message stringlengths 2 22.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 16 109k | cluster float64 1 1 | __index_level_0__ int64 32 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 to N, connected by M railroads.
You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.
The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes. You cannot use gold coins to pay the fare.
There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin. The transaction takes D_i minutes for each gold coin you give. You can exchange any number of gold coins at each exchange counter.
For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.
Constraints
* 2 \leq N \leq 50
* N-1 \leq M \leq 100
* 0 \leq S \leq 10^9
* 1 \leq A_i \leq 50
* 1 \leq B_i,C_i,D_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).
* Each city t=2,...,N can be reached from City 1 with some number of railroads.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N
Output
For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.
Examples
Input
3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5
Output
2
14
Input
4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4
Output
5
5
7
Input
6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1
Output
1
9003
14606
16510
16576
Input
4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7
Output
1
3
5
Input
2 1 0
1 2 1 1
1 1000000000
1 1
Output
1000000001
Submitted Solution:
```
import sys,bisect,string,math,time,functools,random
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
def Golf():*a,=map(int,open(0))
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def GI(V,E,ls=None,Directed=False,index=1):
org_inp=[];g=[[] for i in range(V)]
FromStdin=True if ls==None else False
for i in range(E):
if FromStdin:
inp=LI()
a,b,c=(inp+[1])[:3]
org_inp.append(inp)
else:
#index=0
a,b,c=(list(ls[i])+[1])[:3]
if index==1:a-=1;b-=1
aa=(a,c);bb=(b,c);g[a].append(bb)
if not Directed:g[b].append(aa)
return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0}):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0}) # sample usage
mp=[1]*(w+2);found={}
for i in range(h):
s=input()
for char in search:
if char in s:
found[char]=((i+1)*(w+2)+s.index(char)+1)
mp_def[char]=mp_def[replacement_of_found]
mp+=[1]+[mp_def[j] for j in s]+[1]
mp+=[1]*(w+2)
return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def bit_combination(k,n=2):
rt=[]
for tb in range(n**k):
s=[tb//(n**bt)%n for bt in range(k)];rt+=[s]
return rt
def show(*inp,end='\n'):
if show_flg:print(*inp,end=end)
YN=['YES','NO'];Yn=['Yes','No']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**7)
input=lambda: sys.stdin.readline().rstrip()
class Comb:
def __init__(self,n,mo=10**9+7):
self.fac=[0]*(n+1)
self.inv=[1]*(n+1)
self.fac[0]=1
self.fact(n)
for i in range(1,n+1):
self.fac[i]=i*self.fac[i-1]%mo
self.inv[n]*=i
self.inv[n]%=mo
self.inv[n]=pow(self.inv[n],mo-2,mo)
for i in range(1,n):
self.inv[n-i]=self.inv[n-i+1]*(n-i+1)%mo
return
def fact(self,n):
return self.fac[n]
def invf(self,n):
return self.inv[n]
def comb(self,x,y):
if y<0 or y>x:
return 0
return self.fac[x]*self.inv[x-y]*self.inv[y]%mo
show_flg=False
show_flg=True
## Segment Tree ##
## Initializer Template ##
# Range Sum: sg=SegTree(n)
# Range Minimum: sg=SegTree(n,inf,min,inf)
class SegTree:
def __init__(self,n,init_ls=None,init_val=0,function=lambda a,b:a+b,ide=0):
self.n=n
self.ide_ele=ide_ele=ide
self.num=num=2**(n-1).bit_length()
self.seg=seg=[self.ide_ele]*2*self.num
self.lazy=lazy=[self.ide_ele]*2*self.num
self.segfun=segfun=function
#set_val
if init_ls==None:
for i in range(n):
self.seg[i+self.num-1]=init_val
else:
for i in range(n):
self.seg[i+self.num-1]=init_ls[i]
#build
for i in range(self.num-2,-1,-1):
self.seg[i]=self.segfun(self.seg[2*i+1],self.seg[2*i+2])
def update(self,k,x):
k += self.num-1
self.seg[k] = x
while k:
k = (k-1)//2
self.seg[k] = self.segfun(self.seg[k*2+1],self.seg[k*2+2])
def evaluate(k,l,r): #ι
ε»Άθ©δΎ‘ε¦η
if lazy[k]!=0:
node[k]+=lazy[k]
if(r-l>1):
lazy[2*k+1]+=lazy[k]//2
lazy[2*k+2]+=lazy[k]//2
lazy[k]=0
def query(self,p,q):
if q<=p:
return self.ide_ele
p += self.num-1
q += self.num-2
res=self.ide_ele
while q-p>1:
if p&1 == 0:
res = self.segfun(res,self.seg[p])
if q&1 == 1:
res = self.segfun(res,self.seg[q])
q -= 1
p = p//2
q = (q-1)//2
if p == q:
res = self.segfun(res,self.seg[p])
else:
res = self.segfun(self.segfun(res,self.seg[p]),self.seg[q])
return res
def find_min_index(self,p,q,m):
if q<=p:
return self.ide_ele
p += self.num-1
q += self.num-2
res=self.ide_ele
while q-p>1:
if p&1 == 0:
res = self.segfun(res,self.seg[p])
if q&1 == 1:
res = self.segfun(res,self.seg[q])
q -= 1
p >>= 1
q = (q-1)//2
if p == q:
res = self.segfun(res,self.seg[p])
else:
res = self.segfun(self.segfun(res,self.seg[p]),self.seg[q])
return res
def __str__(self):
# ηι
εγ葨瀺
rt=self.seg[self.num-1:self.num-1+self.n]
return str(rt)
class Tree:
def __init__(self,inp_size=None,init=True):
if init:
self.stdin(inp_size)
return
def stdin(self,inp_size=None):
if inp_size==None:
self.size=int(input())
else:
self.size=inp_size
self.edges,_=GI(self.size,self.size-1)
return
def listin(self,ls):
self.size=len(ls)+1
self.edges,_=GI(self.size,self.size-1,ls)
return
def __str__(self):
return str(self.edges)
def dfs(self,x,func=lambda prv,nx,dist:prv+dist,root_v=0):
q=deque()
q.append(x)
v=[-1]*self.size
v[x]=root_v
while q:
c=q.pop()
for nb,d in self.edges[c]:
if v[nb]==-1:
q.append(nb)
v[nb]=func(v[c],nb,d)
return v
def EulerTour(self,x,func=lambda prv,nx,dist:prv+dist,root_v=0):
q=deque()
q.append((-1,x))
v=[None]*self.size
v[x]=root_v
et=[]
while q:
cb,ce=q.pop()
et.append(ce)
for nb,d in self.edges[ce]:
if v[nb]==None:
q.append((nb,ce))
q.append((ce,nb))
v[nb]=func(v[ce],nb,d)
vid=[[-1,-1]for i in range(self.size)]
for i,j in enumerate(et):
if vid[j][0]==-1:
vid[j][0]=i
else:
vid[j][1]=i
return v,et,vid
def LCA_init(self,depth,et):
self.st=SegTree(self.size*2-1,func=min,ide=inf)
for i,j in enumerate(et):
self.st.update(i,j)
self.LCA_init_stat==True
return
def LCA(self,root,x,y):
if self.LCA_init_stat==False:
depth,et,vid=self.EulerTour(root)
self.LCA_init(depth,et)
return self.st.query(x,y+1)
def dijkstra(edge,st):
