text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
N=int(input())
ans=[]
for i in range(N-1):
tmp=[]
for j in range(i+1,N):
if i%2 != j%2:
tmp.append(1)
else:
tmp.append((i+2)//2+1)
ans.append(tmp)
for i in range(N-1):
print(*ans[i])
```
No
| 95,900 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
res = [[0] * x for x in range(N - 1, 0, -1)]
for x in range(N - 1):
for y in range(N - 1 - x):
b = 0
for k in range(10):
if y + 1 < pow(2, k): break
b = k + 1
res[x][y] = b
for r in res: print(*r)
```
No
| 95,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder's head office consists of N rooms numbered 1 to N. For any two rooms, there is a direct passage connecting these rooms.
For security reasons, Takahashi the president asked you to set a level for every passage, which is a positive integer and must satisfy the following condition:
* For each room i\ (1 \leq i \leq N), if we leave Room i, pass through some passages whose levels are all equal and get back to Room i, the number of times we pass through a passage is always even.
Your task is to set levels to the passages so that the highest level of a passage is minimized.
Constraints
* N is an integer between 2 and 500 (inclusive).
Input
Input is given from Standard Input in the following format:
N
Output
Print one way to set levels to the passages so that the objective is achieved, as follows:
a_{1,2} a_{1,3} ... a_{1,N}
a_{2,3} ... a_{2,N}
.
.
.
a_{N-1,N}
Here a_{i,j} is the level of the passage connecting Room i and Room j.
If there are multiple solutions, any of them will be accepted.
Example
Input
3
Output
1 2
1
Submitted Solution:
```
def f_d():
n = int(input())
for i in range(n-1):
print(" ".join([str(i+1)]*(n-i-1)))
if __name__ == "__main__":
f_d()
```
No
| 95,902 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
"Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
N, X = map(int, input().split())
P = 998244353
A = [[0] * (2*i+1) for i in range(N+1)] # A[i][j] is the number of permutations of 1's and 2's whose length is i and whose sum is j
A[0][0] = 1
B = [0] * (N + 1) # B[i] is the number of permutations of 1's and 2's with length i none of whose consecutive subsequenses is equal to X
for i in range(N):
for j in range(i, 2*i+1):
A[i+1][j+1] += A[i][j]
A[i+1][j+2] += A[i][j]
if A[i+1][j+1] >= P: A[i+1][j+1] -= P
if A[i+1][j+2] >= P: A[i+1][j+2] -= P
# Type 1: Sum is below X
for i in range(N+1):
for j in range(i, min(2*i+1, X)):
B[i] += A[i][j]
# Type 2: Sum is 2X or more
if X % 2:
for i in range(X, N+1):
B[i] += 1
# Type 3: X <= Sum < 2X
for i in range(1, X):
a = X - 1 - 2 * i
if a < 0: continue
for j in range((a+1) // 2, a + 1):
k = j + 2 * i
if k > N: break
B[k] += A[j][a]
if B[k] >= P: B[k] -= P
ans = 0
for i, b in enumerate(B):
ans = (ans + b * A[-1][-1-i]) % P
print(ans)
```
| 95,903 |
Provide a correct Python 3 solution for this coding contest problem.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
"Correct Solution:
```
"""
Writer: SPD_9X2
https://atcoder.jp/contests/tenka1-2019/tasks/tenka1_2019_f
Xが偶数か奇数かは重要な気がする
長さL(<=N)の1,2で構成された文字列の問題、に変換できる
O(L)で求められばおｋ
dpか？
X == 1なら、0,2のみで構成されてればおｋ
X == 2なら、1が2つ以上入っていたら不可能 0 & 1が1つまで
X == 3なら、2を入れる場合→1は入れられないが、2を好きなだけ入れていい
入れない場合→1を2個まで入れられる
X == 4なら、
→やはり偶奇が重要か？
とどく場合は X-1 → X+1の移動は強制
すなわち 座標1に行ってはいけない
初手は2に行くしかない
同様にX+2には行けない
X+3に行くしかない
3には行けないから4に行くしかない
つまり2以外不可？？
無限に移動を続ける場合は少なくともX=奇数 2のみ以外無理
偶数の場合はX以前に奇数回1をはさむ必要あり
a→a+1と移動すると、X+a-1から先は移動できなくなる
よって2Xは絶対超えられない
あとは数え上げにどうやって落とすか
Lは固定でいいと思うんだよな
O(L)で解ければ勝ち
dp ?
総和で場合分け or Lで場合分け
総和で場合分けする場合、当然Lは異なるのでまとめる必要がある
Lの長さの移動の寄与にはNCLを掛ける必要がある
移動するときにLは1増えるんだよな…
nC(l+1) = nCl * (n-l)/(l+1)
移動処理を一気に行えばlは統一できるからおｋ？
総和SはX以上の場合 X+1+2kしかありえない
S-X以前は2のみ進軍可能、以降は自由
対称性？
1と2の数を全探索？ ならばO(N**2)でいける
1がA個,2がB個の時、最初と最後に2をいくつか置く必要がある
→後は残りの位置から1を置く場所を選べばよい
→解けたっぽい？
場合分け頑張ろう
合計がX以下の場合→完全に自由
合計が2X以上の場合→Xが奇数の場合のみ all2が可能。それ以外は死
X以上2X以下の場合
→
S-Xが奇数である必要あり
まず2を(S-X+1)こ消費する（前後に置く分）
あとは自由に順列を作ってどうぞ
"""
def modfac(n, MOD):
f = 1
factorials = [1]
for m in range(1, n + 1):
f *= m
f %= MOD
factorials.append(f)
inv = pow(f, MOD - 2, MOD)
invs = [1] * (n + 1)
invs[n] = inv
for m in range(n, 1, -1):
inv *= m
inv %= MOD
invs[m - 1] = inv
return factorials, invs
def modnCr(n,r,mod,fac,inv): #上で求めたfacとinvsを引数に入れるべし(上の関数で与えたnが計算できる最大のnになる)
return fac[n] * inv[n-r] * inv[r] % mod
N,X = map(int,input().split())
mod = 998244353
fac,inv = modfac(N+10,mod)
ans = 0
for two in range(N+1):
for one in range(N+1):
if one + two > N:
break
dist = one + two*2
zero = modnCr(N,one+two,mod,fac,inv)
now = 0
if dist < X:
now = modnCr(one+two,one,mod,fac,inv) * zero
elif dist == X:
continue
elif dist < 2*X:
if (dist-X) % 2 == 0:
continue
dtwo = two - (dist-X+1)
if dtwo >= 0:
now = modnCr(one+dtwo,one,mod,fac,inv) * zero
elif X % 2 == 1 and one == 0:
now = zero
#print (one,two,now)
ans += now
ans %= mod
print (ans)
```
| 95,904 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
Submitted Solution:
```
MOD = 998244353
MAX_K = 3000
def ceil(a: int, b: int)->int:
'''a/b の切り上げを計算する。
'''
if (a // b) * b == a:
return a//b
return a//b + 1
# __fact[n] = n!
__fact = [0] * (MAX_K+1)
def init_comb():
# initialize __fact
__fact[0] = 1
for n in range(MAX_K):
__fact[n+1] = ((n+1) * __fact[n]) % MOD
def mod_inv(n: int, mod: int)->int:
b, u, v = mod, 1, 0
while b > 0:
t = n // b
n -= t * b
u -= t * v
n, b = b, n
u, v = v, u
return (u+mod) % mod
def comb(n: int, r: int)->int:
'''nCr を計算する
'''
res = __fact[n]
res = (res * mod_inv(__fact[n-r], MOD)) % MOD
res = (res * mod_inv(__fact[r], MOD)) % MOD
return res
def banned_x(N: int, X: int)->int:
# 数列の全体の和を固定して調べる。
# 数列の和の最大値は全て 2 でできている数列の 2*N
res = 1 # 0 だけでできている分
for S in range(1, 2*N+1):
tmp = res
if S < X:
# 1, 2 を合計で k 個利用するとする。
# 1 を p 個、2 を q 個とすると
# p+q=K; p+2q=S;
# より
# p=2K-S; q=S-K
# となる。
# このとき、0, 1, 2 の並べ方は
# comb(N,p) * comb(N-p,q)
# である。
# また、0<=p; 0<=q より S/2<=k<=S が成り立つ。
for k in range(ceil(S, 2), min(S, N)+1):
p, q = 2*k - S, S-k
res += comb(N, p) * comb(N-p, q)
res %= MOD
elif S == X:
# nothing
continue
elif (S-X) & 1:
# 2 を置かなければならない場所が [0, only2)
# と (only2, N]
only2 = (S-(X-1))//2
# 1 が含まれない場合
if 2*only2 >= X-1:
if X & 1:
# 2 の置き方は N 個の位置から S//2 個
# 選ぶ組み合わせ。残りは 0 を置く
res += comb(N, S//2)
res %= MOD
# 1 が含まれる場合
else:
# SS は 1,2 なんでも置いていいゾーンの幅
SS = X-1-2*only2
# k,p,q の変数の意味は S<X の時と同じ
# ただし、2 は 2*only2 の領域にも置くので
# 2 の個数は SS-k+2*only2 になり、
# (N 個の位置から k+2*only2 個の位置を 1,2 用に確保)
# x (k 個の位置から p 個の 1 を置く位置を確保)
for k in range(ceil(SS, 2), min(N-2*only2, SS)+1):
p, q = 2*k-SS, SS-k
res += comb(N, k+2*only2) * comb(k, p)
res %= MOD
# print('S={}, res={}'.format(S, res-tmp))
return res
if __name__ == "__main__":
N, X = map(int, input().split())
init_comb()
ans = banned_x(N, X)
print(ans)
```
No
| 95,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
Submitted Solution:
```
MOD = 998244353
MAX_K = 3000
def ceil(a: int, b: int)->int:
'''a/b の切り上げを計算する。
'''
if (a // b) * b == a:
