text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Tags: brute force, greedy, implementation, math, strings
Correct Solution:
```
s,ans=input(),0
for i in range(len(s)):
f=s.find("bear",i)
if f!=-1:
ans+=len(s)-f-3
print(ans)
```
| 2,800 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Tags: brute force, greedy, implementation, math, strings
Correct Solution:
```
from math import ceil
s=input();n=len(s);ans=0
last_idx=-1
for i in range(n-3):
if s[i:i+4]=='bear':
if i==0:ans+=(n-1)-(i+3)+1;last_idx=i
else:
k=i-last_idx-1
k1=(n-1)-(i+3)+1
ans+=k*k1
ans+=k1
last_idx=i
print(ans)
```
| 2,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Tags: brute force, greedy, implementation, math, strings
Correct Solution:
```
s=input()
if(len(s)<=3):
print(0)
else:
n=len(s)
ans=0
A=0
for i in range(3,n):
if(s[i-3]+s[i-2]+s[i-1]+s[i]=='bear'):
ans+=((i-3)-A+1)*(n-i)
A=i-2
print(ans)
```
| 2,802 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
s = input()
k = 0
for i in range(len(s)):
c = s.find('bear', i)
if c >= 0:
k += len(s) - c - 3
print(k)
```
Yes
| 2,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
s=input().rstrip()
x=list(s)
if len(x)<4:
print(0)
else:
l=[]
q=[]
for i in range(0,len(x)-4+1):
V=x[i:i+4]
if ''.join(V)=="bear":
l.append(i+1)
q.append(i+4)
total=0;
for i in range(0,len(l)):
if i==0:
A=l[i]-0
B=q[i]
total+=(A * (len(x)-B+1));
else:
A=l[i];
B=q[i];
C=l[i-1];
D=A-C
E=len(x)-B+1
total+=(E*D)
print(total)
```
Yes
| 2,804 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
s=str(input())
i=0
n=len(s)
ans=0
count=0
while(i<n):
if s[i:i+4]=="bear":
if count==0:
ans=ans+n-i-4+1
ans=ans+(i-0)*(n-(i+3))
k=i+1
# print(ans)
else:
p=i-k
ans=ans+(p)*(n-i-3)
# print(ans)
ans=ans+n-i-4+1
# print(ans)
k=i+1
count+=1
i=i+4
else:
i+=1
print(ans)
```
Yes
| 2,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
s = input().strip()
p = "bear"
l = len(s)
start = 0
total = 0
while True:
i = s.find(p, start)
if i==-1:
break
prev = (i-start)+1
multiplier = l - (i+3)
total += prev * multiplier
start = i+1
print(total)
```
Yes
| 2,806 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
a=input()
b=0
for i in range(len(a)):
c=a.find('bear',i)
if(c>0):
b+=len(a)-c-3
print(b)
```
No
| 2,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
######### # # # # ##### # ##### # ## # ## # #
"""
PPPPPPP RRRRRRR OOOO VV VV EEEEEEEEEE
PPPPPPPP RRRRRRRR OOOOOO VV VV EE
PPPPPPPPP RRRRRRRRR OOOOOOOO VV VV EE
PPPPPPPP RRRRRRRR OOOOOOOO VV VV EEEEEE
PPPPPPP RRRRRRR OOOOOOOO VV VV EEEEEEE
PP RRRR OOOOOOOO VV VV EEEEEE
PP RR RR OOOOOOOO VV VV EE
PP RR RR OOOOOO VV VV EE
PP RR RR OOOO VVVV EEEEEEEEEE
"""
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
from bisect import bisect_left
def solve():
s = input()
if s[-1] == "\n":
s = s[:-1]
c = 0
a = ""
ans = 0
last = 0
for i in range(len(s)):
a = a + s[i]
# print(a)
if a == "b" or a == "be" or a == "bea":
continue
elif a != "bear":
a = ""
if a == "bear":
# print(len(s), i)
last = i
c += 1
m = (len(s) - i) * (i - 2)
# print(m)
ans += m
a = ""
# print(last, ans)
print(ans - (len(s) - last) * (c - 1))
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
```
No
| 2,808 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
import io
import os
from collections import Counter, OrderedDict
from itertools import permutations as perm
from fractions import Fraction
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
# range = xrange
def nc2(x):
print(x)
v = x*(x-1)
return v//2
s = g()
n = len(s)
r = "bear"
ans = 0
p = []
for i in range(n-3):
if s[i:i+4] == r:
p.append(i)
if len(p) == 0:
print(-1)
exit()
ans += nc2(p[0]+3)
for i in range(1, len(p)):
ans += nc2(p[i]-p[i-1]+2)
ans += nc2(n-1-p[-1])
ans -= len(p)*nc2(2)
ans = nc2(n) - ans
print(ans)
```
No
| 2,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The bear has a string s = s1s2... s|s| (record |s| is the string's length), consisting of lowercase English letters. The bear wants to count the number of such pairs of indices i, j (1 ≤ i ≤ j ≤ |s|), that string x(i, j) = sisi + 1... sj contains at least one string "bear" as a substring.
String x(i, j) contains string "bear", if there is such index k (i ≤ k ≤ j - 3), that sk = b, sk + 1 = e, sk + 2 = a, sk + 3 = r.
Help the bear cope with the given problem.
Input
The first line contains a non-empty string s (1 ≤ |s| ≤ 5000). It is guaranteed that the string only consists of lowercase English letters.
Output
Print a single number — the answer to the problem.
Examples
Input
bearbtear
Output
6
Input
bearaabearc
Output
20
Note
In the first sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9).
In the second sample, the following pairs (i, j) match: (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (1, 11), (2, 10), (2, 11), (3, 10), (3, 11), (4, 10), (4, 11), (5, 10), (5, 11), (6, 10), (6, 11), (7, 10), (7, 11).