# edge=[[(v_to,dist_to_v),...],[],...]
# initialize: def: d=dist(st,i), prev=[previous vertex in minimum path], q[]
n=len(edge)
d=[(0 if st==i else inf) for i in range(n)]
prev=[0]*n
q=[(j,i) for i,j in enumerate(d)]
heapify(q)
# calc
while q:
dist,cur=heappop(q)
for dst,dist in edge[cur]:
alt=d[cur]+dist
if alt<d[dst]:
d[dst]=alt
prev[dst]=cur
heappush(q,(alt,dst))
return d,prev
ans=0
n,m,s=LI()
t_ls=[]
p_ls=[]
k=2501
s=min(k-1,s)
for i in range(m):
u,v,a,b=LI()
p_ls+=[(u,v,a)]
u-=1
v-=1
for j in range(a,k):
t_ls+=[(u*k+j,v*k+j-a,b)]
C,D=[],[]
for i in range(n):
c,d=LI()
for j in range(k):
t_ls+=[(i*k+j,i*k+min(j+c,k-1),d)]
C+=[c]
D+=[d]
gp,_=GI(n,m,ls=p_ls,index=1)
gt,_=GI(n*k,len(t_ls),ls=t_ls,index=0)
dist_t,_=dijkstra(gt,i*k+s)
for i in range(1,n):
ans=inf
for j in range(k):
alt=dist_t[i*k+j]
if alt<ans:
ans=alt
print(ans)
``` | instruction | 0 | 34,031 | 1 | 68,062 |
No | output | 1 | 34,031 | 1 | 68,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 to N, connected by M railroads.
You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.
The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes. You cannot use gold coins to pay the fare.
There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin. The transaction takes D_i minutes for each gold coin you give. You can exchange any number of gold coins at each exchange counter.
For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.
Constraints
* 2 \leq N \leq 50
* N-1 \leq M \leq 100
* 0 \leq S \leq 10^9
* 1 \leq A_i \leq 50
* 1 \leq B_i,C_i,D_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).
* Each city t=2,...,N can be reached from City 1 with some number of railroads.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N
Output
For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.
Examples
Input
3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5
Output
2
14
Input
4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4
Output
5
5
7
Input
6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1
Output
1
9003
14606
16510
16576
Input
4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7
Output
1
3
5
Input
2 1 0
1 2 1 1
1 1000000000
1 1
Output
1000000001
Submitted Solution:
```
from heapq import *
import sys
input = sys.stdin.buffer.readline
n, m, s = map(int, input().split())
lim = 2510
di = [[float("inf")] * lim for _ in range(n)]
di[0][s] = 0
con = [[] for _ in range(n)]
for _ in range(m):
u, v, a, b = map(int, input().split())
con[u - 1].append((v - 1, a, b))
con[v - 1].append((u - 1, a, b))
q = [(0, s, 0)]
cd = [list(map(int, input().split())) for _ in range(n)]
while q:
time, have, cur = heappop(q)
if di[cur][have] < time:
continue
c, d = cd[cur]
if have + c < lim and di[cur][have + c] > time + d:
di[cur][have + c] = time + d
heappush(q, (time + d, have + c, cur))
for nxt, dh, dt in con[cur]:
if have >= dh and di[nxt][have - dh] > time + dt:
di[nxt][have - dh] = time + dt
heappush(q, (time + dt, have - dh, nxt))
for p in di[1:]:
print(min(p))
``` | instruction | 0 | 34,032 | 1 | 68,064 |
No | output | 1 | 34,032 | 1 | 68,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities numbered 1 to N, connected by M railroads.
You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.
The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes. You cannot use gold coins to pay the fare.
There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin. The transaction takes D_i minutes for each gold coin you give. You can exchange any number of gold coins at each exchange counter.
For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.
Constraints
* 2 \leq N \leq 50
* N-1 \leq M \leq 100
* 0 \leq S \leq 10^9
* 1 \leq A_i \leq 50
* 1 \leq B_i,C_i,D_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).
* Each city t=2,...,N can be reached from City 1 with some number of railroads.
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N
Output
For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.