return a//b
return a//b + 1
# __fact[n] = n!
__fact = [0] * (MAX_K+1)
# __inv[n] = (n! の mod MOD に関する逆数)
__inv = [0] * (MAX_K+1)
def init_comb():
# initialize __fact
__fact[0] = 1
__inv[0] = 1
for n in range(MAX_K):
__fact[n+1] = ((n+1) * __fact[n]) % MOD
__inv[n+1] = mod_inv((n+1) * __fact[n], MOD)
def mod_inv(n: int, mod: int)->int:
b, u, v = mod, 1, 0
while b > 0:
t = n // b
n -= t * b
u -= t * v
n, b = b, n
u, v = v, u
return (u+mod) % mod
def comb(n: int, r: int)->int:
'''nCr を計算する
'''
res = __fact[n]
res = (res * __inv[n-r]) % MOD
res = (res * __inv[r]) % MOD
return res
def banned_x(N: int, X: int)->int:
# 数列の全体の和を固定して調べる。
# 数列の和の最大値は全て 2 でできている数列の 2*N
res = 1 # 0 だけでできている分
for S in range(1, 2*N+1):
if S < X:
# 1, 2 を合計で k 個利用するとする。
# 1 を p 個、2 を q 個とすると
# p+q=K; p+2q=S;
# より
# p=2K-S; q=S-K
# となる。
# このとき、0, 1, 2 の並べ方は
# comb(N,p) * comb(N-p,q)
# である。
# また、0<=p; 0<=q より S/2<=k<=S が成り立つ。
for k in range(ceil(S, 2), min(S, N)+1):
p, q = 2*k - S, S-k
res += comb(N, p) * comb(N-p, q)
res %= MOD
elif S == X:
# nothing
continue
elif (S-X) & 1:
# 2 を置かなければならない場所が [0, only2)
# と (only2, N]
only2 = (S-(X-1))//2
# 1 が含まれない場合
if 2*only2 >= X-1:
if X & 1:
# 2 の置き方は N 個の位置から S//2 個
# 選ぶ組み合わせ。残りは 0 を置く
res += comb(N, S//2)
res %= MOD
# 1 が含まれる場合
else:
# SS は 1,2 なんでも置いていいゾーンの幅
SS = X-1-2*only2
# k,p,q の変数の意味は S<X の時と同じ
# ただし、2 は 2*only2 の領域にも置くので
# 2 の個数は SS-k+2*only2 になり、
# (N 個の位置から k+2*only2 個の位置を 1,2 用に確保)
# x (k 個の位置から p 個の 1 を置く位置を確保)
for k in range(ceil(SS, 2), min(N-2*only2, SS)+1):
p, q = 2*k-SS, SS-k
res += comb(N, k+2*only2) * comb(k, p)
res %= MOD
# print('S={}, res={}'.format(S, res-tmp))
return res
if __name__ == "__main__":
N, X = map(int, input().split())
init_comb()
ans = banned_x(N, X)
print(ans)
```
No
| 95,906 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
Submitted Solution:
```
a
```
No
| 95,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number, modulo 998244353, of sequences of length N consisting of 0, 1 and 2 such that none of their contiguous subsequences totals to X.
Constraints
* 1 \leq N \leq 3000
* 1 \leq X \leq 2N
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number, modulo 998244353, of sequences that satisfy the condition.
Examples
Input
3 3
Output
14
Input
8 6
Output
1179
Input
10 1
Output
1024
Input
9 13
Output
18402
Input
314 159
Output
459765451
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
"""
・累積和の到達地点集合ごとに見る
・type1：X未満で終わる
・type2：X+1以上2X未満
・type3：2X以上、これはXが奇数の場合のみ
"""
MOD = 998244353
N,X = map(int,read().split())
def cumprod(arr,MOD):
L = len(arr); Lsq = int(L**.5+1)
arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
for n in range(1,Lsq):
arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
for n in range(1,Lsq):
arr[n] *= arr[n-1,-1]; arr[n] %= MOD
return arr.ravel()[:L]
def make_fact(U,MOD):
x = np.arange(U,dtype=np.int64); x[0] = 1
fact = cumprod(x,MOD)
x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
fact_inv = cumprod(x,MOD)[::-1]
return fact,fact_inv
U = N+100
fact,fact_inv = make_fact(U,MOD)
combN = fact[N] * fact_inv[:N+1] % MOD * fact_inv[N::-1] % MOD
def F1(N,X):
# X未満で終わる場合
def mult(P,Q):
# 多項式の積
dP = len(P) - 1
dQ = len(Q) - 1
if dP < dQ:
dP,dQ = dQ,dP
P,Q = Q,P
R = np.zeros(dP+dQ+1,np.int64)
for n in range(dQ+1):
R[n:n+dP+1] += Q[n] * P % MOD
R %= MOD
return R[:X]
def power(P,n):
if n == 0:
return np.array([1],dtype=np.int64)
Q = power(P,n//2)
Q = mult(Q,Q)
return mult(P,Q) if n&1 else Q
P = np.array([1,1,1],np.int64)
Q = power(P,N)
return Q.sum() % MOD
def F2(N,X):
U = N+100
fact,fact_inv = make_fact(U,MOD)
combN = fact[N] * fact_inv[:N+1] % MOD * fact_inv[N::-1] % MOD
x = np.zeros(N+1,np.int64)
# X+1+2nで終わる場合
for n in range(X):
m = (X-1) - (2+2*n)
if m < 0:
break
# 2+2n -> X-1に含まれる2の回数ごとに
two = np.arange(m//2+1,dtype=np.int64)
one = m - 2*two
coef = fact[one+two] * fact_inv[one] % MOD * fact_inv[two] % MOD
rest = N-one-two-(2*n+2)
ind = rest>=0
rest = rest[ind]; coef = coef[ind]
x[rest] += coef
x %= MOD
return (x * combN % MOD).sum() % MOD
def F3(N,X):
U = N+100
fact,fact_inv = make_fact(U,MOD)
combN = fact[N] * fact_inv[:N+1] % MOD * fact_inv[N::-1] % MOD
# 2X以上の偶数。Xは奇数で、+2を連打している場合
if X%2 == 0:
return 0
if X > N:
return 0
n = N-X+1
return combN[:n].sum() % MOD
answer = (F1(N,X) + F2(N,X) + F3(N,X)) % MOD
print(answer)
```
No
| 95,908 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
N, X = map(int, input().split())
a, p = [1], [1]
for i in range(N):
a.append(a[i] * 2 + 3)
p.append(p[i] * 2 + 1)
def paty(L, n):
if L == 0:
return 0 if n <= 0 else 1
elif n <= 1 + a[L - 1]:
return paty(L-1, n-1)
else:
return p[L-1] + 1 + paty(L-1, n-2-a[L-1])
print(paty(N, X))
```
| 95,909 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
N,X = map(int,input().split())
P = [1]
H = [1]
for i in range(1,51):
P.append(P[i-1]*2+1)
H.append(H[i-1]*2+3)
def r(n,x):
if x<=n:
return 0
if x==H[n]:
return P[n]
if x>=H[n-1]+2:
return P[n-1]+1+r(n-1,x-H[n-1]-2)
else:
return r(n-1,x-1)
print(r(N,X))
```
| 95,910 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
n,x=map(int,input().split())
def p(n,x):
if x<=n:
return 0
if n==0:
return 1
i=0
while 2**(2+i)-3+n-i<=x:
i+=1
if 2**(i+1)-2+n-i==x:
return 2**i-1
else:
return 2**i+p(i-1,x-(2**(i+1)+n-i-2)-1)
print(p(n,x))
```
| 95,911 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
n, x = map(int, input().split())
S, P = [1], [1]
for i in range(n):
S.append(S[i]*2+3)
P.append(P[i]*2+1)
def f(n, x):
if n == 0:
return 0 if x <= 0 else 1
elif x <= 1+S[n-1]:
return f(n-1, x-1)
else:
return P[n-1] + 1 + f(n-1, x-2-S[n-1])
print(f(n, x))
```
| 95,912 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
n,x=map(int,input().split())
p=[1]
a=[1]
for i in range(0,n):
p.append(p[i]*2+1)
a.append(a[i]*2+3)
def f(n,x):
if n==0 and x<=0:
return 0
elif n==0:
return 1
elif x<=1+a[n-1]:
return f(n-1,x-1)
else:
return p[n-1]+f(n-1,x-a[n-1]-2)+1
print(f(n,x))
```
| 95,913 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
N, X = map(int, input().split())
a = [2**(i+2) - 3 for i in range(N+1)] # 層の数
p = [2**(i+1) - 1 for i in range(N+1)] # パティの数
def f(N, X):
if N == 0:
return 1 if X > 0 else 0
elif X <= 1 + a[N-1]:
return f(N-1, X-1)
else:
return p[N-1] + 1 + f(N-1, X-2-a[N-1])
print(f(N, X))
```
| 95,914 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
n,x=map(int,input().split())
a,p=[1],[1]
for i in range(n):
a.append(a[i]*2+3)
p.append(p[i]*2+1)
def f(n,x):
if n==0:
return 0 if x<=0 else 1
elif x<=a[n-1]+1:
return f(n-1,x-1)
else:
return p[n-1]+1+f(n-1,x-2-a[n-1])
print(f(n,x))
```
| 95,915 |
Provide a correct Python 3 solution for this coding contest problem.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
"Correct Solution:
```
N, X = map(int, input().split())
def count(N, X):
if N == 0:
return 1
if X == 1:
return 0
elif 1 < X < 2**(N+1) - 1:
return count(N-1, X-1)
elif X == 2**(N+1) - 1:
return 2**N
elif 2**(N+1) - 1 < X:
return 2**N + count(N-1, X-2**(N+1)+1)
print(count(N, X))
```
| 95,916 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
N, X = map(int, input().split())
a, p = [1], [1]
for i in range(N):
a.append(a[i] * 2 + 3)
p.append(p[i] * 2 + 1)
def f(N, X): # X <= 0 や X > a_N を許容し解説本文から簡略化
if N == 0:
return 0 if X <= 0 else 1
elif X <= 1 + a[N-1]:
return f(N-1, X-1)
else:
return p[N-1] + 1 + f(N-1, X-2-a[N-1])
print(f(N, X))
```
Yes
| 95,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
def f(N, X):
m = 2 ** (N + 1) - 1
if X <= N:
return 0
elif N == 0:
return 1
elif X < m:
return f(N-1, X-1)
else:
return 2 ** N + f(N-1, X-m)
N, X = map(int, input().split())
print(f(N, X))
```
Yes
| 95,918 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
def hamb(n,x):
if x<=n:return 0
elif x==1 and n==0:return 1
elif x<=s[n-1]+1:return hamb(n-1,x-1)
elif x==s[n]:return p[n]
else:return hamb(n-1,x-2-s[n-1])+p[n-1]+1
n,x=map(int,input().split())
s=[0]*(n+1)
p=[0]*(n+1)
s[0]=1
p[0]=1
for i in range(n):
s[i+1]=3+2*s[i]
p[i+1]=1+2*p[i]
print(hamb(n,x))
```
Yes
| 95,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
n,x=map(int, input().split())
b=[1]*51
p=[1]*51
for i in range(50):
b[i+1]=2*b[i]+3
p[i+1]=2*p[i]+1
def eat(l,x):
if x==0:
return 0
if l==0:
return 1
if x<b[l-1]+2:
ans=eat(l-1,x-1)
else:
ans=p[l-1]+1+eat(l-1,x-2-b[l-1])
return ans
print(eat(n,x))
```
Yes
| 95,920 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
N, X = map(int, input().split())
patty = [1]
buns = [0]
for i in range(0, N):
patty.append(patty[i] * 2 + 1)
buns.append(buns[i] * 2 + 2)
# print(patty)
# print(buns)
burger_high = patty[N] + buns[N]
def dfs(now_runrun, now_burger_high, dimension):
if now_runrun == 1:
return 0
elif now_runrun == now_burger_high // 2 + 1:
return patty[dimension - 1] + 1
elif now_runrun == now_burger_high:
return patty[dimension]
elif now_runrun <= now_burger_high // 2:
return dfs(now_runrun - 1, now_burger_high // 2, dimension - 1)
else:
return dfs(now_runrun - (patty[dimension - 1] + buns[dimension - 1]) - 2,
now_burger_high // 2, dimension - 1) + patty[dimension - 1] + 1
print(dfs(X, burger_high, N))
```
No
| 95,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import math
N,X = list(map(int, input().rstrip().split()))
#-----
## レベルnバーガーの、層の数
def material(n):
return 2**(n+2)-3
## レベルnバーガーの、Pattyの数
def patty(n):
return 2**(n+1)-1
cnt=0
while N > 0:
half = material(N) // 2 + 1
if X == half:
cnt = patty(N-1) + 1
break
elif X == 1:
break
elif X > half:
cnt += patty(N-1) + 1
N -= 1
X -= half
elif 1 < X < half:
N -= 1
X -= 1
## last
if N == 0 and X == 1:
cnt += 1
print(cnt)
```
No
| 95,922 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
n,x=map(int,input().split())
l,p=[1],[1]
for i in range(n):
l+=[l[-1]*2+3]
p+=[p[-1]*2+1]
x-=1
a=0
while x>0:
t=l[n]//2
if x>=t:
a+=p[n-1]+1
x-=t
x-=1
n-=1
print(a)
```
No
| 95,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In some other world, today is Christmas.
Mr. Takaha decides to make a multi-dimensional burger in his party. A level-L burger (L is an integer greater than or equal to 0) is the following thing:
* A level-0 burger is a patty.
* A level-L burger (L \geq 1) is a bun, a level-(L-1) burger, a patty, another level-(L-1) burger and another bun, stacked vertically in this order from the bottom.
For example, a level-1 burger and a level-2 burger look like `BPPPB` and `BBPPPBPBPPPBB` (rotated 90 degrees), where `B` and `P` stands for a bun and a patty.
The burger Mr. Takaha will make is a level-N burger. Lunlun the Dachshund will eat X layers from the bottom of this burger (a layer is a patty or a bun). How many patties will she eat?
Constraints
* 1 \leq N \leq 50
* 1 \leq X \leq ( the total number of layers in a level-N burger )
* N and X are integers.
Input
Input is given from Standard Input in the following format:
N X
Output
Print the number of patties in the bottom-most X layers from the bottom of a level-N burger.
Examples
Input
2 7
Output
4
Input
1 1
Output
0
Input
50 4321098765432109
Output
2160549382716056
Submitted Solution:
```
n, x = map(int, input().split())
def pi(n):
return 2**(n+1)-1
def ai(n):
return 2**(n+2) - 3
def f(n, x):
if N == 0:
return 0 if X <= 0 else 1
elif (1 < x) and (x <= 1 + ai(n-1)):
return f(n - 1, x - 1)
elif x == 2 + ai(n - 1):
return pi(n - 1) + 1
elif (2 + ai(n - 1) < x) and (x <= 2 + 2 * ai(n - 1)):
return pi(n-1) + 1 + f(n-1, x-2-ai(n-1))
elif x >= ai(n):
return pi(n)
print(f(n, x))
```
No
| 95,924 |
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
"Correct Solution:
```
"""
https://atcoder.jp/contests/arc097/tasks/arc097_d
始点と終点関係ある？
→逆にしても問題はない
黒い葉は完全に無視できる(落とせる)
よって、葉はすべて白
葉以外は白黒どっちもあり得る
すべての葉をめぐる最短経路？
ある葉からスタートするのは自明っぽい？
ある白からスタートするのはそう
→自分を塗ってからdfsする
→全方位木dp?