Submitted Solution:
```
from sys import stdin ,stdout
from os import path
if(path.exists('input.txt')):
stdin = open("input.txt","r")
x=stdin.readline()
k=0
for i in range(len(x)):
t=x.find('bear',i)
if t>=0:
k+=len(x)-t-3
stdout.write(str(k)+'\n')
```
No
| 2,810 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
from sys import stdin
test = stdin.readlines()
n = int(test[0])
matrix = [test[i+1].split() for i in range(n)]
dot = 0
for i in range(n):
dot ^= matrix[i][i] == '1'
out = []
for q in range(int(test[n + 1])):
query = test[n + q + 2].split()
if len(query) == 1:
out.append(dot)
else:
dot ^= 1
print(''.join(map(str, out)))
```
| 2,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n = int(input())
lis=list(list(map(int, input().split())) for _ in range(n))
u=0
for i in range(n):
for j in range(n):
if i==j:
u^=lis[i][j]
ans =[]
k = int(input())
for i in range(k):
s = input()
if s[0]=='3':
ans.append(str(u))
else:
u^=1
print(''.join(ans))
```
| 2,812 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
def main():
from sys import stdin, stdout
from functools import reduce
from operator import xor
n = int(int(input()))
mat = [list(map(int, input().split())) for _ in range(n)]
ans, q, a = reduce(xor, [mat[i][i] for i in range(n)]), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
if __name__ == '__main__':
main()
```
| 2,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
from functools import reduce
from operator import xor
def main():
n = int(int(input()))
mat = [list(map(int, input().split())) for _ in range(n)]
ans, q, a = reduce(xor, [mat[i][i] for i in range(n)]), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
if __name__ == '__main__':
main()
```
| 2,814 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
from sys import stdin
test = stdin.readlines()
n = int(test[0])
dot = 0
j = 0
for i in range(n):
if test[i+1][j] == '1':
dot ^= 1
j += 2
out = []
for q in range(int(test[n + 1])):
query = test[n + q + 2].split()
if len(query) == 1:
out.append(dot)
else:
dot ^= 1
print(''.join(map(str, out)))
```
| 2,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
def main():
from sys import stdin
from operator import xor
from functools import reduce
x, res = reduce(xor, (input()[i] == '1' for i in range(0, int(input()) * 2, 2))), []
input()
for s in stdin.read().splitlines():
if s == '3':
res.append("01"[x])
else:
x ^= True
print(''.join(res))
if __name__ == "__main__":
main()
```
| 2,816 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
from sys import stdin, stdout
from functools import reduce
from operator import xor
def arr_inp(n):
return [int(x) for x in input().split()]
class Matrix:
def __init__(self, r, c, mat=None):
self.r, self.c = r, c
if mat != None:
self.mat = mat
else:
self.mat = [[0 for i in range(c)] for j in range(r)]
def __add__(self, other):
mat0 = Matrix(self.r, self.c)
for i in range(self.r):
for j in range(self.c):
mat0.mat[i][j] = self.mat[i][j] + other.mat[i][j]
return mat0.mat
def __mul__(self, other):
mat0 = Matrix(self.r, other.c)
for i in range(self.r):
for j in range(other.c):
for k in range(self.c):
mat0.mat[i][j] += self.mat[i][k] * other.mat[k][j]
return mat0.mat
def trace(self):
res = 0
for i in range(self.r):
res += self.mat[i][i]
return res % 2
def dot_mul(self, other):
res = 0
for i in range(self.r):
for j in range(self.c):
res += self.mat[i][j] * other.mat[j][i]
return res % 2
def rotate(self):
mat0 = Matrix(self.c, self.r)
for i in range(self.r):
for j in range(self.c):
mat0.mat[j][self.r - (i + 1)] = self.mat[i][j]
self.mat, self.r, self.c = mat0.mat.copy(), self.c, self.r
return self.mat
def reflect(self):
mat0 = Matrix(self.r, self.c)
for i in range(self.r):
for j in range(self.c):
mat0.mat[i][self.c - (j + 1)] = self.mat[i][j]
self.mat = mat0.mat.copy()
return self.mat
n = int(int(input()))
mat = Matrix(n, n, [arr_inp(1) for _ in range(n)])
ans, q, a = mat.trace(), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
```
| 2,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Tags: implementation, math
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
n=Int()
ans=0
for i in range(n):
a=array()
ans+=a[i]
#print(ans)
for _ in range(Int()):
s=input()
if(s=='3'):
print(ans%2,end="")
else:
t,c=map(int,s.split())
ans+=1
```
| 2,818 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
from sys import stdin,stdout
cnt = 0
arr = []
for i in range(int(stdin.readline())):
if stdin.readline().split()[i]=='1':
arr.append(1)
cnt += 1
else:
arr.append(0)
stdin.readline()
for i in stdin:
if len(i)<3:
stdout.write(str(cnt%2))
else:
_,j = map(int,i.split())
j -= 1
if arr[j]:
arr[j] = 0
cnt -= 1
else:
arr[j] = 1
cnt += 1
```
Yes
| 2,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
from sys import stdin
input = stdin.readline
n = int(input())
lis=list(list(map(int, input().split())) for _ in range(n))
u=0
for i in range(n):
for j in range(n):
if i==j:
u^=lis[i][j]
ans =[]
k = int(input())
for i in range(k):
s = input()
if s[0]=='3':
ans.append(str(u))
else:
u^=1
print(''.join(ans))
```
Yes
| 2,820 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
from sys import stdin, stdout
from functools import reduce
from operator import xor
from threading import Thread, stack_size
def arr_inp(n):
return [int(x) for x in input().split()]
def main():
n = int(int(input()))
mat = [list(map(int, input().split())) for _ in range(n)]