Examples
Input
3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5
Output
2
14
Input
4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4
Output
5
5
7
Input
6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1
Output
1
9003
14606
16510
16576
Input
4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7
Output
1
3
5
Input
2 1 0
1 2 1 1
1 1000000000
1 1
Output
1000000001
Submitted Solution:
```
# -*- coding: utf-8 -*-
##################################
# University of Wisconsin-Madison
# Author: Yaqi Zhang
##################################
# standard library
import collections
import heapq
import sys
from functools import lru_cache
sys.setrecursionlimit(10**9)
Input = lambda: sys.stdin.readline().rstrip()
IntRead = lambda: int(Input())
LineRead = lambda: list(map(int, Input().split()))
INF = 10**18
def main():
N, M, S = LineRead()
S = min(S, 50 * 50)
# state: start at 1, for 2 ... N
graph = [[] for _ in range(N + 1)]
cs = [-1]
ds = [-1]
for _ in range(M):
u, v, a, b = LineRead()
graph[u].append((v, a, b))
graph[v].append((u, a, b))
for _ in range(N):
c, d = LineRead()
cs.append(c)
ds.append(d)
def dijkstra(start, end):
pq = [(0, start, S)]
dist = {}
dist[(start, S)] = 0
# print("Hello")
while pq:
t, u, s = heapq.heappop(pq)
# print(t, u, s)
if u == end:
return t
if t < dist[u, s]:
continue
for v, a, b in graph[u]:
if s >= a:
if (v, s - a) not in dist or dist[v, s - a] > t + b:
dist[v, s - a] = t + b
heapq.heappush(pq, (t + b, v, s - a))
# exchange
c = cs[u]
d = ds[u]
if s >= 2500:
continue
if (u, s + c) not in dist or dist[u, s + c] > t + d:
dist[u, s + c] = t + d
heapq.heappush(pq, (t + d, u, s + c))
for i in range(2, N + 1):
print(dijkstra(1, i))
if __name__ == "__main__":
main()
``` | instruction | 0 | 34,033 | 1 | 68,066 |
No | output | 1 | 34,033 | 1 | 68,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,836 | 1 | 69,672 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
n = int(input())
a = []
for i in range(0, n):
a.append([])
for i in range(0, n-1):
u, v = map(int, input().split())
a[u-1].append(v-1)
a[v-1].append(u-1)
parent = []
t1 = []
t2 = []
ff = []
for i in range(0, n):
ff.append(True)
ff[0] = False
parent.append([0, -1, 0])
t1.append(0)
count = 1
while len(t1) > 0:
for i in range(0, len(t1)):
for j in range(0, len(a[t1[i]])):
if ff[a[t1[i]][j]]:
t2.append(a[t1[i]][j])
parent.append([count, t1[i], a[t1[i]][j]])
ff[a[t1[i]][j]] = False
count += 1
t1, t2 = t2, t1
t2 = []
parent.sort()
cost = []
for i in range(0, n):
cost.append(0)
cost[0] = 1
for i in range(1, n):
if parent[i][1] == 0:
cost[parent[i][2]] = cost[parent[i][1]]/len(a[parent[i][1]])
else:
cost[parent[i][2]] = cost[parent[i][1]]/(len(a[parent[i][1]])-1)
ans = 0
if len(a[0]) > 0:
ans += 1
for i in range(1, n):
if len(a[i]) > 1:
ans += cost[i]
print(ans)
``` | output | 1 | 34,836 | 1 | 69,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,837 | 1 | 69,674 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
import queue
n = int(input().strip())
graph = [[] for _ in range(n)]
probabilities = [1.0]*n
dist = [0]*n
visited = [False]*n
for _ in range(n-1):
a, b = map(int, input().split())
graph[a-1].append(b-1)
graph[b-1].append(a-1)
q = queue.Queue()
q.put(0)
while not q.empty():
top = q.get()
count = 0
visited[top] = True
for i in graph[top]:
if not visited[i]:
count += 1
for i in graph[top]:
if not visited[i]:
dist[i] = dist[top] + 1
probabilities[i] = probabilities[top]/count
q.put(i)
sum_ = 0
for i in range(1, n):
if len(graph[i]) == 1:
sum_ += dist[i]*probabilities[i]
print("{0:.15f}".format(sum_))
``` | output | 1 | 34,837 | 1 | 69,675 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,838 | 1 | 69,676 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
from collections import deque
def edge_input(m):
edges=[]
for i in range(m):
edges.append(list(map(int,input().split())))
return edges
def convert(edges):
l={i:[] for i in range(1,n+1)}
for i in edges:
c1,c2=i[0],i[1]
l[c1].append(c2)
l[c2].append(c1)
return l
n=int(input())
l=convert(edge_input(n-1))
v={i:False for i in l}
d={i:-1 for i in l}
p={i:1 for i in l}
end=deque([])
def bfs(x):
d[x]=0
q=deque([x])
v[x]=True
while len(q)!=0:
temp=q.popleft()
c=0
for i in l[temp]:
if v[i]:
continue
d[i]=d[temp]+1
c+=1
q.append(i)
if c==0:
end.append(temp)
else:
for i in l[temp]:
if v[i]:
continue
v[i]=True
p[i]=p[temp]*(1/c)
return d
d=bfs(1)
ans=0
#print(d)
#print(end)
#print(p)
for i in end:
ans+=p[i]*d[i]
print(ans)
``` | output | 1 | 34,838 | 1 | 69,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,839 | 1 | 69,678 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
# ls = list(map(int, input().split()))
#n=int(input())
#arr = list(map(int, input().split()))
#for _ in range(int(input())):
n=int(input())
g=[[] for i in range(100000)]
ans=0.0
q=[]
for i in range(n-1):
u,v = map(int, input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
q.append((0,1,0))
v=[0]*n
while q:
node,prob,h=q.pop()
v[node]=1
adj=0
for i in g[node]:
if v[i]==0:
adj+=1
if adj==0:
ans+=prob*h
else:
for i in g[node]:
if v[i]==0:
q.append((i,prob/adj,h+1))
v[i]=1
print("{0:.20f}".format(ans))
``` | output | 1 | 34,839 | 1 | 69,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,840 | 1 | 69,680 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
from queue import Queue
from collections import defaultdict
ans = 0
def dfs(i, p, graph, prob, d):
global ans
l = len(graph[i])
if l == 1 and p != -1:
ans += prob * d
return
if p == -1:
newProb = prob / l
