 93,476 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Osaki
Osaki
English text is not available in this practice contest.
The Yamanote Line is a circular railway line laid in the 23 wards of Tokyo. The total route distance is 34.5km, and one lap takes about one hour. There are 29 stations in total. The line color is Uguisu color. The peak congestion rate exceeds 200%, making it one of the busiest railway lines in Japan. One train runs every three minutes during the busiest hours, and people who come to Tokyo for the first time are amazed at the sight.
Mr. Tetsuko loves the Yamanote Line and loves genuine railways. One day, she came up with the following question while reading her favorite book, the JR timetable. "How many vehicles are used a day on the Yamanote Line?"
She tried to figure out the minimum number of vehicles required for operation from the timetable. However, the number of trains was so large that she couldn't count it by herself. So she turned to you, a good programmer, for help.
Your job is to write a program to find the minimum number of cars required to operate the Yamanote Line from a given timetable. Since the Yamanote Line is a circular line, the timetable is often written with "Osaki Station" as the starting and ending stations for convenience. Therefore, only the departure and arrival times of each train at Osaki Station are recorded in the timetable given by her.
Although this cannot happen on the actual Yamanote Line, in order to simplify the situation setting, we will consider that the train can depart from Osaki Station immediately after it arrives at Osaki Station. In addition, there are cases where the time taken by Mr. Tetsuko is incorrect, or the train added by Mr. Tetsuko's delusion is mixed in the timetable, but you can not see them, so to the last The number must be calculated for the time as written.
If you write a program that works, she may invite you to a date on the train. However, it is up to you to accept or decline the invitation.
Input
The input consists of multiple datasets. Each dataset has the following format.
> n
> hh: mm: ss hh: mm: ss
> hh: mm: ss hh: mm: ss
> ...
> hh: mm: ss hh: mm: ss
The integer n in the first line is the number of trains included in the timetable. This value is guaranteed not to exceed 10,000. The departure time and arrival time of each train at Osaki Station are given in this order on the n lines from the 2nd line to the n + 1st line, and the departure time and arrival time are separated by a single space. Each time is expressed in the format hh: mm: ss, where hh is the hour, mm is the minute, and ss is the second. The range of each value is 0 ≤ hh <24, 0 ≤ mm <60, 0 ≤ ss <60. All of these numbers are prefixed with 0s as needed to be two digits.
Trains that run across 24:00 at night are not included. Therefore, the departure time is always before the arrival time.
The end of input is indicated by n = 0. It is not included in the dataset.
Output
For each dataset, output the minimum number of vehicles required on one line.
Sample Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output for the Sample Input
1
3
Example
Input
3
05:47:15 09:54:40
12:12:59 12:13:00
16:30:20 21:18:53
6
00:00:00 03:00:00
01:00:00 03:00:00
02:00:00 03:00:00
03:00:00 04:00:00
03:00:00 05:00:00
03:00:00 06:00:00
0
Output
1
3
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
times = [[0] * 2 for i in range(n)]
for i in range(n):
l, a = input().split()
lh, lm, ls = map(int, l.split(":"))
ah, am, ass = map(int, a.split(":"))
times[i][0] = lh * 3600 + lm * 60 + ls
times[i][1] = ah * 3600 + am * 60 + ass
times.sort(key=lambda x: x[1])
ans = 1
arrive = times[0][1]
idx = 1
for i in range(1, n):
if times[idx][1] < arrive:
arrive = times[idx][1]
idx += 1
if times[i][0] < arrive:
arrive = times[i][1]
ans += 1
print(ans)
```
No
| 93,477 |
Provide a correct Python 3 solution for this coding contest problem.
Open Binary and Object Group organizes a programming contest every year. Mr. Hex belongs to this group and joins the judge team of the contest. This year, he created a geometric problem with its solution for the contest. The problem required a set of points forming a line-symmetric polygon for the input. Preparing the input for this problem was also his task. The input was expected to cover all edge cases, so he spent much time and attention to make them satisfactory.
However, since he worked with lots of care and for a long time, he got tired before he finished. So He might have made mistakes - there might be polygons not meeting the condition. It was not reasonable to prepare the input again from scratch. The judge team thus decided to find all line-asymmetric polygons in his input and fix them as soon as possible. They asked a programmer, just you, to write a program to find incorrect polygons.
You can assume the following:
* Edges of the polygon must not cross or touch each other except for the end points of adjacent edges.
* It is acceptable for the polygon to have adjacent three vertexes on a line, but in such a case, there must be the vertex symmetric to each of them.
Input
The input consists of a set of points in the following format.
N
x1 y1
x2 y2
...
xN yN
The first line of the input contains an integer N (3 ≤ N ≤ 1000), which denotes the number of points. The following N lines describe each point. The i-th line contains two integers x1, y1 (-10000 ≤ xi, yi ≤ 10000), which denote the coordinates of the i-th point.