ans, q, a = reduce(xor, [mat[i][i] for i in range(n)]), int(input()), []
queries = [stdin.readline() for i in range(q)]
for query in queries:
if query[0] == '3':
a.append(str(ans))
else:
ans ^= 1
print(''.join(a))
if __name__ == '__main__':
stack_size(102400000)
thread = Thread(target=main)
thread.start()
```
Yes
| 2,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
import os,sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n=int(input())
mat=[]
for i in range(n):
l=list(map(int,input().split()))
mat.append(l)
summ=0
for i in range(n):
for j in range(n):
if(i==j):
summ+=mat[i][j]
q=int(input())
for i in range(q):
p=list(map(int,input().split()))
if(p[0]==3):
print(summ%2,end="")
else:
summ+=1
```
Yes
| 2,822 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
from operator import and_, xor
from functools import reduce
from itertools import chain
from sys import stdin
input = stdin.readline
n = int(input())
l = list(chain(*list(list(map(int, input().split())) for _ in range(n))))
q = int(input())
commands = list(list(map(int, input().split())) for _ in range(q))
output = []
for i in range(q):
if commands[i][0] == 3:
ans = 0
for i in range(n):
ans += sum([*map(and_, l[i*n:(1+i)*n], l[i::n])]) % 2
ans %= 2
output.append(ans)
if commands[i][0] == 2:
col = commands[i][1] - 1
l[col::n] = [*map(lambda v : 1 - v, l[col::n])]
if commands[i][0] == 1:
row = commands[i][1] - 1
l[row*n:(row+1)*n] = [*map(lambda v : 1 - v, l[row*n:(row+1)*n])]
print(''.join([*map(str, output)]))
```
No
| 2,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
from operator import and_, xor
from functools import reduce
from itertools import chain
from sys import stdin
input = stdin.readline
n = int(input())
l = list(chain(*list(list(map(int, input().split())) for _ in range(n))))
q = int(input())
commands = list(list(map(int, input().split())) for _ in range(q))
output = list()
for i in range(q):
if commands[i][0] == 3:
ans = 0
for i in range(n):
ans += sum([*map(and_, l[i*n:(1+i)*n], l[i::n])]) % 2
ans %= 2
output += [ans]
if commands[i][0] == 2:
col = commands[i][1] - 1
l[col::n] = [*map(lambda v : 1 - v, l[col::n])]
if commands[i][0] == 1:
row = commands[i][1] - 1
l[row*n:(row+1)*n] = [*map(lambda v : 1 - v, l[row*n:(row+1)*n])]
print(''.join([*map(str, l)]))
```
No
| 2,824 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
def main():
from sys import stdin
from operator import xor
from functools import reduce
x, res = reduce(xor, (input()[i] == '1' for i in range(0, int(input()) * 2, 2))), []
for s in stdin.read().splitlines():
if s == '3':
res.append("01"[x])
else:
x ^= True
print(''.join(res))
if __name__ == "__main__":
main()
```
No
| 2,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.
The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.
Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:
<image>
The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.
However, there is much more to the homework. Chris has to process q queries; each query can be one of the following:
1. given a row index i, flip all the values in the i-th row in A;
2. given a column index i, flip all the values in the i-th column in A;
3. find the unusual square of A.
To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.
Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?
Input
The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.
The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following:
* 1 i — flip the values of the i-th row;
* 2 i — flip the values of the i-th column;
* 3 — output the unusual square of A.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.
Examples
Input
3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3
Output
01001
Submitted Solution:
```
current = False
n = int(input())
for i in range(n):
v = [bool for i in input().split()]
if v[i] == True:
current = not current
output = ""
q = int(input())
for i in range(q):
query = [int(x) for x in input().split()]
if query[0] == 3:
output += str(int(current))
# print(int(current))
else:
current = not current
print(output)
```
No
| 2,826 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
# Input
n = int(input())
v = list(map(int, input().split()))
# Sorting V
sorted_v = sorted(v)
# Prefix Sum Arrays
prefix_v = [0]
sorted_prefix = [0]
sum_v = 0
sums = 0
# Prefix For Unsorted Values
for x in v:
sums += x
prefix_v.append(sums)
# Prefix For Sorted Values
for x in sorted_v:
sum_v += x
sorted_prefix.append(sum_v)
for _ in range(int(input())):
question = list(map(int, input().split()))
q = question[0]
# For Question 2
if q == 2:
point1 = question[1] - 1
point2 = question[2]
sums = 0
print(sorted_prefix[point2] - sorted_prefix[point1])
# For Question 1
else:
point1 = question[1] - 1
point2 = question[2]
print(prefix_v[point2] - prefix_v[point1])
```
| 2,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
I = lambda:map(int,input().split())
n = int(input())
v = [0] + list(I())
u = sorted(v)
for i in range(1,n + 1):
v[i] += v[i-1]
u[i] += u[i-1]
for _ in range(int(input())):
t,l,r = I()
if t == 1:
print(v[r] - v[l-1])
else:
print(u[r] - u[l-1])
```
| 2,828 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
from itertools import accumulate
n = int(input())
v = [0]+list(map(int,input().split()))
ls = sorted(v)
v = list(accumulate(v))
ls = list(accumulate(ls))
m = int(input())
for i in range(m):
s = 0
t,l,r = map(int,input().split())