else:
newProb = prob / (l - 1)
for j in graph[i]:
if j != p:
dfs(j, i, graph, newProb, d + 1)
n = int(input())
graph = defaultdict(list)
for _ in range(n - 1):
u, v = map(int, input().split())
graph[u].append(v)
graph[v].append(u)
ans = 0
q = Queue()
q.put([1, -1, 1, 0])
while not q.empty() and n != 1:
t = q.get()
i, p, prob, d = t[0], t[1], t[2], t[3]
# print(i)
l = len(graph[i])
if l == 1 and p != -1:
ans += prob * d
continue
# print(i, l)
if p == -1:
newProb = prob / l
else:
newProb = prob / (l - 1)
for j in graph[i]:
if j != p:
q.put([j, i, newProb, d + 1])
# dfs(1, -1, graph, 1, 0)
print(ans)
``` | output | 1 | 34,840 | 1 | 69,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,841 | 1 | 69,682 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 30/04/2020
from sys import stdin,stdout
from math import gcd, ceil, sqrt
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
def bfs(d, n):
queue = [[1, 1, 0]]
res = 0
visited = [0] * (n + 1)
while len(queue):
cur = queue.pop(0)
visited[cur[0]] = 1
flag = 0
for i in d[cur[0]]:
if not visited[i]:
prop = cur[1] / len(d[cur[0]])
queue.append([i, prop, cur[2] + 1])
flag = 1
d[i].remove(cur[0])
if flag == 0:
res += cur[1] * cur[2]
return res
n = ii1()
d = {}
if n>1:
for i in range(n - 1):
a, b = iia()
d.setdefault(a, []).append(b)
d.setdefault(b, []).append(a)
print(bfs(d, n))
else:
print(0)
``` | output | 1 | 34,841 | 1 | 69,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,842 | 1 | 69,684 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(10**7)
import threading
threading.stack_size(2**26)
class SOLVE:
def dfs(self, child, parent, node):
total = 0
for v in node[child]:
if v != parent:
total += self.dfs(v, child, node) + 1
return ((total / (len(node[child]) - (1 if parent else 0))) if total else 0)
def solve(self):
R = stdin.readline
#f = open('input.txt');R = f.readline
W = stdout.write
n = int(R())
node = [[] for i in range(n+1)]
for _ in range(n-1):
u, v = [int(x) for x in R().split()]
node[u].append(v)
node[v].append(u)
print(self.dfs(1, 0, node))
#
return 0
def main():
s = SOLVE()
s.solve()
threading.Thread(target=main).start()
``` | output | 1 | 34,842 | 1 | 69,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2. | instruction | 0 | 34,843 | 1 | 69,686 |
Tags: dfs and similar, dp, graphs, probabilities, trees
Correct Solution:
```
from sys import stdin
inp = stdin.readline
n, ans, = int(inp()), 0
g = {v: [] for v in range(1, n + 1)}
visited = [0] * (n + 1)
q = [(1, 1, 0)]
for _ in range(n - 1):
a, b = map(int, input().split())
g[a].append(b)
g[b].append(a)
while q:
v, cnt, h = q.pop()
visited[v] = 1
adj = 0
for i in g[v]:
if not visited[i]: adj += 1
if adj == 0: ans += cnt * h
else:
for i in g[v]:
if not visited[i]: q.append((i, cnt / adj, h + 1))
print(ans)
``` | output | 1 | 34,843 | 1 | 69,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
#https://codeforces.com/problemset/problem/839/C
from collections import defaultdict
class Graph:
def __init__(self):
self.dict = defaultdict(set)
def main():
n_cities = int(input())
g = Graph()
s = []
visited = set()
for i in range(0,n_cities-1):
u, v = (int(j) for j in input().split())
g.dict[u].add(v)
g.dict[v].add(u)
#dfs
e = 0
s.append((1,0,1))
while s:
city, l, p = s.pop()
neighbors = [c for c in g.dict[city] if c not in visited]
n = len(neighbors)
if n == 0:
e += l*p
else:
new_p = p/n
for neighbor in neighbors:
s.append((neighbor, l+1, new_p))
visited.add(city)
print(e)
if __name__ == "__main__":
main()
``` | instruction | 0 | 34,844 | 1 | 69,688 |
Yes | output | 1 | 34,844 | 1 | 69,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
input = __import__("sys").stdin.readline
n = int(input())
e = [[] for i in range(n+1)]
for i in range(n-1):
u, v = map(int, input().split())
e[u].append(v)
e[v].append(u)
used = [False for i in range(n+1)]
sol = 0
q = [(1, 0, 1)]
used[1] = True
while len(q):
u, d, p = q.pop()
#print(u, d, p)
if len(e[u]) == 1:
sol += d*p
for v in e[u]:
if not used[v]:
np = p / (len(e[u])-(u!=1))
q.append((v, d+1, np))
used[v] = True
print(sol)
``` | instruction | 0 | 34,845 | 1 | 69,690 |
Yes | output | 1 | 34,845 | 1 | 69,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
import math
import sys
def bfs(start):
q=[start]
used[start]=0
while q!=[]:
rt=0
for u in g[q[0]]:
if used[u]==-1:
rt+=1
if rt==0:
E[0]+=used[q[0]]/ver[q[0]]
ver[q[0]]=0
for u in g[q[0]]:
if used[u]==-1:
used[u]=used[q[0]]+1
ver[u]=ver[q[0]]*rt
q.append(u)
del(q[0])
n=int(input())
g=[[] for i in range(n)]
for i in range(n-1):
u,v=map(int,input().split())
v-=1;u-=1
g[u].append(v)
g[v].append(u)
used=[-1]*n
ver=[0]*n
ver[0]=1
E=[0]
bfs(0)
print("%.12f"%E[0])
``` | instruction | 0 | 34,846 | 1 | 69,692 |
Yes | output | 1 | 34,846 | 1 | 69,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
import sys
n = int(input())
if n==1:
print(0)
exit()
d = {}
for _ in range(n-1):
a,b = map(int,input().split())
if a not in d:
d[a]=set()
d[a].add(b)
if b not in d:
d[b]=set()
d[b].add(a)
x = [None]*(n+1)
for node in d:
if len(d[node])==1 and node!=1:
x[node]=0
done = False
visited = [None]*(n+1)
unvisited = set([1])
traverseOrder = [1]
while unvisited:
node = unvisited.pop()
visited[node]=True
for neighbour in d[node]:
if visited[neighbour]==None:
unvisited.add(neighbour)
traverseOrder+=[neighbour]
visited[neighbour]=True
for node in traverseOrder[::-1]:
if len(d[node])==1 and node!=1:
x[node]=0
else:
cost=0
count=0
for child in d[node]:
if x[child]!=None:
cost+=x[child]
count+=1.