行きのみで帰らない場所が1つ存在するはず
→始点と終点のパスとそこから生える木って感じのイメージ
足踏み(停止)の回数を極力少なくしたい
→始点でも終点でもない場合、 e本の辺がつながってたらe回訪問は確定
e^colorが黒でない場合は+1
→つまり、各頂点で消費する回数は始点・終点でない限り常に一定？
終点はまず葉→始点も葉でおｋ
始点は足踏みなので1秒つかう 終点も1秒使う
パス上の点は、e-1回訪問(ここが根本的に違う！)
つまり、あるパスがあって その長さがL(両端除く)の時、最大でL減る
ただし、パス上にあっても回数が減らない点もある
B=0 W=1とするか
(color^E) == 1 の点は、E-1回訪問になると足踏みがいらないから2減る
つまり、そのような点が最大数あるパスを求めればいい
"""
import sys
sys.setrecursionlimit(200000)
from collections import deque
MM = 0
def dfs(v,p):
global MM
if v != p and linknum[v] == 1:
return 0
cl = []
for nex in lis[v]:
if nex != p and exist[nex]:
tmp = dfs(nex,v)
cl.append(tmp)
cl.sort()
cl.reverse()
if len(cl) == 1:
MM = max(MM , cl[0])
else:
if (linknum[v]+c[v])%2 == 1:
MM = max(MM , cl[0] + cl[1] + 2)
else:
MM = max(MM , cl[0] + cl[1])
if (linknum[v]+c[v]) % 2 == 1:
return cl[0]+ 2
else:
return cl[0]
N = int(input())
lis = [ [] for i in range(N) ]
linknum = [0] * N
for i in range(N-1):
u,v = map(int,input().split())
u -= 1
v -= 1
lis[u].append(v)
lis[v].append(u)
linknum[u] += 1
linknum[v] += 1
ctmp = input()
c = []
for i in ctmp:
if i == "B":
c.append(0)
else:
c.append(1)
exist = [True] * N
q = deque([])
for i in range(N):
if linknum[i] <= 1 and c[i] == 0:
q.append(i)
while len(q) > 0:
now = q.popleft()
exist[now] = False
linknum[now] = 0
for nex in lis[now]:
linknum[nex] -= 1
if linknum[nex] == 1 and c[nex] == 0:
q.append(nex)
#print (exist)
start = None
for i in range(N):
if exist[i]:
start = i
break
else:
print (0)
sys.exit()
if linknum[start] == 0:
print (1)
sys.exit()
ans = 0
#print (linknum)
for i in range(N):
if exist[i]:
if (linknum[i] + c[i]) % 2 == 0:
ans += linknum[i]
else:
ans += linknum[i] + 1
MM = 0
pick = dfs(start,start)
print (ans - MM)
```
| 95,925 |
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
"Correct Solution:
```
from collections import deque
def first_cut(links, colors):
tmp_links = links.copy()
for v, neighbors in tmp_links.items():
while len(neighbors) == 1 and colors[v]:
del links[v]
par = neighbors.pop()
links[par].remove(v)
v = par
neighbors = links[par]
return links
def diameter(links, flags):
def dfs(s):
fs = flags[s]
d, v = 0, 0
q = deque(sorted((fs + flags[v], v, s) for v in links[s]))
while q:
d, v, a = q.popleft()
for u in links[v]:
if u == a:
continue
fu = flags[u]
if fu:
q.append((d + 1, u, v))
else:
q.appendleft((d, u, v))
return d, v
s = next(iter(links))
_, t = dfs(s)
d, _ = dfs(t)
return d
def solve(links, colors):
if all(colors):
return 0
links = first_cut(links, colors)
k = len(links)
if k == 1:
return 1
flags = {v: colors[v] ^ (len(link) % 2 == 0) for v, link in links.items()}
euler_tour = 2 * (k - 1) + sum(flags.values())
return euler_tour - 2 * diameter(links, flags)
n = int(input())
links = {i: set() for i in range(n)}
for _ in range(n - 1):
x, y = map(int, input().split())
x -= 1
y -= 1
links[x].add(y)
links[y].add(x)
colors = [c == 'B' for c in input()]
print(solve(links, colors))
```
| 95,926 |
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
data = read().split()
m = map(int,data[:-1])
XY = zip(m,m)
C = [0] + [1 * (x == ord('W')) for x in data[-1]]
graph = [set() for _ in range(N+1)]
for x,y in XY:
graph[x].add(y)
graph[y].add(x)
deg = [len(x) for x in graph]
b_leaf = [i for i in range(N+1) if deg[i] == 1 and not C[i]]
while b_leaf:
x = b_leaf.pop()
for y in graph[x]:
graph[y].remove(x)
deg[y] -= 1
if deg[y] == 1 and not C[y]:
b_leaf.append(y)
deg[x] = 0
graph[x].clear()
V = set(i for i,E in enumerate(graph) if E)
if len(V) <= 2:
print(sum(C))
exit()
root = None
for v in V:
if deg[v] > 1:
root = v
break
parent = [0] * (N+1)
order = []
stack = [root]
while stack:
x = stack.pop()
order.append(x)
for y in graph[x]:
if y == parent[x]:
continue
parent[y] = x
stack.append(y)
"""
・Euler tourの状態から、パスをひとつ除く
・葉と葉を結ぶとしてよい。通ると2コスト安くなる点がある
"""
for i in range(N+1):
C[i] ^= (1°[i])
cost_full = sum(deg) + sum(C)
# C[i] == 1 となる i を最もたくさん含むパスを探す
opt = 0
dp = [0] * (N+1)
for v in order[::-1]:
if deg[v] == 1:
continue
arr = sorted((dp[w] for w in graph[v] if w != parent[v]),reverse = True) + [0]
x = arr[0] + arr[1] + C[v]
if opt < x:
opt = x
dp[v] = arr[0] + C[v]
answer = cost_full - 2 * opt
print(answer)
```
| 95,927 |
Provide a correct Python 3 solution for this coding contest problem.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
"Correct Solution:
```
import sys
from collections import deque
input=sys.stdin.readline
N=int(input())
edge=[[] for i in range(N)]
for i in range(N-1):
x,y=map(int,input().split())
edge[x-1].append(y-1)
edge[y-1].append(x-1)
c=input()[:N]
deg=[len(edge[i]) for i in range(N)]
leaf=set([])
for i in range(N):
if deg[i]==1 and c[i]=="B":
leaf.add(i)
ban=set([])
while leaf:
v=leaf.pop()
ban.add(v)
deg[v]=0
for nv in edge[v]:
deg[nv]-=1
if deg[nv]==1 and c[nv]=="B":
leaf.add(nv)
for i in range(N):
edge[i]=[nv for nv in edge[i] if nv not in ban]
root=-1
for i in range(N):
if i not in ban:
root=i
parent=[-2]*N
deq=deque([(root,-1)])
node=[]
while deq:
v,pv=deq.popleft()
parent[v]=pv
node.append(v)
for nv in edge[v]:
if nv!=pv:
deq.append((nv,v))
node=node[::-1]
for i in range(N):
edge[i]=[nv for nv in edge[i] if nv!=parent[i]]
check=True
for i in range(N):
check&=(deg[i]<=0)
if check:
print(int(c[root]=="W"))
exit()
cond=[0]*N
for v in range(N):
if (deg[v]%2==1 and c[v]=="B") or (deg[v]%2==0 and c[v]=="W"):
cond[v]+=1
else:
cond[v]-=1
lower=[0]*N
for v in node:
res=0
for nv in edge[v]:
res=max(res,lower[nv])
res+=1+cond[v]
lower[v]=res
upper=[0]*N
node=node[::-1]
for v in node:
n=len(edge[v])
if n>1:
left=[0]*n
right=[0]*n
for i in range(n-1):
nv=edge[v][i]
left[i]=max(left[i-1],lower[nv]+2+cond[v])
nv=edge[v][-1]
upper[nv]=left[n-2]+cond[nv]
right[n-1]=lower[nv]+2+cond[v]
for i in range(n-2,0,-1):
nv=edge[v][i]
upper[nv]=max(left[i-1],right[i+1])+cond[nv]
right[i]=max(right[i+1],lower[nv]+2+cond[v])
if edge[v][0]!=pv:
nv=edge[v][0]
upper[nv]=right[1]+cond[nv]
if v!=root:
for nv in edge[v]:
upper[nv]=max(upper[nv],upper[v]+1+cond[nv])
base=sum(deg[i] for i in range(N))+sum(cond[i]==1 for i in range(N))
#print(deg)
#print(base)
#print(lower)
#print(upper)
#print(base)
print(base-max(max(upper),max(lower)))
```
| 95,928 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
Submitted Solution:
```
import sys
from collections import deque
input=sys.stdin.readline
N=int(input())
edge=[[] for i in range(N)]
for i in range(N-1):
x,y=map(int,input().split())
edge[x-1].append(y-1)
edge[y-1].append(x-1)
c=input()[:N]
deg=[len(edge[i]) for i in range(N)]
leaf=set([])
for i in range(N):
if deg[i]==1 and c[i]=="B":
leaf.add(i)
ban=set([])
while leaf:
v=leaf.pop()
ban.add(v)
deg[v]=0
for nv in edge[v]:
deg[nv]-=1
if deg[nv]==1 and c[nv]=="B":
leaf.add(nv)
for i in range(N):
edge[i]=[nv for nv in edge[i] if nv not in ban]
root=-1
for i in range(N):
if i not in ban:
root=i
parent=[-2]*N
deq=deque([(root,-1)])
node=[]
while deq:
v,pv=deq.popleft()
parent[v]=pv
node.append(v)
for nv in edge[v]:
if nv!=pv:
deq.append((nv,v))
node=node[::-1]
for i in range(N):
edge[i]=[nv for nv in edge[i] if nv!=parent[i]]
check=True
for i in range(N):
check&=(deg[i]<=0)
if check:
print(int(c[root]=="W"))
exit()
cond=[0]*N
for v in range(N):
if (deg[v]%2==1 and c[v]=="B") or (deg[v]%2==0 and c[v]=="W"):
cond[v]+=1
else:
cond[v]-=1
lower=[0]*N
for v in node:
res=0
for nv in edge[v]:
res=max(res,lower[nv])
res+=1+cond[v]
lower[v]=res
upper=[0]*N
for v in node:
n=len(edge[v])
if n>1:
left=[0]*n
right=[0]*n
for i in range(n-1):
nv=edge[v][i]
left[i]=max(left[i-1],lower[nv]+2+cond[v])
nv=edge[v][-1]
upper[nv]=left[n-2]+cond[nv]
right[n-1]=lower[nv]+2+cond[v]
for i in range(n-2,0,-1):
nv=edge[v][i]
upper[nv]=max(left[i-1],right[i+1])+cond[nv]
right[i]=max(right[i+1],lower[nv]+2+cond[v])
if edge[v][0]!=pv:
nv=edge[v][0]
upper[nv]=right[1]+cond[nv]
if v!=root:
for nv in edge[v]:
upper[nv]=max(upper[nv],upper[v]+1+cond[nv])
base=sum(deg[i] for i in range(N))+sum(cond[i]==1 for i in range(N))
#print(deg)
#print(base)
#print(lower)
#print(upper)
#print(base)
print(base-max(max(upper),max(lower)))
```
No
| 95,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
Submitted Solution:
```
"""
https://atcoder.jp/contests/arc097/tasks/arc097_d
始点と終点関係ある？
→逆にしても問題はない
黒い葉は完全に無視できる(落とせる)
よって、葉はすべて白
葉以外は白黒どっちもあり得る
すべての葉をめぐる最短経路？
ある葉からスタートするのは自明っぽい？
ある白からスタートするのはそう
→自分を塗ってからdfsする
→全方位木dp?