if t==1:
print(v[r]-v[l-1])
elif t==2:
print(ls[r]-ls[l-1])
```
| 2,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
n = int(input())
v = list(map(int, input().split(' ')))
u = sorted(v)
s_v = [0]
s_u = [0]
for i in range(1, n + 1):
s_v.append(s_v[i - 1] + v[i - 1])
s_u.append(s_u[i - 1] + u[i - 1])
ans = []
for _ in range(int(input())):
t, l, r = map(int, input().split(' '))
if t == 1:
tt = s_v[r] - s_v[l - 1]
else:
tt = s_u[r] - s_u[l - 1]
ans.append(str(tt))
print('\n'.join(ans))
```
| 2,830 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
n=int(input())
arr=list(map(int, input().split()))
arranged=sorted(arr)
sum1=sum(arr)
sum2=sum(arranged)
prefix_sum1=[arr[0]]
for i in range(1,n):
prefix_sum1.append(arr[i]+prefix_sum1[-1])
suffix_sum1=[arr[-1]]
for i in range(n-2,-1,-1):
suffix_sum1.append(arr[i]+suffix_sum1[-1])
prefix_sum2=[arranged[0]]
for i in range(1,n):
prefix_sum2.append(arranged[i]+prefix_sum2[-1])
suffix_sum2=[arranged[-1]]
for i in range(n-2,-1,-1):
suffix_sum2.append(arranged[i]+suffix_sum2[-1])
m=int(input())
for i in range(m):
type, l, r = map(int, input().split())
if type==1:
if r-l+1==n:
print(sum1)
elif l==1:
print(sum1-(suffix_sum1[n-r-1]))
elif r==n:
print(sum1-(prefix_sum1[l-2]))
elif r==n:
print(sum1-(prefix_sum1[l-2]+suffix_sum1[n-r-1]))
else:
print(sum1-(prefix_sum1[l-2]+suffix_sum1[n-r-1]))
else:
if r-l+1==n:
print(sum2)
elif l==1:
print(sum2-(suffix_sum2[n-r-1]))
elif r==n:
print(sum2-(prefix_sum2[l-2]))
elif r==n:
print(sum2-(prefix_sum2[l-2]+suffix_sum2[n-r-1]))
else:
print(sum2-(prefix_sum2[l-2]+suffix_sum2[n-r-1]))
```
| 2,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
n=int(input())
v=list(map(int,input().split()))
l1=[0]*(n+1)
l2=[0]*(n+1)
l1[1]=v[0]
for i in range(2,n+1):
l1[i]=l1[i-1]+v[i-1]
v.sort()
l2[1]=v[0]
for i in range(2,n+1):
l2[i]=l2[i-1]+v[i-1]
m=int(input())
for i in range(m):
t,l,r=map(int,input().split())
if t==1:
print(l1[r]-l1[l-1])
else:
print(l2[r]-l2[l-1])
```
| 2,832 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
n=int(input())
v=list(map(int,input().split()))
u=[]
for i in range(n):
u.append(v[i])
u.sort()
for i in range(1,n):
v[i]+=v[i-1]
for i in range(1,n):
u[i]+=u[i-1]
c=[0]
d=[0]
for i in range(n):
c.append(v[i])
d.append(u[i])
q=int(input())
for i in range(q):
a,l,r=map(int,input().split())
if a==1:
print(c[r]-c[l-1])
else:
print(d[r]-d[l-1])
```
| 2,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Tags: dp, implementation, sortings
Correct Solution:
```
# Kuriyama mari stones
from typing import List
class Solution:
@classmethod
def sum_array(cls, numbers: List[int]) -> List[int]:
sum_numbers = [0]
for index, num in enumerate(numbers):
prev = sum_numbers[index]
next = numbers[index]
sum_numbers.append(prev + next)
return sum_numbers
@classmethod
def sorted_sum_array(cls, numbers: List[int]) -> List[int]:
sorted_numbers = sorted(numbers)
sorted_sum_numbers = cls.sum_array(sorted_numbers)
return sorted_sum_numbers
@classmethod
def query(
cls,
sum_numbers: List[int],
sorted_sum_numbers: List[int],
qtype: int,
low: int,
high: int,
) -> int:
if qtype == 1:
return sum_numbers[high] - sum_numbers[low - 1]
else:
return sorted_sum_numbers[high] - sorted_sum_numbers[low - 1]
if __name__ == "__main__":
num = int(input())
numbers = list(map(lambda x: int(x), input().split()))
sum_numbers = Solution.sum_array(numbers)
sorted_sum_numbers = Solution.sorted_sum_array(numbers)
num_questions = int(input())
for i in range(num_questions):
[qtype, low, high] = list(map(lambda x: int(x), input().split()))
answer = Solution.query(sum_numbers, sorted_sum_numbers, qtype, low, high)
print(answer)
```
| 2,834 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
"Codeforces Round #339 (Div. 2)"
"B. Gena's Code"
# y=int(input())
# # a=list(map(int,input().split()))
# a=list(input().split())
# nz=0
# nb=''
# z=0
# # print(len(str(z)))
# for i in a:
# if i=='0':
# z=1
# break
# else:
# s='1'
# l=(len(i)-1)
# qz='0'*l
# s+=qz
# if s==i:
# nz+=l
# else:
# nb=i
# if nb=='':
# nb='1'
# ans=nb+('0'*nz)
# if z==1:
# ans='0'
# print(ans)
"Codeforces Round #177 (Div. 2)"
"B. Polo the Penguin and Matrix"
# n,m,d=map(int,input().split())
# a=[]
# for i in range(n):
# b=list(map(int,input().split()))
# a.extend(b)
# a.sort()
# fa=a[0]
# f=0
# c=(a[len(a)//2]-fa)//d
# moves=0
# for i in a:
# if (i-fa)%d>0:
# f=-1
# moves+=abs(int((i-fa)/d)-c)
# if f==-1:
# print(-1)
# else:
# print(moves)
"Codeforces Round #264 (Div. 2)"
"B. Caisa and Pylons"
# y=int(input())
# a=list(map(int,input().split()))
# mini=0
# p=-a[0]
# for i in range(1,y):
# if p<mini:
# mini=p
# p=p+a[i-1]-a[i]
# if p<mini:
# mini=p
# if mini<0:
# print(-1*mini)
# else:
# print(0)
"Codeforces Beta Round #79 (Div. 2 Only)"
"B. Sum of Digits"
# y=input()
# def sumofdigits(s):
# ans=0
# for i in s:
# ans+=int(i)
# return ans
# n=0
# while len(y)>1:
# n+=1
# y=str(sumofdigits(y))
# print(n)
"Codeforces Beta Round #70 (Div. 2)"
"B. Easter Eggs"
# y=int(input())
# y=y-7
# s="ROYGBIV"
# sq=["G","B","I","V"]
# i=0
# while y>0:
# if i==4:
# i=0
# s+=sq[i]
# i+=1
# y-=1
# print(s)
"Codeforces Round #386 (Div. 2)"
"B. Decoding"
# y=int(input())
# sq=input()
# i=y%2
# s=""
# s+=sq[0]
# for j in range(1,y):
# if i==0:
# s+=sq[j]
# else:
# s=sq[j]+s
# i=1-i
# print(s)
"Codeforces Round #280 (Div. 2)"
"B. Vanya and Lanterns"