x[node]=1+cost/count
print(x[1])
``` | instruction | 0 | 34,847 | 1 | 69,694 |
Yes | output | 1 | 34,847 | 1 | 69,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**6)
n = int(input())
def recur(pos):
global visited
global edges
global leaf
visited[pos] = True
count = 0
if edges[pos][1]:
size = float(len(edges[pos][1]))
for node in edges[pos][1]:
if not visited[node]:
edges[node][2] /= size
edges[node][0] = edges[pos][0] + 1.0
recur(node)
count += 1
if count == 0:
leaf.append(pos)
visited = [False]*(n+1)
edges = [[0.0,[],1.0] for i in range(n+1)]
leaf = []
for i in range(n-1):
a,b = map(int,input().split())
edges[a][1].append(b)
edges[b][1].append(a)
visited[0] = True
recur(1)
s = sum([edges[idx][0]*edges[idx][2] for idx in leaf])
print(s)
``` | instruction | 0 | 34,848 | 1 | 69,696 |
No | output | 1 | 34,848 | 1 | 69,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
def get(cur_root):
res = 1
length = len(graph[cur_root])
if length > 0:
temp = 0
for el in graph[cur_root]:
temp += get(el)
res += temp/length
return res
n = int(input())
graph = [[] for _ in range(n+1)]
root_pretendents = set(range(1,n+1))
for _ in range(n-1):
a, b = map(int, input().split(' '))
root_pretendents -= {b}
graph[a].append(b)
root = root_pretendents.pop()
print(get(root)-1)
``` | instruction | 0 | 34,849 | 1 | 69,698 |
No | output | 1 | 34,849 | 1 | 69,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
from collections import deque, defaultdict
sz = int(1e5)
tree = [0] * (sz + 1)
for i in range(sz + 1):
tree[i] = []
vis = [False] * (sz + 1)
dis = [0] * sz
def addEdge(a: int, b: int):
global tree
tree[a].append(b)
tree[b].append(a)
def bfs(node: int):
global dis, vis
qu = deque()
qu.append((node, 0))
dis[0] = 0
while qu:
p = qu[0]
qu.popleft()
vis[p[0]] = True
for child in tree[p[0]]:
if not vis[child]:
dis[child] = dis[p[0]] + 1
qu.append((child, p[0]))
# Driver Code
d = defaultdict(int)
n = int(input())
for i in range(n-1):
a, b = map(int, input().split(" "))
a-=1
b-=1
d[a] += 1
d[b] += 1
addEdge(a, b)
bfs(0)
q = []
for i, j in d.items():
if j == 1:
q.append(i)
s = 0
for i in range(len(q)):
s+= dis[q[i]]
print(s/len(q))
``` | instruction | 0 | 34,850 | 1 | 69,700 |
No | output | 1 | 34,850 | 1 | 69,701 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n cities and n - 1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they canβt see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link <https://en.wikipedia.org/wiki/Expected_value>.
Input
The first line contains a single integer n (1 β€ n β€ 100000) β number of cities.
Then n - 1 lines follow. The i-th line of these lines contains two integers ui and vi (1 β€ ui, vi β€ n, ui β vi) β the cities connected by the i-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output
Print a number β the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
4
1 2
1 3
2 4
Output
1.500000000000000
Input
5
1 2
1 3
3 4
2 5
Output
2.000000000000000
Note
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
Submitted Solution:
```
from collections import deque
n = int(input())
g = [[] for _ in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split())
g[a].append(b)
g[b].append(a)
q = deque()
q.append(1)
par = [0] * (n + 1)
rank = [0] * (n + 1)
while q:
v = q.pop()
for x in g[v]:
if x == par[v]:
continue
par[x] = v
rank[x] = rank[v] + 1
q.append(x)
#print(g, rank)
res = 0
cnt = 0
for i in range(2, n + 1):
if len(g[i]) == 1:
y = rank[i]
res += y
cnt += 1
print(res / cnt)
``` | instruction | 0 | 34,851 | 1 | 69,702 |
No | output | 1 | 34,851 | 1 | 69,703 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,011 | 1 | 70,022 |
"Correct Solution:
```
N, M = map(int, input().split())
l = []
for i in range(M):
l.extend(list(map(int, input().split())))
for i in range(1,N+1):
print(l.count(i))
``` | output | 1 | 35,011 | 1 | 70,023 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,012 | 1 | 70,024 |
"Correct Solution:
```
N,M= map(int,input().split())
NN = [0]*N
for i in range(M):
a,b = map(int,input().split())
NN[a-1] += 1
NN[b-1] += 1
for _ in NN:
print(_)
``` | output | 1 | 35,012 | 1 | 70,025 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,013 | 1 | 70,026 |
"Correct Solution:
```
n, m = map(int, input().split())
a = sum([input().split() for _ in range(m)], [])
[print(a.count(str(i + 1))) for i in range(n)]
``` | output | 1 | 35,013 | 1 | 70,027 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,014 | 1 | 70,028 |
"Correct Solution:
```
n,m=[int(x) for x in input().split()]
r=[input() for _ in range(m)]
r=' '.join(r).split()
for i in range(1,n+1):
print(r.count(str(i)))
``` | output | 1 | 35,014 | 1 | 70,029 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,015 | 1 | 70,030 |
"Correct Solution:
```
n,m = map(int,input().split())
a = []
for i in range(m):
a += map(int,input().split())
for j in range(1,n+1):
print(a.count(j))
``` | output | 1 | 35,015 | 1 | 70,031 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,016 | 1 | 70,032 |
"Correct Solution:
```
N,M=map(int,input().split())
d=[0]*N
for i in range(M):
a,b=map(int,input().split())
d[a-1]+=1
d[b-1]+=1
for i in d:print(i)