Note that, although the points are the vertexes of a polygon, they are given in an artibrary order, not necessarily clockwise or counterclockwise.
Output
Output "Yes" in a line if the points can form a line-symmetric polygon, otherwise output "No".
Examples
Input
4
0 1
1 0
0 0
1 1
Output
Yes
Input
4
0 1
1 -1
0 0
1 1
Output
No
Input
9
-1 1
0 1
1 1
-1 0
0 0
1 0
-1 -1
0 -1
1 -1
Output
No
Input
3
-1 -1
0 0
1 1
Output
No
Input
4
0 2
0 0
-1 0
1 0
Output
Yes
"Correct Solution:
```
from collections import defaultdict
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def gcd(m, n):
while n:
m, n = n, m % n
return m
def solve():
N = int(readline())
P = [list(map(int, readline().split())) for i in range(N)]
ok = 0
x0, y0 = P[0]; x1, y1 = P[1]
for i in range(2, N):
x2, y2 = P[i]
if (x1 - x0) * (y2 - y0) != (x2 - x0) * (y1 - y0):
ok = 1
break
if not ok:
write("No\n")
return
A = defaultdict(int)
B = defaultdict(int)
for i in range(N):
xi, yi = P[i]
for j in range(i):
xj, yj = P[j]
dx = xi - xj; dy = yi - yj
g = gcd(abs(dx), abs(dy))
dx //= g; dy //= g
if dy < 0:
dx = -dx; dy = -dy
if dx == 0:
p = xi * dy - yi * dx; q = dy
g = gcd(abs(p), abs(q))
p //= g; q //= g
if q < 0:
q = -q; p = -p
key = (p, 0, q, dx, dy)
else:
p = yi * dx - xi * dy; q = dx
g = gcd(abs(p), abs(q))
p //= g; q //= g
if q < 0:
q = -q; p = -p
key = (0, p, q, dx, dy)
A[key] += 1
gx = -dy; gy = dx
if gy < 0:
gx = -gx; gy = -gy
if dy == 0:
p = dy * (yi + yj) + dx * (xi + xj)
q = 2 * dx
g = gcd(abs(p), abs(q))
p //= g; q //= g
if q < 0:
q = -q; p = -p
key = (p, 0, q, gx, gy)
else:
p = dx * (xi + xj) + dy * (yi + yj)
q = 2 * dy
g = gcd(abs(p), abs(q))
p //= g; q //= g
if q < 0:
q = -q; p = -p
key = (0, p, q, gx, gy)
B[key] += 1
ok = 0
if N % 2 == 0:
for k, v in B.items():
if 2*v == N:
ok = 1
break
if 2*v == N-2:
if A[k] == 1:
ok = 1
break
else:
R = []
for k, v in B.items():
if 2*v+1 == N:
R.append(k)
ok = 0
for x0, y0, z0, dx, dy in R:
cnt = 0
for x, y in P:
if dy*(x*z0 - x0) == dx*(y*z0 - y0):
cnt += 1
if cnt == 1:
ok = 1
break
if ok:
write("Yes\n")
else:
write("No\n")
solve()
```
| 93,478 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
n = int(input())
ans = 0
tmp = 1
while n > tmp:
tmp *= 3
ans+=1
print(ans)
```
| 93,479 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
n=int(input())
import math
ans=math.ceil(math.log(n,3))
print(ans)
```
| 93,480 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
n = int(input())
ans = 0
while n > 1:
ans += 1
n = n // 3 + int(bool(n % 3))
print(ans)
```
| 93,481 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
N = int(input())
ans = 0
while N > 1:
N = (N+2) // 3
ans += 1
print (ans)
```
| 93,482 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())
N = inp()
if N <= 3:
print(1)
else:
ans = 1
while True:
N = (N+2)//3
ans += 1
if N<= 3:
break
print(ans)
```
| 93,483 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import math
print(math.ceil(math.log(int(input()),3)))
```
| 93,484 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
import math
def getN():
return int(input())
def getlist():
return list(map(int, input().split()))
def solve(n):
cnt = 0
while(True):
n = math.ceil(n / 3)
cnt += 1
if n == 1:
return cnt
n = getN()
print(solve(n))
```
| 93,485 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
i = 0
while n > 1:
n = math.ceil(n/3)
i += 1
rr.append(i)
break
return '\n'.join(map(str, rr))
print(main())
```
| 93,486 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
x = int(input())
count = 1
val = 3
while(val < x):
val *= 3
count += 1
print(count)
```
Yes