# n,l=map(int,input().split())
# a=list(map(int,input().split()))
# a.sort()
# max=float(a[0])
# for i in range(1,n):
# m=(a[i]-a[i-1])/2
# if max<m:
# max=m
# m=l-a[-1]
# if max<m:
# max=m
# print(max)
"Codeforces Round #248 (Div. 2)"
"B. Kuriyama Mirai's Stones"
y=int(input())
a=list(map(int,input().split()))
t1=int(input())
a1=[0]
for i in range(y):
a1.append(a1[-1]+a[i])
a.sort()
a2=[0]
for i in range(y):
a2.append(a2[-1]+a[i])
for i in range(t1):
t,l,r=map(int,input().split())
if t==1:
print(a1[r]-a1[l-1])
else:
print(a2[r]-a2[l-1])
```
Yes
| 2,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
n=int(input())
arr=input().split()
brr=sorted(arr,key=int)
for i in range(1,n):
arr[i]=int(arr[i-1])+int(arr[i])
for i in range(1,n):
brr[i]=int(brr[i-1])+int(brr[i])
m=int(input())
for i in range(m):
ty,l,r=map(int,input().split())
if ty==1:
if l-2>=0:
print(int(arr[r-1])-int(arr[l-2]))
else:
print(int(arr[r-1]))
else:
if l-2>=0:
print(int(brr[r-1])-int(brr[l-2]))
else:
print(int(brr[r-1]))
```
Yes
| 2,836 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
n= int(input())
mas1 = list(map(int,input().split(" ")))
mas2 = sorted(mas1)
for i in range(1,n):
mas1[i]+=mas1[i-1]
mas2[i]+=mas2[i-1]
m = int(input())
mas1.append(0)
mas2.append(0)
print("\n".join(map(str, (mas1[b-1]-mas1[a-2] if s==1 else mas2[b-1]-mas2[a-2] for s, a, b in (map(int,input().split(" ")) for i in range(m))))))
```
Yes
| 2,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
n = int(input())
arr = [0]
for x in input().split():
arr.append(int(x))
ac = [0]
for i in range(1,n+1):
ac.append(arr[i] + ac[i-1])
arr.sort()
acSorted = [0]
for i in range(1,n+1):
acSorted.append(arr[i] + acSorted[i-1])
m = int(input())
for i in range(m):
op, l, r = map(int, input().split())
if (op == 1):
print(ac[r] - ac[l-1])
else:
print(acSorted[r] - acSorted[l-1])
# 1481936527735
```
Yes
| 2,838 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
n = int(input())
data1 = input().split();
x = 0
for x in range(len(data1)):
data1[x] = int(data1[x])
data2 = data1[:]
data2.sort();
k = int(input())
def query1(fr, to):
sum = 0
for x in data1[fr:to+1]:
sum += x
print("outs: ",sum)
x = 0
def query2(fr, to):
sum = 0
for x in data2[fr:to+1]:
sum += x
print("outs: ",sum)
while x < k:
data3 = input().split();
if(int(data3[0]) is 1):
query1(int(data3[1])-1, int(data3[2])-1)
else:
query2(int(data3[1])-1, int(data3[2])-1)
x+=1
```
No
| 2,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
def rat(b,x,y):print(b[y]-b[x-1])
a=int(input())
b=[0]+list(map(int,input().split()))
c=sorted(b.copy())
for i in range(1,a+1):b[i]+=b[i-1];c[i]+=c[i-1]
for _ in " "*int(input()):
x,y,z=map(int,input().split())
if x==1:rat(b,y,z)
else:rat(c,x,y)
```
No
| 2,840 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
def lista_acumulada(lista):
lista[0] = int(lista[0])
i = 1
while i < len(lista):
lista[i] = int(lista[i]) + int(lista[i - 1])
i += 1
return lista
n = int(input())
entrada = input()
m = int(input())
questoes = []
res = []
i = 0
while i < m:
questoes.append(input().split())
i += 1
acumulada1 = lista_acumulada(entrada.split())
acumulada2 = lista_acumulada(sorted(entrada.split()))
for e in questoes:
l = int(e[1])
r = int(e[2])
if e[0] == '1':
if l == 1:
res.append(acumulada1[r - 1])
else:
res.append(acumulada1[r - 1] - acumulada1[l - 2]) # na entrada os indices (l, r) são contados a partir do 1
else:
if l == 1:
res.append(acumulada2[r - 1])
else:
res.append(acumulada2[r - 1] - acumulada2[l - 2])
for e in res:
print(e)
```
No
| 2,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuriyama Mirai has killed many monsters and got many (namely n) stones. She numbers the stones from 1 to n. The cost of the i-th stone is vi. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
2. Let ui be the cost of the i-th cheapest stone (the cost that will be on the i-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, l and r (1 ≤ l ≤ r ≤ n), and you should tell her <image>.
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The second line contains n integers: v1, v2, ..., vn (1 ≤ vi ≤ 109) — costs of the stones.
The third line contains an integer m (1 ≤ m ≤ 105) — the number of Kuriyama Mirai's questions. Then follow m lines, each line contains three integers type, l and r (1 ≤ l ≤ r ≤ n; 1 ≤ type ≤ 2), describing a question. If type equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output
Print m lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Examples
Input
6
6 4 2 7 2 7
3
2 3 6
1 3 4
1 1 6
Output
24
9
28
Input
4
5 5 2 3
10
1 2 4
2 1 4
1 1 1
2 1 4
2 1 2
1 1 1
1 3 3
1 1 3
1 4 4
1 2 2
Output
10
15
5
15
5
5
2
12
3
5
Note
Please note that the answers to the questions may overflow 32-bit integer type.
Submitted Solution:
```
def sumLisNum(lis):
sumLis = [0] * len(lis)
sumLis[0] = lis[0]
for i in range(1, len(lis)):
sumLis[i] = sumLis[i - 1] + lis[i]
return sumLis
n = int(input())
v = list(map(int, input().split()))
u = v[:]
u.sort()
sumV = sumLisNum(v)
sumU = sumLisNum(u)
m = int(input())
print(sumV)
print(sumU)
for i in range(m):
t, l, r = map(int, input().split())
if t == 1:
print(sumV[r - 1] - sumV[l - 1] + v[l - 1])
else:
print(sumU[r - 1] - sumU[l - 1] + u[l - 1])
```
No
| 2,842 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serega loves fun. However, everyone has fun in the unique manner. Serega has fun by solving query problems. One day Fedor came up with such a problem.