``` | output | 1 | 35,016 | 1 | 70,033 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,017 | 1 | 70,034 |
"Correct Solution:
```
N,M=map(int,input().split())
city=[0]*N
for i in range(M):
road=list(map(int,input().split()))
for j in road:
city[j-1]+=1
for i in city:
print(i)
``` | output | 1 | 35,017 | 1 | 70,035 |
Provide a correct Python 3 solution for this coding contest problem.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2 | instruction | 0 | 35,018 | 1 | 70,036 |
"Correct Solution:
```
# coding: utf-8
N, M = map(int, input().split())
lst=[]
for _ in range(M):
lst += input().split()
for i in range(1, N+1):
print(lst.count(str(i)))
``` | output | 1 | 35,018 | 1 | 70,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
n,m=map(int,input().split())
c=[0]*n
for _ in range(m):
a,b=[int(j) for j in input().split()]
c[a-1]+=1
c[b-1]+=1
for i in c:
print(i)
``` | instruction | 0 | 35,019 | 1 | 70,038 |
Yes | output | 1 | 35,019 | 1 | 70,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
n,m = map(int,input().split())
x =[0]*n
for i in range(m):
a,b = map(int,input().split())
x[a-1]+=1
x[b-1]+=1
for i in range(n):
print(x[i])
``` | instruction | 0 | 35,020 | 1 | 70,040 |
Yes | output | 1 | 35,020 | 1 | 70,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
n,m=map(int,input().split())
l=[]
for i in range(m):l+=list(map(int,input().split()))
for j in range(1,n+1):print(l.count(j))
``` | instruction | 0 | 35,021 | 1 | 70,042 |
Yes | output | 1 | 35,021 | 1 | 70,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
N,_,*R=open(0).read().split()
for i in range(int(N)):print(R.count(str(-~i)))
``` | instruction | 0 | 35,022 | 1 | 70,044 |
Yes | output | 1 | 35,022 | 1 | 70,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
N,M = map(int, input().strip())
G = [[] for _ in range(N)]
for _ in range(M):
a,b = map(int, input().strip().split(' '))
G[a].append(b)
G[b].append(a)
for i in range(N):
print(len(G[i]))
``` | instruction | 0 | 35,023 | 1 | 70,046 |
No | output | 1 | 35,023 | 1 | 70,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
from sys import stdin
from itertools import chain
from collections import Counter
N, M = [int(_) for _ in stdin.readline().rstrip().split()]
AB = [list(map(int, stdin.readline().rstrip().split())) for _ in range(M)]
AB = Counter(chain.from_iterable(AB))
for v in AB.values():
print(v)
``` | instruction | 0 | 35,024 | 1 | 70,048 |
No | output | 1 | 35,024 | 1 | 70,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
n,m = map(int, input().split())
bridge = []
for i in range(n):
bridge.append(0)
for i in range(m):
a,b = map(int,input().split())
bridge[a-1] += 1
bridge[b-1] += 1
for i in bridge:
if i > 0:
print(i)
``` | instruction | 0 | 35,025 | 1 | 70,050 |
No | output | 1 | 35,025 | 1 | 70,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N cities and M roads. The i-th road (1β€iβ€M) connects two cities a_i and b_i (1β€a_i,b_iβ€N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
Constraints
* 2β€N,Mβ€50
* 1β€a_i,b_iβ€N
* a_i β b_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N M
a_1 b_1
:
a_M b_M
Output
Print the answer in N lines. In the i-th line (1β€iβ€N), print the number of roads connected to city i.
Examples
Input
4 3
1 2
2 3
1 4
Output
2
2
1
1
Input
2 5
1 2
2 1
1 2
2 1
1 2
Output
5
5
Input
8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8
Output
3
3
2
2
2
1
1
2
Submitted Solution:
```
N, M = map(int, input().split())
R = [list(map(int, input().split())) for _ in range(N)]
for i in range(1,N+1):
print(R.count(i))
``` | instruction | 0 | 35,026 | 1 | 70,052 |
No | output | 1 | 35,026 | 1 | 70,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,316 | 1 | 70,632 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
n = int(input())
for i in range(n):
a, b, p = map(int, input().split())
s = input()
t = len(s)
l = [0]
for j in range(t - 1):
if s[t - 2 - j] == s[t - 1 - j] and j != 0:
l.append(l[j])
elif s[t - 2 - j] == 'B':
l.append(l[j] + b)
else:
l.append(l[j] + a)
l.reverse()
v = len(l) - 1
for k in range(len(l)):
if p >= l[k]:
v = k
break;
print(v + 1)
``` | output | 1 | 35,316 | 1 | 70,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,317 | 1 | 70,634 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
for __ in range(int(input())):
a,b,p = [int(x) for x in input().split()]
s = input()
if p < a and p < b:
print(len(s))
else:
n = len(s)
cost = [-1]*n
if s[0] == 'A':
cost[0]=a
else:
cost[0]=b
ind = []
ind.append(0)
for i in range(1,n):
if s[i-1] == s[i]:
cost[i]=0;
else:
if s[i] == 'B':
cost[i] = b
else:
cost[i] = a
ind.append(i)
if cost[n-1] == -1:
cost[n-1] = 0
for i in range(n-3,-1,-1):
cost[i] += cost[i+1]
f=0
for i in ind:
if p >= cost[i]:
f=1
print(i+1)
break
if f == 0:
print(n)
``` | output | 1 | 35,317 | 1 | 70,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,318 | 1 | 70,636 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
from sys import stdin
input = stdin.readline
def mp():return map(int,input().split())
def it():return int(input())
for _ in range(it()):
a,b,p=mp()
s=input()
s=s[:len(s)-1]
if len(s)==1:
print(1)
continue
m=0
ans=len(s)
for i in range(len(s)-2,-1,-1):
if (i==0 or s[i]!=s[i-1]):
if s[i]=="A":
m+=a