行きのみで帰らない場所が1つ存在するはず
→始点と終点のパスとそこから生える木って感じのイメージ
足踏み(停止)の回数を極力少なくしたい
→始点でも終点でもない場合、 e本の辺がつながってたらe回訪問は確定
e^colorが黒でない場合は+1
→つまり、各頂点で消費する回数は始点・終点でない限り常に一定？
終点はまず葉→始点も葉でおｋ
始点は足踏みなので1秒つかう 終点も1秒使う
パス上の点は、e-1回訪問(ここが根本的に違う！)
つまり、あるパスがあって その長さがL(両端除く)の時、最大でL減る
ただし、パス上にあっても回数が減らない点もある
B=0 W=1とするか
(color^E) == 1 の点は、E-1回訪問になると足踏みがいらないから2減る
つまり、そのような点が最大数あるパスを求めればいい
"""
import sys
sys.setrecursionlimit(200000)
from collections import deque
MM = 0
def dfs(v,p):
global MM
if v != p and linknum[v] == 1:
return 0
cl = []
for nex in lis[v]:
if nex != p and exist[nex]:
tmp = dfs(nex,v)
cl.append(tmp)
cl.sort()
cl.reverse()
if len(cl) == 1:
MM = max(MM , cl[0])
else:
if (linknum[v]+c[v])%2 == 1:
MM = max(MM , cl[0] + cl[1] + 2)
else:
MM = max(MM , cl[0] + cl[1])
if (linknum[v]+c[v]) % 2 == 1:
return cl[0]+ 2
else:
return cl[0]
N = int(input())
lis = [ [] for i in range(N) ]
linknum = [0] * N
for i in range(N-1):
u,v = map(int,input().split())
u -= 1
v -= 1
lis[u].append(v)
lis[v].append(u)
linknum[u] += 1
linknum[v] += 1
ctmp = input()
c = []
for i in ctmp:
if i == "B":
c.append(0)
else:
c.append(1)
exist = [True] * N
q = deque([])
for i in range(N):
if linknum[i] <= 1 and c[i] == 0:
q.append(i)
while len(q) > 0:
now = q.popleft()
exist[now] = False
linknum[now] = 0
for nex in lis[now]:
linknum[nex] -= 1
if linknum[nex] == 1 and c[nex] == 0:
q.append(nex)
#print (exist)
start = None
for i in range(N):
if exist[i]:
start = i
break
else:
print (0)
sys.exit()
if linknum[start] == 0:
print (1)
sys.exit()
ans = 0
#print (linknum)
for i in range(N):
if exist[i]:
if (linknum[i] + c[i]) % 2 == 0:
ans += linknum[i]
else:
ans += linknum[i] + 1
MM = 0
pick = dfs(start,start)
print (ans - MM)
```
No
| 95,930 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
Submitted Solution:
```
from collections import deque
def first_cut(links, colors):
tmp_links = links.copy()
for v, neighbors in tmp_links.items():
while len(neighbors) == 1 and colors[v]:
del links[v]
par = neighbors.pop()
links[par].remove(v)
v = par
neighbors = links[par]
return links
def diameter(links):
flags = {v: colors[v] ^ (len(link) % 2 == 0) for v, link in links.items()}
s = next(iter(links))
fs = flags[s]
v, d = 0, 0
q = deque(sorted((fs + flags[v], v, s) for v in links[s]))
while q:
d, v, a = q.popleft()
for u in links[v]:
if u == a:
continue
fu = flags[u]
if fu:
q.append((d + 1, u, v))
else:
q.appendleft((d, u, v))
t = v
ft = flags[t]
q = deque(sorted((ft + flags[v], v, t) for v in links[s]))
while q:
d, v, a = q.popleft()
for u in links[v]:
if u == a:
continue
fu = flags[u]
if fu:
q.append((d + 1, u, v))
else:
q.appendleft((d, u, v))
return d
def solve(links, colors):
if all(colors):
return 0
if sum(colors) == len(colors) - 1:
return 1
links = first_cut(links, colors)
k = len(links)
euler_tour = 2 * (k - 1) + sum(colors[v] ^ (len(link) % 2 == 0) for v, link in links.items())
return euler_tour - 2 * diameter(links)
n = int(input())
links = {i: set() for i in range(n)}
for _ in range(n - 1):
x, y = map(int, input().split())
x -= 1
y -= 1
links[x].add(y)
links[y].add(x)
colors = [c == 'B' for c in input()]
print(solve(links, colors))
```
No
| 95,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a tree with N vertices numbered 1 through N. The i-th edge connects Vertex x_i and y_i. Each vertex is painted white or black. The initial color of Vertex i is represented by a letter c_i. c_i = `W` represents the vertex is white; c_i = `B` represents the vertex is black.
A cat will walk along this tree. More specifically, she performs one of the following in one second repeatedly:
* Choose a vertex that is adjacent to the vertex where she is currently, and move to that vertex. Then, invert the color of the destination vertex.
* Invert the color of the vertex where she is currently.
The cat's objective is to paint all the vertices black. She may start and end performing actions at any vertex. At least how many seconds does it takes for the cat to achieve her objective?
Constraints
* 1 ≤ N ≤ 10^5
* 1 ≤ x_i,y_i ≤ N (1 ≤ i ≤ N-1)
* The given graph is a tree.
* c_i = `W` or c_i = `B`.
Input
Input is given from Standard Input in the following format:
N
x_1 y_1
x_2 y_2
:
x_{N-1} y_{N-1}
c_1c_2..c_N
Output
Print the minimum number of seconds required to achieve the objective.
Examples
Input
5
1 2
2 3
2 4
4 5
WBBWW
Output
5
Input
6
3 1
4 5
2 6
6 1
3 4
WWBWBB
Output
7
Input
1
B
Output
0
Input
20
2 19
5 13
6 4
15 6
12 19
13 19
3 11
8 3
3 20
16 13
7 14
3 17
7 8
10 20
11 9
8 18
8 2
10 1
6 13
WBWBWBBWWWBBWWBBBBBW
Output
21
Submitted Solution:
```
input = __import__('sys').stdin.readline
MIS = lambda: map(int,input().split())
n = int(input())
adj = [set() for i in range(n+1)]
white = [False]
for i in range(n-1):
a, b = MIS()
adj[a].add(b); adj[b].add(a)
s = input().rstrip()
for i in range(n): white.append(s[i] == 'W')
# Corner cases
if sum(white) <= 1: print(sum(white)); exit()
# Remove black leaves
stack = [i for i in range(1,n+1) if len(adj[i]) == 1 and not white[i]]
while stack:
p = stack.pop()
pa = min(adj[p])
adj[pa].remove(p); adj[p].clear()
if len(adj[pa]) == 1 and not white[pa]: stack.append(pa)
# Simulate Euler tour
root = 1
while not adj[root]: root+= 1
tadj = [[] for i in range(n+1)]
vis = [False]*(n+1); vis[root] = True
stack = [root]
while stack:
p = stack.pop()
for q in adj[p]:
if vis[q]: continue
vis[q] = True; tadj[p].append(q)
stack.append(q)
for i in range(1, n+1):
if adj[i] and len(tadj[i])%2 == 0: white[i] = not white[i]
white[root] = not white[root]
# Diameter
def bfs(v):
stack = [v]
dist = [-1]*(n+1); dist[v] = white[v]
while stack:
p = stack.pop()
for q in adj[p]:
if dist[q] != -1: continue
dist[q] = dist[p] + white[q]
stack.append(q)
return dist
numedge = sum(map(len, adj))
d1 = bfs(root)
u = d1.index(max(d1))
d2 = bfs(u)
diam = max(d2)
print(numedge - diam)
```
No
| 95,932 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has come up with the following problem.