You are given an array a consisting of n positive integers and queries to it. The queries can be of two types:
1. Make a unit cyclic shift to the right on the segment from l to r (both borders inclusive). That is rearrange elements of the array in the following manner:a[l], a[l + 1], ..., a[r - 1], a[r] → a[r], a[l], a[l + 1], ..., a[r - 1].
2. Count how many numbers equal to k are on the segment from l to r (both borders inclusive).
Fedor hurried to see Serega enjoy the problem and Serega solved it really quickly. Let's see, can you solve it?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of elements of the array. The second line contains n integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ n).
The third line contains a single integer q (1 ≤ q ≤ 105) — the number of queries. The next q lines contain the queries.
As you need to respond to the queries online, the queries will be encoded. A query of the first type will be given in format: 1 l'i r'i. A query of the second type will be given in format: 2 l'i r'i k'i. All the number in input are integer. They satisfy the constraints: 1 ≤ l'i, r'i, k'i ≤ n.
To decode the queries from the data given in input, you need to perform the following transformations:
li = ((l'i + lastans - 1) mod n) + 1; ri = ((r'i + lastans - 1) mod n) + 1; ki = ((k'i + lastans - 1) mod n) + 1.
Where lastans is the last reply to the query of the 2-nd type (initially, lastans = 0). If after transformation li is greater than ri, you must swap these values.
Output
For each query of the 2-nd type print the answer on a single line.
Examples
Input
7
6 6 2 7 4 2 5
7
1 3 6
2 2 4 2
2 2 4 7
2 2 2 5
1 2 6
1 1 4
2 1 7 3
Output
2
1
0
0
Input
8
8 4 2 2 7 7 8 8
8
1 8 8
2 8 1 7
1 8 1
1 7 3
2 8 8 3
1 1 4
1 2 7
1 4 5
Output
2
0
Submitted Solution:
```
n =int(input())
arr = [int(x) for x in input().split()]
big = 10**5
freq = [0]*(big + 1)
for i in arr:
freq[i] += 1
dp = {}
for a in range(big+1):
if(a==0):
val = 0
elif(a==1):
val = freq[1]
else:
val = max(dp[a-1], dp[a-2]+a*freq[a])
dp[a] = val
print(dp[big])
```
No
| 2,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
def f(k):
return k*(k-1)//2
n,m=map(int,input().split())
print(f(n//m+1)*(n%m)+f(n//m)*(m-n%m),f(n-m+1))
```
| 2,844 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
n, m = tuple(map(int, input().split()))
n1 = int(n / m)
mod = n % m
print((n1 + 1) * n1 // 2 * mod + n1 * (n1 - 1) // 2 * (m - mod), (n - m + 1) * (n - m) // 2)
```
| 2,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
# coding: utf-8
n, m = [int(i) for i in input().split()]
if n%m==0:
kmin = (n//m)*(n//m-1)//2*m
kmax = (n-m+1)*(n-m)//2
else:
t = n//m
kmin = t*(t+1)//2*(n%m)+t*(t-1)//2*(m-n%m)
kmax = (n-m+1)*(n-m)//2
print(kmin,kmax)
```
| 2,846 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
w=input().split(" ")
n=int(w[0])
m=int(w[1])
formax=n-m
maxx=int((formax)*(formax+1)//2)
a=n//m
b=n%m
minn=int((b*(a*(a+1)/2))+((m-b)*(a*(a-1)/2)))
print(str(minn)+" "+str(maxx))
```
| 2,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
n, m = map(int, input().split())
print(((n // m) * (n // m - 1)) // 2 * (m - n % m) + (n % m) * ((n // m + 1) * (n // m)) // 2,
(n - m + 1) * (n - m) // 2)
```
| 2,848 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
n, m = map(int, input().split())
#pentru doua gramezi este optim sa muti dintr-una in cealalta pentru a maximiza C(a,2)+C(b,2)
#a+b == n. graficul apare ca al unei functii de gradul 2 cu fundul in jos
#cum treci la 3/4.. variabile?
def P(x):
return x*(x-1)//2
print(P(n//m)*(m - n%m) + P(n//m+1)*(n%m), end = ' ')
print(P(n-m+1))
```
| 2,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
n, m = (int(s) for s in input().split(" "))
res2 = (n-m+1) * (n-m) // 2
res1 = (n//m - 1) * (n//m) * (m - n%m) //2 + (n//m ) * (n//m + 1) * (n%m)//2
print(res1, res2)
```
| 2,850 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Tags: combinatorics, constructive algorithms, greedy, math
Correct Solution:
```
n,k=map(int,input().split())
x=n-(k-1)
kmax = (x*(x-1))//2
t = n//k
y = n%k
kmin = (k-y)*((t*(t-1))//2) + y*(((t+1)*t)//2)
print(kmin,kmax)
```
| 2,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
n,m=map(int,input().split())
mxc=n%m
mnc=m-mxc
mn=n//m
mx=mn+1
print(mxc*mx*(mx-1)//2 + mnc*mn*(mn-1)//2, (n-m)*(n-m+1)//2)
```
Yes
| 2,852 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
n,m=map(int,input().split())
av=n//m
mi=av*(av-1)//2
mi*=m
mi+=av*(n%m)
ma=(n-m)*(n-m+1)//2
print(mi,ma)
```
Yes
| 2,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
def friends_in_team(amount):
if amount <= 1:
return 0
return (amount * (amount - 1)) // 2
n, m = (int(x) for x in input().split())
a = friends_in_team(n // m) * (m - n % m)
a += friends_in_team(n // m + 1) * (n % m)
b = friends_in_team(n - m + 1)
print('{} {}'.format(min(a, b), max(a, b)))
```
Yes
| 2,854 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
n, m = (int(x) for x in input().split())
q, r = n // m, n % m
kmin = r * (q + 1) * q // 2 + (m - r) * q * (q - 1) // 2
kmax = (n - m + 1) * (n - m) // 2
print(kmin, kmax)
```
Yes
| 2,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
n, m = map(int, input().split())
div = n//m
mod = n%m
mini = ((div-1) * (m-mod) + (div + 1) * (mod))*div/2
maxi = (n-(m+1)) * (n-m) / 2
print(int(mini), int(maxi))
```
No
| 2,856 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
w=input().split(" ")
n=int(w[0])
m=int(w[1])
formax=n-m
maxx=int((formax)*(formax+1)/2)
a=n//m
b=n%m
minn=int((b*(a*(a+1)/2))+((m-b)*(a*(a-1)/2)))
print(str(minn)+" "+str(maxx))
```
No
| 2,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
def comb(p): return ((p/2)*(p-1)) if (p%2==0) else (((p-1)/2)*p)
def max_pair(n,m): print(int(comb(n-m+1)),end="");
def min_pair(n,m):
val,mod,sm=n//m,n%m,0
for i in range(m):
if mod:
sm+=comb(val+1)
mod-=1
else: sm+=comb(val)
print(int(sm),end=" ")
n,m=input().split(" ")
n,m=int(n),int(m)
if n==m:
print("0 0")
else:
min_pair(n,m)
max_pair(n,m)
```
No
| 2,858 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n participants of the competition were split into m teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input
The only line of input contains two integers n and m, separated by a single space (1 ≤ m ≤ n ≤ 109) — the number of participants and the number of teams respectively.