else:
m+=b
if m<=p:
ans=i+1
print(ans)
``` | output | 1 | 35,318 | 1 | 70,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,319 | 1 | 70,638 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
def main():
t = int(input())
for _ in range(t):
a,b,p = map(int , input().split())
s = list(input())
n = len(s)
s.append('P')
cost = 0
i = 0;j = 0
while i < n - 1:
j = i
while j < n - 1 and s[j] == s[i]:
j += 1
cost += a if s[i] == 'A' else b
i = j
i = 0;j = 0
while i < n - 1:
if cost > p:
j = i
while j < n - 1 and s[j] == s[i]:
j += 1
cost -= a if s[i] == 'A' else b
i = j
else:
break
print(i + 1)
return
if __name__ == "__main__":
main()
``` | output | 1 | 35,319 | 1 | 70,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,320 | 1 | 70,640 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
for i in range(int(input())):
a, b, p = map(int, input().split())
string = input()
crossroads = len(string)
money, answer = 0, crossroads
for i in range(crossroads-1):
if (crossroads-2-i) == 0:
if string[0] == "A":
money += a
else:
money += b
if money <= p:
answer = crossroads-1-i
print(answer)
break
if (string[crossroads-2-i] != string[crossroads-3-i]):
if string[crossroads-2-i] == "A":
money += a
else:
money += b
if money <= p:
answer = crossroads-1-i
else:
print(answer)
break
``` | output | 1 | 35,320 | 1 | 70,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,321 | 1 | 70,642 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
a,b,p = list(map(int, input().split()))
s = input()
price = {'A':a, 'B':b}
route = []
prev = None
idx = 1
start = 0
for ch in s:
if prev == None:
prev = ch
start = idx
idx+=1
continue
if prev!=ch:
route.append( [start, price[prev]])
prev = ch
start = idx
idx+=1
if start<len(s):
route.append([start, price[prev]])
route.reverse()
cost = 0
result = len(s)
prev = None
for item in route:
cost+= item[1]
if cost>p:
break
prev = item
if prev != None:
result = prev[0]
print(result)
``` | output | 1 | 35,321 | 1 | 70,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,322 | 1 | 70,644 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
a, b, p = map(int, input().split())
s = list(str(input()))
n = len(s)
s.reverse()
A = []
temp = s[1]
cnt = 0
for i in range(1, n):
if s[i] != temp:
if temp == 'A':
A.append((a, i))
else:
A.append((b, i))
temp = s[i]
else:
cnt += 1
else:
if temp == 'A':
A.append((a, i+1))
else:
A.append((b, i+1))
#print(A)
m = 0
pre_i = 1
for c, i in A:
m += c
if m > p:
print(n-pre_i+1)
break
else:
pre_i = i
else:
print(1)
``` | output | 1 | 35,322 | 1 | 70,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,323 | 1 | 70,646 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
T = int(input())
result =[]
for i in range(0,T):
list11 = list(map(int, input().split()))
a = list11[0]
b = list11[1]
p = list11[2]
cost = 0
st = input()
length = len(st)
st = st[::-1]
s = st[1]
res = 0
for j in range(2, length):
if(s != st[j]):
if(s=='A'):
cost = cost+a
if(cost>p):
break
else:
res = j-1
if(s=='B'):
cost = cost+b
if(cost>p):
break
else:
res = j-1
s = st[j]
if(j==length-1):
if(st[j]=='A'):
cost = cost+a
if(cost>p):
break
else:
res = length-1
if(st[j]=='B'):
cost = cost+b
if(cost>p):
break
else:
res = length-1
if(length==2):
if(st[1]=='A'):
if(a>p):
res = 0
else:
res = 1
if(st[1]=='B'):
if(b>p):
res = 0
else:
res = 1
result.append(length-res)
for element in result:
print(element)
``` | output | 1 | 35,323 | 1 | 70,647 |
Provide tags and a correct Python 2 solution for this coding contest problem.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6 | instruction | 0 | 35,324 | 1 | 70,648 |
Tags: binary search, dp, greedy, strings
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr('\n'.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
def check(s,n,mid,d):
c=1
cost=0
for i in range(mid+1,n):
if s[i]!=s[i-1]:
c=1
cost+=d[s[i-1]]
else:
c+=1
if c>1:
cost+=d[s[-1]]
return cost
for t in range(ni()):
a,b,p=li()
s=raw_input().strip()
n=len(s)
d={}
d['A']=a
d['B']=b
l=0
r=n-1
ans=n-1
while l<=r:
mid=(l+r)/2
if check(s,n,mid,d)<=p:
r=mid-1
ans=mid
else:
l=mid+1
print ans+1
``` | output | 1 | 35,324 | 1 | 70,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
for T in range(int(input())):
b, a, p = map(int, input().split(" "))
s = " " + input()[::-1]
#print(s)
cost = 0
i_min = 1
pre = 'A' if s[2] == 'B' else 'B'
while i_min < len(s) - 1:
j = i_min + 1
if pre == 'A':
while j < len(s):
if j == len(s)-1 or (s[j] == 'B' and s[j+1] == 'A'):
break
j += 1
if cost + a <= p:
cost += a
i_min = j
pre = 'B'
else:
break
else:
while j < len(s):
if j == len(s)-1 or (s[j] == 'A' and s[j+1] == 'B'):
break
j += 1
if cost + b <= p:
cost += b
i_min = j
pre = 'A'
else:
break
#print(i_min, cost)
print(len(s) - i_min)
``` | instruction | 0 | 35,325 | 1 | 70,650 |
Yes | output | 1 | 35,325 | 1 | 70,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
if __name__ == '__main__':
for _ in range(int(input())):
*a,c = map(int,input().split())
str = input()
i = len(str)-2
j = str[-2] =='B'
while i>=0 and c-a[j]>=0:
c -=a[j]
j ^= 1
i = str.rfind('AB'[j], 0, i)
print(i+2)
``` | instruction | 0 | 35,326 | 1 | 70,652 |
Yes | output | 1 | 35,326 | 1 | 70,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