| 93,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
from math import log
n = int(input())
print(int(log(n - 1, 3)) + 1)
```
Yes
| 93,488 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
i = 1
while True:
if pow(3, i) >= N:
print(i)
break
i += 1
```
Yes
| 93,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque, OrderedDict
from copy import deepcopy
from fractions import gcd
from functools import lru_cache, reduce
from math import ceil, floor
from sys import setrecursionlimit
import heapq
import itertools
import operator
# globals
inf = float('inf')
N = 0
def set_inputs():
global N
N = get_int()
return
def main():
setrecursionlimit(100000)
set_inputs()
# ----------MAIN----------
for i in range(100):
if 3 ** i >= N:
print(i)
return
return
def get_int():
return int(input())
def get_float():
return float(input())
def get_str():
return input().strip()
def get_li():
return [int(i) for i in input().split()]
def get_lf():
return [float(f) for f in input().split()]
def get_lc():
return list(input().strip())
def get_data(n, types, sep=None):
if len(types) == 1:
return [types[0](input()) for _ in range(n)]
return list(zip(*(
[t(x) for t, x in zip(types, input().split(sep=sep))]
for _ in range(n)
)))
if __name__ == '__main__':
main()
```
Yes
| 93,490 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
n = int(input())
c = 0
while n:
n //= 3
c += 1
print(c)
```
No
| 93,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
i = 0
while True:
N //= 3
i += 1
if N == 0:
print(i)
break
```
No
| 93,492 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
n = 1
ans = 1
while True:
n *= 3
if n > N:
print(ans)
break
ans+=1
```
No
| 93,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 ≤ N ≤ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
#coding:utf-8
import sys
import copy
def main():
line = sys.stdin.readline()
while line.split() != ["0","0"]:
analy(line)
line = sys.stdin.readline()
return
def analy(line):
field = [int(n) for n in line.split()]
num = int(sys.stdin.readline())
data = []
for i in range(num):
a = sys.stdin.readline()
b = [int(n) for n in a.split()]
data.append(b)
print (compute(field, data))
def compute(field, ng):
list = []
for i in range(field[1]):
for j in range(field[0]):
if i == 0 :
if [j+1,i+1] in ng:
list.append(0)
elif j == 0:
#左がない
list.append(1)
else:
list.append(list[j-1])
else:
if [j+1,i+1] in ng:
list[j] = 0
elif j != 0:
#no left the same.
list[j] = list[j-1] + list[j]
return list[len(list)-1]
if __name__ == "__main__":
main()
```
No
| 93,494 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 ≤ N ≤ 50
* 1 ≤ Q ≤ 50
* -500 ≤ xi, yi, zi ≤ 500
* 1 ≤ ri ≤ 1,000
* 1 ≤ li ≤ 1016
* -500 ≤ sxj, syj, szj ≤ 500
* -500 ≤ dxj, dyj, dzj ≤ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
"Correct Solution:
```
import math
def sub(a, b):
return [x-y for x,y in zip(a,b)]
def dot(a, b):
return sum(x*y for x,y in zip(a,b))
def cross(a, b):
return [a[1]*b[2]-a[2]*b[1], a[2]*b[0]-a[0]*b[2], a[0]*b[1]-a[1]*b[0]]
def abs2(a):
return sum([x*x for x in a])
def deg(a, b):
return (math.acos(dot(a,b)/math.sqrt(abs2(a)*abs2(b))))
N,Q = map(int,input().split())
ls = [list(map(int,input().split())) for i in range(N)]
for i in range(Q):
sx,sy,sz,dx,dy,dz = map(int,input().split())
cost = 0
s = [sx,sy,sz]
d = [dx,dy,dz]
sd = sub(d,s)
ds = sub(s,d)
for v in ls:
if abs(deg(sub(v[:3],s),sd)) < math.pi/2 and abs(deg(sub(v[:3],d), ds)) < math.pi/2:
if abs2(cross(sub(v[:3],s),sd)) <= abs2(sd)*v[3]**2:
cost += v[4]
else:
if min(abs2(sub(v[:3],s)), abs2(sub(v[:3],d))) <= v[3]**2:
cost += v[4]
print(cost)