Output
The only line of the output should contain two integers kmin and kmax — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Examples
Input
5 1
Output
10 10
Input
3 2
Output
1 1
Input
6 3
Output
3 6
Note
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
Submitted Solution:
```
import math
def combine(x,y):
if x==y:
return 1
elif y>x:
return 0
else:
ans = math.factorial(x)/(math.factorial(x-y)*math.factorial(y))
return ans
n,m=map(int,input().split())
#min
group=[]
check=n
later=m
a=n%m
while a!=0:
check-=a
later-=1
group.append(a)
a=check%later
group.append(a)
k=check//later
v=combine(k,2)
ans=0
ans+=(v*later)
for t in group:
v=combine(t,2)
ans+=(v)
print (int(ans),end=' ')
#max
x=n-(m-1)
v=combine(x,2)
print (int(v))
```
No
| 2,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Tags: combinatorics, dfs and similar, graphs, trees
Correct Solution:
```
from queue import Queue
import sys
cost = []
#coo
def readarray(): return map(int, input().split(' '))
n = int(input())
graph = [[] for i in range(n)]
for i in range(n - 1):
u, v, c = readarray()
u, v = u - 1, v - 1
cost.append(c)
graph[u].append((v, i))
graph[v].append((u, i))
order = []
used = [0] * n
q = [0] * (n + n)
qh = qt = 0
used[qh] = 1
qh += 1
while qt < qh:
v = q[qt]
qt += 1
order.append(v)
for (to, e) in graph[v]:
if used[to]:
continue
used[to] = 1
q[qh] = to
qh += 1
order.reverse()
sz = [0 for x in range(n)]
for v in order:
sz[v] = 1
for (to, e) in graph[v]:
sz[v] += sz[to]
"""
sz = [0] * n
sys.setrecursionlimit(100505)
def dfs(v, p):
sz[v] = 1
for (to, e) in graph[v]:
if to != p:
dfs(to, v)
sz[v] += sz[to]
dfs(0, -1)
"""
distanceSum = 0.0
edgeMult = [0] * n
for v in range(n):
for (to, e) in graph[v]:
x = min(sz[v], sz[to])
edgeMult[e] = x
distanceSum += 1.0 * cost[e] * x * (n - x)
distanceSum /= 2.0
queryCnt = int(input())
ans = []
for i in range(queryCnt):
x, y = readarray()
x -= 1
distanceSum -= 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
cost[x] = y
distanceSum += 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
ans.append('%.10lf' % (distanceSum / n / (n - 1) * 6.0))
print('\n'.join(ans))
```
| 2,860 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Tags: combinatorics, dfs and similar, graphs, trees
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
@bootstrap
def dfs(r,p):
if len(g[r])==1 and p!=-1:
yield 1
res=1
for child in g[r]:
if child!=p:
tmp=yield(dfs(child,r))
cnt[d[tuple(sorted([r,child]))]]=tmp*(n-tmp)
res+=tmp
yield res
t=1
for i in range(t):
n=N()
edg=[]
d={}
cnt=[0]*(n-1)
g=[[] for i in range(n)]
for i in range(n-1):
a,b,l=RL()
a-=1
b-=1
g[a].append(b)
g[b].append(a)
if a>b: a,b=b,a
d[a,b]=i
edg.append(l)
dfs(0,-1)
#print(cnt)
ans=sum(edg[i]*cnt[i] for i in range(n-1))
ans=ans*6/(n*(n-1))
#print(ans)
q=N()
for i in range(q):
c,l=RL()
dec=edg[c-1]-l
edg[c-1]=l
#print(dec)
ans-=dec*cnt[c-1]*6/((n-1)*n)
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
| 2,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Tags: combinatorics, dfs and similar, graphs, trees
Correct Solution:
```
import sys
sys.setrecursionlimit(1500)
MAX = 100005;
g = [[] for _ in range(MAX)]
vis = [False] * MAX
dp = [0] * MAX
prod = [0] * MAX
edges = []
order = []
def dfs(st):
stack = []
stack.append((st, -1))
vis[st] = True
while stack:
st, parent = stack.pop()
order.append(st)
vis[st] = True
if (st == parent): continue;
for i in g[st]:
if (vis[i[0]]): continue;
stack.append((i[0], st))
n = int(input())
for i in range(n-1):
a, b, w = map(int, sys.stdin.readline().split(' '))
g[a].append([b, w])
g[b].append([a,w])
edges.append([[a, b], w])
dfs(1);
order.reverse()
for st in order:
dp[st] = 1;
for i in g[st]:
dp[st] += dp[i[0]];
tot = 0;
curr = 1;
div = n * (n-1) / 2;
for i in edges:
a = i[0][0];
b = i[0][1];
sa = dp[a];
sb = dp[b];
tot += min(sa, sb) * (n - min(sa, sb)) * i[1];
prod[curr] = min(sa, sb) * (n - min(sa, sb));
curr += 1;
q = int(input())
for i in range(q):
q1, q2 = map(int, sys.stdin.readline().split(' '))
tot -= prod[q1] * edges[q1-1][1];
edges[q1-1][1] = q2;
tot += prod[q1] * edges[q1-1][1];
sys.stdout.write(str(tot*3/div)+"\n")