def minimum_index(a, b, p, s):
expense = 0
j = len(s) - 2
while j >= 0:
i = j
while i > 0 and s[i] == s[i-1]:
i -= 1
delta = a if s[j] == 'A' else b
if expense + delta > p:
return j + 2
expense += delta
j = i - 1
return 1
if __name__ == '__main__':
T = int(input())
for t in range(T):
a, b, p = list(map(int, input().split()))
s = input()
print(minimum_index(a, b, p, s))
'''
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
2
1
3
1
6
'''
``` | instruction | 0 | 35,327 | 1 | 70,654 |
Yes | output | 1 | 35,327 | 1 | 70,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b, p = map(int, input().split())
r = {'A': a, 'B': b}
s = input()
n = len(s)
rs = [0] * n
next = ''
for i in range(n - 1):
j = n - i - 2
rs[j] = rs[j + 1]
if next != s[j]:
rs[j] += r[s[j]]
next = s[j]
for i in range(n):
if rs[i] <= p:
break
print(i + 1)
``` | instruction | 0 | 35,328 | 1 | 70,656 |
Yes | output | 1 | 35,328 | 1 | 70,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
t = int(input())
while t>0:
a,b,p = map(int,input().split())
bus = input()
l= len(bus)
cost=0
flag=0
if l==2:
if bus[0]=="A":
if p>=a:
print("1")
else:
print("2")
else:
if p>=b:
print("1")
else:
print("2")
else:
if bus[l-2]=="B":
cost+=b
else:
cost+=a
if p<cost:
print(l)
else:
for j in range(l-3,-1,-1):
if bus[j]==bus[j+1]:
continue
else:
if bus[j]=="B":
cost+=b
else:
cost+=a
if cost>=p:
break
if flag==0 and j==0:
print("1")
else:
ans=j+2
print(ans)
t-=1
``` | instruction | 0 | 35,329 | 1 | 70,658 |
No | output | 1 | 35,329 | 1 | 70,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a long party Petya decided to return home, but he turned out to be at the opposite end of the town from his home. There are n crossroads in the line in the town, and there is either the bus or the tram station at each crossroad.
The crossroads are represented as a string s of length n, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad. Currently Petya is at the first crossroad (which corresponds to s_1) and his goal is to get to the last crossroad (which corresponds to s_n).
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a bus station, one can pay a roubles for the bus ticket, and go from i-th crossroad to the j-th crossroad by the bus (it is not necessary to have a bus station at the j-th crossroad). Formally, paying a roubles Petya can go from i to j if s_t = A for all i β€ t < j.
If for two crossroads i and j for all crossroads i, i+1, β¦, j-1 there is a tram station, one can pay b roubles for the tram ticket, and go from i-th crossroad to the j-th crossroad by the tram (it is not necessary to have a tram station at the j-th crossroad). Formally, paying b roubles Petya can go from i to j if s_t = B for all i β€ t < j.
For example, if s="AABBBAB", a=4 and b=3 then Petya needs:
<image>
* buy one bus ticket to get from 1 to 3,
* buy one tram ticket to get from 3 to 6,
* buy one bus ticket to get from 6 to 7.
Thus, in total he needs to spend 4+3+4=11 roubles. Please note that the type of the stop at the last crossroad (i.e. the character s_n) does not affect the final expense.
Now Petya is at the first crossroad, and he wants to get to the n-th crossroad. After the party he has left with p roubles. He's decided to go to some station on foot, and then go to home using only public transport.
Help him to choose the closest crossroad i to go on foot the first, so he has enough money to get from the i-th crossroad to the n-th, using only tram and bus tickets.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 β€ t β€ 10^4).
The first line of each test case consists of three integers a, b, p (1 β€ a, b, p β€ 10^5) β the cost of bus ticket, the cost of tram ticket and the amount of money Petya has.
The second line of each test case consists of one string s, where s_i = A, if there is a bus station at i-th crossroad, and s_i = B, if there is a tram station at i-th crossroad (2 β€ |s| β€ 10^5).
It is guaranteed, that the sum of the length of strings s by all test cases in one test doesn't exceed 10^5.
Output
For each test case print one number β the minimal index i of a crossroad Petya should go on foot. The rest of the path (i.e. from i to n he should use public transport).
Example
Input
5
2 2 1
BB
1 1 1
AB
3 2 8
AABBBBAABB
5 3 4
BBBBB
2 1 1
ABABAB
Output
2
1
3
1
6
Submitted Solution:
```
import sys
input = sys.stdin.readline
def main():
t = int(input())
for _ in range(t):
A, B, P = [int(x) for x in input().split()]
S = input().strip()
if P < A and P < B:
print(len(S))
continue
c = S[0]
cost = []
if c == 'A':
cost.append([1, A])
else:
cost.append([1, B])
for i, s in enumerate(S):
if s != c:
c = s
if c == 'A':
cost.append([i + 1, A])
else:
cost.append([i + 1, B])
if len(cost) == 1:
if cost[0][1] > P:
print(len(S))
else:
print(1)
continue
for i in range(len(cost) - 2, -1, -1):
if cost[i + 1][0] == len(S):
if cost[i][1] > P:
print(len(S))
break
continue
cost[i][1] += cost[i + 1][1]
if cost[i][1] > P:
print(cost[i + 1][0])
break
else:
print(1)
if __name__ == '__main__':
main()
``` | instruction | 0 | 35,330 | 1 | 70,660 |
No | output | 1 | 35,330 | 1 | 70,661 |
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