> You are given a sequence d of length N. Find the number of the undirected graphs with N vertices labeled 1,2,...,N satisfying the following conditions, modulo 10^{9} + 7:
>
> * The graph is simple and connected.
> * The degree of Vertex i is d_i.
>
When 2 \leq N, 1 \leq d_i \leq N-1, {\rm Σ} d_i = 2(N-1), it can be proved that the answer to the problem is \frac{(N-2)!}{(d_{1} -1)!(d_{2} - 1)! ... (d_{N}-1)!}.
Snuke is wondering what the answer is when 3 \leq N, 1 \leq d_i \leq N-1, { \rm Σ} d_i = 2N. Solve the problem under this condition for him.
Constraints
* 3 \leq N \leq 300
* 1 \leq d_i \leq N-1
* { \rm Σ} d_i = 2N
Input
Input is given from Standard Input in the following format:
N
d_1 d_2 ... d_{N}
Output
Print the answer.
Examples
Input
5
1 2 2 3 2
Output
6
Input
16
2 1 3 1 2 1 4 1 1 2 1 1 3 2 4 3
Output
555275958
Submitted Solution:
```
import sys
import numpy as np
import numba
from numba import njit
i8 = numba.int64
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
MOD = 10**9 + 7
@njit((i8, ), cache=True)
def fact_table(N):
inv = np.empty(N, np.int64)
inv[0] = 0
inv[1] = 1
for n in range(2, N):
q, r = divmod(MOD, n)
inv[n] = inv[r] * (-q) % MOD
fact = np.empty(N, np.int64)
fact[0] = 1
for n in range(1, N):
fact[n] = n * fact[n - 1] % MOD
fact_inv = np.empty(N, np.int64)
fact_inv[0] = 1
for n in range(1, N):
fact_inv[n] = fact_inv[n - 1] * inv[n] % MOD
return fact, fact_inv, inv
@njit((i8[::1], ), cache=True)
def main(A):
# key：サイクルに選んだ個数、(d-2)の和
# value：選んでいないもの：1/(d-1)!、選んだもの：1/(d-2)! の積の和
N = len(A)
U = N + 10
dp = np.zeros((U, U), np.int64)
dp[0, 0] = 1
fact, fact_inv, inv = fact_table(10**3)
for d in A:
newdp = np.zeros_like(dp)
# 選ばない場合の遷移
newdp += dp * fact_inv[d - 1] % MOD
# 選ぶ場合の遷移
if d >= 2:
newdp[1:, d - 2:U] += dp[:-1, 0:U + 2 - d] * fact_inv[d - 2] % MOD
dp = newdp % MOD
ret = 0
for n in range(3, N + 1):
for a in range(N + 1):
x = a * fact[n - 1] % MOD * fact[N - n - 1] % MOD
x = x * inv[2] % MOD * dp[n, a] % MOD
ret += x
return ret % MOD
A = np.array(read().split(), np.int64)[1:]
print(main(A))
```
No
| 95,933 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
x,a,b=map(int,input().split())
print("delicious"if a>=b else "safe" if b-a<=x else "dangerous")
```
| 95,934 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
X, A, B = list(map(int, input().split()))
print("delicious" if A >= B else "safe" if B - A <= X else "dangerous")
```
| 95,935 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
x,a,b=map(int,input().split())
if a-b>=0:
print("delicious")
elif -x<=a-b<0:
print("safe")
else:
print("dangerous")
```
| 95,936 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
X,A,B=map(int,input().split())
print(["delicious" if -A+B<=0 else "safe","dangerous"][X<-A+B])
```
| 95,937 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
x, a, b = [int(s) for s in input().split()]
print("delicious" if b <= a else ("safe" if b <= a + x else "dangerous"))
```
| 95,938 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
x,a,b = map(int, input().split())
if b > a+x:
print('dangerous')
elif b > a:
print('safe')
else:
print('delicious')
```
| 95,939 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
a,b,c=map(int,input().split());print('delicious' if c<=b else 'safe' if c<=a+b else 'dangerous')
```
| 95,940 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
"Correct Solution:
```
x,a,b=map(int,input().split())
if b<=a:print('delicious')
elif 0<(b-a)<=x:print('safe')
else:print('dangerous')
```
| 95,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
x,a,b = map(int, input().split())
if a+x < b:
print("dangerous")
elif a < b:
print("safe")
else:
print("delicious")
```
Yes
| 95,942 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
x,a,b = [int(i) for i in input().split()]
print('delicious' if x >= b else 'safe' if x >= (b-a) else 'dangerous')
```
Yes
| 95,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
X,A,B=map(int,input().split());print("dsdaaenflgeie cr io ou us s"[(B<=A)+(B<=A+X)::3])
```
Yes
| 95,944 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
a,b,c=map(int,input().split())
if c-b>a:
print("dangerous")
else:
print("safe"if b<c else"delicious")
```
Yes
| 95,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
x,a,b=map(int,input().split())
if a>=b:
print('delicious')
elif x>b-a:
print('safe')
else:
print('dangerous')
```
No
| 95,946 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
[a,b,c]=[int(x) for x in input().split()]
if b<c:
print("delicious")
elif a+b<c:
print("safe")
else:
print("dangerous")
```
No
| 95,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
x,a,b = map(int, input().split())
if a-b >= 0:
print("delicious")
elif 0 > a-b and abs(a-b) <= x+1:
print("safe")
else:
print("dangerous")
```
No
| 95,948 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though.
Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious.
Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it.
Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
Constraints
* 1 ≤ X,A,B ≤ 10^9
Input
Input is given from Standard Input in the following format:
X A B
Output
Print `delicious` if Takahashi found the food delicious; print `safe` if he neither found it delicious nor got a stomachache; print `dangerous` if he got a stomachache.
Examples
Input
4 3 6
Output
safe
Input
6 5 1
Output
delicious
Input
3 7 12
Output
dangerous
Submitted Solution:
```
x, a, b = map(int, input().split())
if b-a<=0:
print('delicious')
elif b-a <= x:
print('safe')
else:
print('dangerouse')
```
No
| 95,949 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
A, B, C = input().split(",")
print(A, B, C)
```
| 95,950 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
s = input().replace(',',' ')
print(s)
```
| 95,951 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
s=input()
s_2=s.replace(",", " ")
print(s_2)
```
| 95,952 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
a=input().split(",")
print(" ".join(a))
```
| 95,953 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
L = input().split(',')
print(' '.join(L))
```
| 95,954 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
a,b,c = input().split(sep=",")
print(a,b,c)
```
| 95,955 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
S = input().split(',')
print(*S,sep=' ')
```
| 95,956 |
Provide a correct Python 3 solution for this coding contest problem.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
"Correct Solution:
```
s=input()
print(s[0:5],s[6:13],s[14:19])
```
| 95,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
s = str(input())
print(s.replace(",", " "))
```
Yes
| 95,958 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
S = input()
print(S[:5], S[6:13], S[14:])
```
Yes
| 95,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
s = input().replace(',', ' ', 2)
print(s)
```
Yes
| 95,960 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
a = input()
A = a.replace(","," ")
print(A)
```
Yes
| 95,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
s = input()
print(s.replace(' ', ','))
```
No
| 95,962 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
[a, b] = map(int, input().split())
if a*3==b:
print(1)
elif b%3==0:
print(b*3+1)
else:
print(b*3)
```
No
| 95,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
def check(x, y, K, S):
if x + y > S:
return False
for z in range(K+1):
if x + y + z == S:
return True
return False
K, S = list(map(int, input('').split()))
count = 0
for x in range(K+1):
for y in range(K+1):
if check(x, y, K, S):
count += 1
print(count)
```
No
| 95,964 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a New Year's gift, Dolphin received a string s of length 19.
The string s has the following format: `[five lowercase English letters],[seven lowercase English letters],[five lowercase English letters]`.
Dolphin wants to convert the comma-separated string s into a space-separated string.
Write a program to perform the conversion for him.
Constraints
* The length of s is 19.
* The sixth and fourteenth characters in s are `,`.
* The other characters in s are lowercase English letters.
Input
The input is given from Standard Input in the following format:
s
Output
Print the string after the conversion.