```
| 93,495 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 ≤ N ≤ 50
* 1 ≤ Q ≤ 50
* -500 ≤ xi, yi, zi ≤ 500
* 1 ≤ ri ≤ 1,000
* 1 ≤ li ≤ 1016
* -500 ≤ sxj, syj, szj ≤ 500
* -500 ≤ dxj, dyj, dzj ≤ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
n = I()
s = 0
ans = "YES"
for i in range(n):
k = input()
if k == "Un":
s -= 1
else:
s += 1
if s < 0:
ans = "NO"
if s:
print("NO")
else:
print(ans)
return
#B
def B():
return
#C
def C():
n,q = LI()
p = LIR(n)
for _ in range(q):
x0,y0,z0,s,t,u = LI()
a,b,c = (s-x0,t-y0,u-z0)
ans = 0
su = a**2+b**2+c**2
for xr,yr,zr,r,l in p:
x = (su-a**2)/su*x0-a/su*(b*y0+c*z0-a*xr-b*yr-c*zr)
y = (su-b**2)/su*y0-b/su*(a*x0+c*z0-a*xr-b*yr-c*zr)
z = (su-c**2)/su*z0-c/su*(b*y0+a*x0-a*xr-b*yr-c*zr)
if min(x0,s)-1e-4 <= x <= max(x0,s)+1e-4 and min(y0,t)-1e-4 <= y <= max(y0,t)+1e-4 and min(z0,u)-1e-4 <= z <= max(z0,u)+1e-4:
x -= xr
y -= yr
z -= zr
r2 = x**2+y**2+z**2
if r2 <= r**2+1e-4:
ans += l
print(ans)
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
C()
```
| 93,496 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 ≤ N ≤ 50
* 1 ≤ Q ≤ 50
* -500 ≤ xi, yi, zi ≤ 500
* 1 ≤ ri ≤ 1,000
* 1 ≤ li ≤ 1016
* -500 ≤ sxj, syj, szj ≤ 500
* -500 ≤ dxj, dyj, dzj ≤ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
"Correct Solution:
```
N, Q = map(int, input().split())
x = [None] * N
y = [None] * N
z = [None] * N
r = [None] * N
l = [None] * N
for i in range(N):
x[i], y[i], z[i], r[i], l[i] = map(int, input().split())
for _ in range(Q):
ans = 0
sx, sy, sz, dx, dy, dz = map(int, input().split())
vx = dx - sx
vy = dy - sy
vz = dz - sz
for i in range(N):
t = 0
t += vx * (x[i] - sx)
t += vy * (y[i] - sy)
t += vz * (z[i] - sz)
t /= vx**2 + vy**2 + vz**2
len2 = 0
len2 += (sx + vx * t - x[i])**2
len2 += (sy + vy * t - y[i])**2
len2 += (sz + vz * t - z[i])**2
if 0 < t < 1 and len2 <= r[i]**2 + 1e-9:
ans += l[i]
print(ans)
```
| 93,497 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 ≤ N ≤ 50
* 1 ≤ Q ≤ 50
* -500 ≤ xi, yi, zi ≤ 500
* 1 ≤ ri ≤ 1,000
* 1 ≤ li ≤ 1016
* -500 ≤ sxj, syj, szj ≤ 500
* -500 ≤ dxj, dyj, dzj ≤ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,q = LI()
a = [LI() for _ in range(n)]
b = [LI() for _ in range(q)]
rr = []
def k(a,b):
return sum([(a[i]-b[i]) ** 2 for i in range(3)]) ** 0.5
def f(a,b,c,r):
ab = k(a,b)
ac = k(a,c)
bc = k(b,c)
if ac <= r or bc <= r:
return True
at = (ac ** 2 - r ** 2)
bt = (bc ** 2 - r ** 2)
t = max(at,0) ** 0.5 + max(bt,0) ** 0.5
return ab >= t - eps
for x1,y1,z1,x2,y2,z2 in b:
tr = 0
ta = (x1,y1,z1)
tb = (x2,y2,z2)
for x,y,z,r,l in a:
if f(ta,tb,(x,y,z),r):
tr += l
rr.append(tr)
return '\n'.join(map(str,rr))
print(main())
```
| 93,498 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 ≤ N ≤ 50
* 1 ≤ Q ≤ 50
* -500 ≤ xi, yi, zi ≤ 500
* 1 ≤ ri ≤ 1,000
* 1 ≤ li ≤ 1016
* -500 ≤ sxj, syj, szj ≤ 500
* -500 ≤ dxj, dyj, dzj ≤ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
"Correct Solution:
```
eps = 1.0 / 10**10
def LI(): return [int(x) for x in input().split()]
def main():
n,q = LI()
na = [LI() for _ in range(n)]
qa = [LI() for _ in range(q)]
rr = []
def k(a,b):
return sum([(a[i]-b[i]) ** 2 for i in range(3)]) ** 0.5
def f(a,b,c,r):
ab = k(a,b)
ac = k(a,c)
bc = k(b,c)
if ac <= r or bc <= r:
return True
at = (ac ** 2 - r ** 2) ** 0.5
bt = (bc ** 2 - r ** 2) ** 0.5
return ab >= at + bt - eps
for x1,y1,z1,x2,y2,z2 in qa:
tr = 0
for x,y,z,r,l in na:
if f((x1,y1,z1),(x2,y2,z2),(x,y,z),r):
tr += l
rr.append(tr)
return '\n'.join(map(str,rr))
print(main())
```
| 93,499 |
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