Examples
Input
happy,newyear,enjoy
Output
happy newyear enjoy
Input
haiku,atcoder,tasks
Output
haiku atcoder tasks
Input
abcde,fghihgf,edcba
Output
abcde fghihgf edcba
Submitted Solution:
```
a = input()
a[5] = " "
a[13] = " "
print(a[0]+a[1]+a[2]+a[3]+a[4]+a[5]+a[6]+a[7]+a[8]+a[9]+a[10]+a[11]+a[12]+a[13]+a[14]+a[15]+a[16]+a[17]+a[18])
```
No
| 95,965 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
def bfs(s: str, n: int, i: int, j: int, k: int):
return k+int(s[j:]) if i==n-1 else bfs(s,n,i+1,i+1,k+int(s[j:i+1]))+bfs(s,n,i+1,j,k)
s=input()
n=len(s)
print(bfs(s,n,0,0,0))
```
| 95,966 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
S = input()
l = len(S)
ans = 0
for i in range(l):
for j in range(i+1,l+1):
temp = int(S[i:j])
ans += temp * max(1,pow(2,i-1)) * max(1,pow(2,l-j-1))
print(ans)
```
| 95,967 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
s = input()
n = len(s)
ans = 0
for i in range(2**(n-1)):
key = ""
for j in range(n-1):
key = key + s[j]
if (i>>j)&1:
key = key + "+"
key = key + s[n-1]
ans += eval(key)
print(ans)
```
| 95,968 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
S = input()
L = len(S)-1
a = 0
for i in range(2**L):
T = S[0]
for j in range(L):
T+=(i>>j&1)*"+"+S[1+j]
a+=eval(T)
print(a)
```
| 95,969 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
s = input()
ans = 0
for i in range(2**(len(s)-1)):
t = 0
for j in range(len(s)-1):
if (i >> j)&1:
ans += int(s[t:j+1])
t = j+1
ans += int(s[t:])
print(ans)
```
| 95,970 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
s = input()
l = len(s)
def dfs(i, f):
if i == l - 1:
return sum(list(map(int, f.split('+'))))
return dfs(i + 1, f + s[i + 1]) + dfs(i + 1, f + '+' + s[i + 1])
print(dfs(0, s[0]))
```
| 95,971 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
s=input()
l=len(s)
def dfs(i,f):
if i==l-1:
return eval(f)
else:
return dfs(i+1, f+s[i+1]) + dfs(i+1,f+'+'+s[i+1])
print(dfs(0,s[0]))
```
| 95,972 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
"Correct Solution:
```
S = input()
def func(S, i):
if len(S) <= i:
return eval(S)
S1 = S[:i] + '+' + S[i:]
return func(S, i + 1) + func(S1, i + 2)
print(func(S, 1))
```
| 95,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
c,*s=input();n=len(s);print(sum(eval(c+''.join('+'*(b>>i&1)+s[i]for i in range(n)))for b in range(2**n)))
```
Yes
| 95,974 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
S = input()
L = len(S)
def dfs(i, f):
if i == L-1: return sum(map(int, f.split('+')))
return dfs(i+1, f+S[i+1]) + dfs(i+1, f+'+'+S[i+1])
print(dfs(0, S[0]))
```
Yes
| 95,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
def dfs(i, f):
if i == n - 1:
return eval(f)
return dfs(i + 1, f + s[i + 1]) + \
dfs(i + 1, f + "+" + s[i + 1])
s = input()
n = len(s)
print(dfs(0, s[0]))
```
Yes
| 95,976 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
s = input()
ans = 0
for i in range(0, len(s)):
for j in range(i+1, len(s)+1):
ans += int(s[i:j]) * max(1, 2**(i-1)) * max(1, 2**(len(s)-j-1))
print(ans)
```
Yes
| 95,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
S = list(input())
cnt = 0
for i in range(1 << len(S)-1):
f = S[0]
for j, b in enumerate(format(i, 'b').zfill(len(S)-1)):
if(b == '1'):
f += '+'
f += S[j+1]
cnt += sum(map(int,f.split('+')))
print(cnt)
```
No
| 95,978 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
print(dfs(0,s[0]))
```
No
| 95,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
import sys
import copy
S = input()
op_list = ["+"]*(len(S)-1)
sum = 0
for i in range(2**len(op_list)):
for j in range(len(op_list)):
if (i >> j) & 1:
op_list[j] = ""
s = ""
for k in range(len(op_list)):
s = s + S[k] + op_list[k]
sum = sum + eval(s)
print(sum)
```
No
| 95,980 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion.
All strings that can be obtained in this way can be evaluated as formulas.
Evaluate all possible formulas, and print the sum of the results.
Constraints
* 1 \leq |S| \leq 10
* All letters in S are digits between `1` and `9`, inclusive.
Input
The input is given from Standard Input in the following format:
S
Output
Print the sum of the evaluated value over all possible formulas.
Examples
Input
125
Output
176
Input
9999999999
Output
12656242944
Submitted Solution:
```
s = input()
op_count = len(s) - 1
op_list = []
for i in range(2**op_count):
op = [''] * op_count
for j in range(op_count):
if (i >> j) & 1:
op[op_count - 1 - j] = '+'
op_list.append(op)
ans = 0
for i in range(2 * op_count):
cur = 1
t = s
for j in range(op_count):
if op_list[i][j] == '+':
t = t[:j + cur] + "+" + t[j + cur:]
cur += 1
ans += eval(t)
print(ans)
```
No
| 95,981 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
while True:
try:
L = input().split()
except:
break
S = []
for i in L:
if i not in "+-*/":
S.append(int(i))
elif i == "+":
S.append(S.pop()+S.pop())
elif i == "-":
t = S.pop()
S.append(S.pop()-t)
elif i == "*":
S.append(S.pop()*S.pop())
else:
t = S.pop()
S.append(S.pop() / t)
print("{:.6f}".format(S[0]))
```
| 95,982 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
def get_input():
while True:
try:
yield ''.join(input())
except EOFError:
break
N = list(get_input())
for l in range(len(N)):
p = N[l].split()
s = []
for i in range(len(p)):
if p[i] == "+":
a = float(s.pop())
b = float(s.pop())
s.append(str(b+a))
elif p[i] == "-":
a = float(s.pop())
b = float(s.pop())
s.append(str(b-a))
elif p[i] == "*":
a = float(s.pop())
b = float(s.pop())
s.append(str(b*a))
elif p[i] == "/":
a = float(s.pop())
b = float(s.pop())
s.append(str(b/a))
else:
s.append(p[i])
print(s[0])
```
| 95,983 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
import re
while True:
try:
f = input().split()
except:
break
stack = []
for c in f:
if re.match("-*[0-9]", c) is None:
b, a = str(stack.pop()), str(stack.pop())
stack.append(float(eval(a+c+b)))
else:
stack.append(c)
print(stack[0])
```
| 95,984 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
d = []
while True:
try:
lst = list(input().split())
for i in lst:
if i == '+':
num = d.pop()
d[-1] = d[-1] + num
elif i == '-':
num = d.pop()
d[-1] = d[-1] - num
elif i == '*':
num = d.pop()
d[-1] = d[-1] * num
elif i == '/':
num = d.pop()
d[-1] = d[-1] / num
else:
d.append(int(i))
print(d.pop())
except EOFError:
break
```
| 95,985 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
while True :
try :
lst = list(input().split())
except EOFError :
break
stack = []
for i in lst :
if i == '+' :
b = stack.pop(-1)
a = stack.pop(-1)
stack.append(a+b)
elif i == '-' :
b = stack.pop(-1)
a = stack.pop(-1)
stack.append(a-b)
elif i == '*' :
b = stack.pop(-1)
a = stack.pop(-1)
stack.append(a*b)
elif i == '/' :
b = stack.pop(-1)
a = stack.pop(-1)
stack.append(a/b)
else :
stack.append(int(i))
print('{:.8f}'.format(stack[0]))
```
| 95,986 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
while True:
try:
a = input().split()
stack = []
for inp in a:
if inp == "+":
x = stack.pop()
y = stack.pop()
stack.append(x+y)
elif inp == "-":
x = stack.pop()
y = stack.pop()
stack.append(y-x)
elif inp == "*":
x = stack.pop()
y = stack.pop()
stack.append(x*y)
elif inp == "/":
x = stack.pop()
y = stack.pop()
stack.append(y/x)
else:
stack.append(float(inp))
print(*stack)
except:
break
```
| 95,987 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
import sys
def isNum(a) :
try:
int(a)
except:
return False
return True
for line in sys.stdin:
stack = []
task = line.strip().split(" ") ;
for i in range(0, len(task) ) :
if isNum(task[i]) :
stack.append(task[i])
elif len(stack) != 0 :
stack.append( str( eval( stack.pop(-2) + task[i] + stack.pop(-1) ) ) )
print("{:6f}".format(float(stack[0])))
```
| 95,988 |
Provide a correct Python 3 solution for this coding contest problem.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
"Correct Solution:
```
def invpol(f):
s = []
for c in f:
if c == '+':
a = s.pop()
s.append(s.pop() + a)
elif c == '-':
a = s.pop()
s.append(s.pop() - a)
elif c == '*':
a = s.pop()
s.append(s.pop() * a)
elif c == '/':
a = s.pop()
s.append(s.pop() / a)
else:
s.append(int(c))
return(s[0])
while True:
try:
f = input().strip().split()
print("%.6f" % invpol(f))
except EOFError:
break
```
| 95,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
import os
import math
for s in sys.stdin:
lst = s.split()
# stack
S = []
for c in lst:
if c == '+':
a = S.pop()
b = S.pop()
S.append(b + a)
elif c == '-':
a = S.pop()
b = S.pop()
S.append(b - a)
elif c == '*':
a = S.pop()
b = S.pop()
S.append(b * a)
elif c == '/':
a = S.pop()
b = S.pop()
S.append(b / a)
else:
S.append(float(c))
print(S[0])
```
Yes
| 95,990 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
import sys
f = sys.stdin
from collections import deque
import operator
func = {'+':operator.add,'-':operator.sub,'*':operator.mul,'/':operator.truediv }
for line in f:
stack = deque()
for element in line.strip().split():
if element in func:
b = stack.pop()
a = stack.pop()
stack.append(func[element](a, b))
else:
stack.append(float(element))
print('{:.6f}'.format(stack[0]))
```
Yes
| 95,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
while(1):
try:
a = [i for i in input().split()]
stack = []
l = len(stack)
for i in a:
if i == "+":
x = stack[l-2]
y = stack[l-1]
stack.pop()
stack.pop()
stack.append(x+y)
elif i == "-":
x = stack[l-2]
y = stack[l-1]
stack.pop()
stack.pop()
stack.append(x-y)
elif i == "*":
x = stack[l-2]
y = stack[l-1]
stack.pop()
stack.pop()
stack.append(x*y)
elif i == "/":
x = stack[l-2]
y = stack[l-1]
stack.pop()
stack.pop()
stack.append(x/y)
else:
stack.append(float(i))
print("{:.6f}".format(stack[0]))
except EOFError:
break
```
Yes
| 95,992 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
while 1:
try:s=list(input().split())
except:break
a=[]
for x in s:
if x in ['+','-','*','/']:
c,b=str(a.pop()),str(a.pop())
a+=[float(eval(b+x+c))]
else:a+=[float(x)]
print(*a)
```
Yes
| 95,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
import sys
def line():return sys.stdin.readline().strip()
ops = {"+":lambda a,b:b + a,
"-":lambda a,b:b - a,
"*":lambda a,b:b * a,
"/":lambda a,b:b / a}
while True:
stack = []
for s in line().split():
if s in ops:
stack.append(ops[s](stack.pop(),stack.pop()))
else:
stack.append(int(s))
print(stack[0])
```
No
| 95,994 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
s=list();[[[s.append(eval('{2}{1}{0}'.format(s.pop(),i,s.pop()))if(i in'+-*/')else i)for i in input().split(' ')]and print(s.pop())]for x in __import__('itertools').count()]
```
No
| 95,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
#!/usr/bin/env python3
import sys
def calc_inverse_polish_notation_string(s):
tokens = [token for token in s.split(' ') if token != '']
stack = []
for token in tokens:
if token in ['+', '-', '*', '/']:
val2 = stack.pop()
val1 = stack.pop()
result = eval("%s %s %s" % (val1, token, val2))
stack.append(result)
else:
stack.append(token)
return stack[0] # if len(stack) != 1, it should throw an exception
def main():
for line in sys.stdin.readlines():
print(calc_inverse_polish_notation_string(line))
if __name__ == "__main__":
main()
```
No
| 95,996 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr .: Peter, I've finally done it.
Peter: See you again? What kind of silly invention is this time?
Dr .: I finally came up with a revolutionary way to process mathematical formulas on a computer. Look at this table.
Ordinary notation | Dr.'s "breakthrough" notation
--- | ---
1 + 2 | 1 2 +
3 * 4 + 7 | 3 4 * 7 +
10 / (2 --12) | 10 2 12-/
(3-4) * (7 + 2 * 3) | 3 4 --7 2 3 * + *
Peter: Yeah.
Dr .: Fufufu. This alone won't tell you what it means to an inexperienced person. It's important from here.
Peter: I mean ...
Dr .: You know that computers have a data structure called a stack. Look, that's "first in, then out".
Peter: Yes. I know, that ...
Dr .: This groundbreaking notation uses that stack. For example, this 10 2 12-/, but process as follows.
Processing target | 10 | 2 | 12 |-| /
--- | --- | --- | --- | --- | ---
| ↓ | ↓ | ↓ | ↓ 2-12 | ↓ 10 / -10
Stack | |.
---
..
Ten
| ..
---
2
Ten
| 12
---
2
Ten
| ..
---
-Ten
Ten
| ..
---
..
-1
Dr .: How is it? You don't have to worry about parentheses or operator precedence, right? The word order is also "10 divided by 2 minus 12", which is somewhat similar to his Far Eastern island nation, Japanese. With this epoch-making invention, our laboratory is safe. Fafafa.
Peter: I mean, Doctor. I learned this in the basic course at the University of Aizu when I was in Japan. Everyone had a simple program called "Reverse Polish Notation".
Dr .: ...
So, instead of Peter, I decided to teach this program to the doctor. Create a program that inputs the formula written in "Reverse Polish Notation" and outputs the calculation result.
input
Given multiple datasets. For each dataset, a formula in Reverse Polish Notation (a string of up to 80 characters with integers and arithmetic symbols separated by one blank character (half-width)) is given on one line. No formula is given that divides a value by 0 or a value that is as close to 0 as possible.
The number of datasets does not exceed 50.
output
Output the calculation result (real number) on one line for each data set. The calculation result may include an error of 0.00001 or less.
Example
Input
10 2 12 - /
3 4 - 7 2 3 * + *
-1 -2 3 + +
Output
-1.000000
-13.000000
0.000000
Submitted Solution:
```
import sys
f = sys.stdin
from collections import deque
import operator
func = {'+':operator.add,'-':operator.sub,'*':operator.mul,'/':operator.div }
for line in f:
stack = deque()
for element in line.strip().split():
if element.isdigit():
stack.append(float(element))
else:
b = stack.pop()
a = stack.pop()
stack.append(func[element](a, b))
print('{:.6f}'.format(stack[0]))
```
No
| 95,997 |
Provide a correct Python 3 solution for this coding contest problem.
There is an ice cream shop named Ten Ice Cream. At this store, we always have 10 types of ice cream on the shelves. The store manager creates a daily graph showing how well ice cream is selling for reference in product development.
For such a store manager, you decided to create a program that displays the number of each ice cream sold in a graph.
Enter the total number of ice creams sold in a day and the number of ice creams sold, and create a program that outputs as many * (half-width asterisks) as the number sold for each type of ice cream. However, the type of ice cream is represented by an integer from 0 to 9. Also, for ice cream with zero sales, one- (half-width hyphen) is output.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
c1
c2
::
cn
The first line gives the total number of ice creams sold per day n (1 ≤ n ≤ 10000). The next n lines are given the i-th ice cream type ci (0 ≤ ci ≤ 9).
The number of datasets does not exceed 20.
Output
For each input dataset, the number of sales is output in numerical order of each ice cream type.
Example
Input
15
2
6
7
0
1
9
8
7
3
8
9
4
8
2
2
3
9
1
5
0
Output
*
*
***
*
*
-
*
**
***
**
-
*
-
-
-
*
-
-
-
*
"Correct Solution:
```
while True:
N=int(input())
A=0
B=0
C=0
D=0
E=0
F=0
G=0
H=0
I=0
J=0
i=0
j=0
L=[]
if N==0:
break
else:
L=[int(input()) for _ in range (N)]
for i in range (N):
if L[i]==0:
A+=1
elif L[i]==1:
B+=1
elif L[i]==2:
C+=1
elif L[i]==3:
D+=1
elif L[i]==4:
E+=1
elif L[i]==5:
F+=1
elif L[i]==6:
G+=1
elif L[i]==7:
H+=1
elif L[i]==8:
I+=1
else:
J+=1
S=[A,B,C,D,E,F,G,H,I,J]
for j in range(10):
T=S[j]
if T==0:
print("-")
else:
k=0
print("*"*T)
```
| 95,998 |
Provide a correct Python 3 solution for this coding contest problem.
There is an ice cream shop named Ten Ice Cream. At this store, we always have 10 types of ice cream on the shelves. The store manager creates a daily graph showing how well ice cream is selling for reference in product development.
For such a store manager, you decided to create a program that displays the number of each ice cream sold in a graph.
Enter the total number of ice creams sold in a day and the number of ice creams sold, and create a program that outputs as many * (half-width asterisks) as the number sold for each type of ice cream. However, the type of ice cream is represented by an integer from 0 to 9. Also, for ice cream with zero sales, one- (half-width hyphen) is output.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
c1
c2
::
cn
The first line gives the total number of ice creams sold per day n (1 ≤ n ≤ 10000). The next n lines are given the i-th ice cream type ci (0 ≤ ci ≤ 9).
The number of datasets does not exceed 20.
Output
For each input dataset, the number of sales is output in numerical order of each ice cream type.
Example
Input
15
2
6
7
0
1
9
8
7
3
8
9
4
8
2
2
3
9
1
5
0
Output
*
*
***
*
*
-
*
**
***
**
-
*
-
-
-
*
-
-
-
*
"Correct Solution:
```
while True:
n= int(input())
if n== 0: break
c= [int(input()) for _ in range(n)]
for i in range(10):
cc= c.count(i)
print('-' if cc== 0 else cc*"*")
```
| 95,999 |
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