```
| 2,862 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Tags: combinatorics, dfs and similar, graphs, trees
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
sys.setrecursionlimit(10**5)
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
@bootstrap
def dfs(u,p):
global ans
for j in adj[u]:
if j!=p:
yield dfs(j,u)
sub[u]+=sub[j]
p1=sub[u]
p2=n-sub[u]
d1[(u,p)]=p1*p2*(p1+p2-2)
d1[(p, u)] = p1 * p2 * (p1 + p2 - 2)
ans+=(p1*p2*(p1+p2-2))*d[(u,p)]
yield
n=int(input())
adj=[[] for i in range(n+1)]
edges=[]
d=defaultdict(lambda:0)
d1=defaultdict(lambda:0)
ans=0
for i in range(n-1):
u,v,l=map(int,input().split())
d[(u,v)]=l
d[(v,u)]=l
edges.append([u,v])
adj[u].append(v)
adj[v].append(u)
sub=[1 for i in range(n+1)]
dfs(1,0)
val=n*(n-1)*(n-2)
for q in range(int(input())):
nu,new=map(int,input().split())
edge=edges[nu-1]
u,v=edge[0],edge[1]
old=d[(u,v)]
tot=d1[(u,v)]
ans+=tot*(new-old)
d[(u,v)]=new
d[(v,u)]=new
print((6*ans)/val)
```
| 2,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Tags: combinatorics, dfs and similar, graphs, trees
Correct Solution:
```
from queue import Queue
import sys
cost = []
def readarray(): return map(int, input().split(' '))
n = int(input())
graph = [[] for i in range(n)]
for i in range(n - 1):
u, v, c = readarray()
u, v = u - 1, v - 1
cost.append(c)
graph[u].append((v, i))
graph[v].append((u, i))
order = []
used = [0] * n
q = [0] * (n + n)
qh = qt = 0
used[qh] = 1
qh += 1
while qt < qh:
v = q[qt]
qt += 1
order.append(v)
for (to, e) in graph[v]:
if used[to]:
continue
used[to] = 1
q[qh] = to
qh += 1
order.reverse()
sz = [0 for x in range(n)]
for v in order:
sz[v] = 1
for (to, e) in graph[v]:
sz[v] += sz[to]
"""
sz = [0] * n
sys.setrecursionlimit(100505)
def dfs(v, p):
sz[v] = 1
for (to, e) in graph[v]:
if to != p:
dfs(to, v)
sz[v] += sz[to]
dfs(0, -1)
"""
distanceSum = 0.0
edgeMult = [0] * n
for v in range(n):
for (to, e) in graph[v]:
x = min(sz[v], sz[to])
edgeMult[e] = x
distanceSum += 1.0 * cost[e] * x * (n - x)
distanceSum /= 2.0
queryCnt = int(input())
ans = []
for i in range(queryCnt):
x, y = readarray()
x -= 1
distanceSum -= 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
cost[x] = y
distanceSum += 1.0 * cost[x] * edgeMult[x] * (n - edgeMult[x])
ans.append('%.10lf' % (distanceSum / n / (n - 1) * 6.0))
print('\n'.join(ans))
```
| 2,864 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Submitted Solution:
```
import sys
MAX = 100005;
g = [[] for _ in range(MAX)]
vis = [False] * MAX
dp = [0] * MAX
prod = [0] * MAX
edges = []
def dfs(st, parent = -1):
vis[st] = True
if (st == parent): return;
for i in g[st]:
if (vis[i[0]]): continue;
dfs(i[0], st)
dp[st] = 1;
for i in g[st]:
dp[st] += dp[i[0]];
n = int(input())
for i in range(n-1):
a, b, w = map(int, sys.stdin.readline().split(' '))
g[a].append([b, w])
g[b].append([a,w])
edges.append([[a, b], w])
dfs(1);
tot = 0;
curr = 1;
div = n * (n-1) / 2;
for i in edges:
a = i[0][0];
b = i[0][1];
sa = dp[a];
sb = dp[b];
tot += min(sa, sb) * (n - min(sa, sb)) * i[1];
prod[curr] = min(sa, sb) * (n - min(sa, sb));
curr += 1;
q = int(input())
for i in range(q):
q1, q2 = map(int, sys.stdin.readline().split(' '))
tot -= prod[q1] * edges[q1-1][1];
edges[q1-1][1] = q2;
tot += prod[q1] * edges[q1-1][1];
sys.stdout.write(str(tot*3/div))
```
No
| 2,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Submitted Solution:
```
n = int(input())
ni = [list(map(int,input().split())) for i in range(n-1)]
q = int(input())
qi = [list(map(int,input().split())) for i in range(q)]
summ = 0
for num in range(n-1):
summ += ni[num][2]*(n-1)
for num in range(q):
summ -= (ni[qi[num][0]-1][2] - qi[num][1])*(n-1)
ni[qi[num][0]-1][2] = qi[num][1]
print(float(summ)/((n)*(n-1)*(n-2)))
```
No
| 2,866 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Submitted Solution:
```
import sys
sys.setrecursionlimit(1500)
MAX = 100005;
g = [[] for _ in range(MAX)]
vis = [False] * MAX
dp = [0] * MAX
prod = [0] * MAX
edges = []
def dfs(st):
stack = []
stack.append((st, -1))
vis[st] = True
while stack:
st, parent = stack.pop()
vis[st] = True
if (st == parent): continue;
for i in g[st]:
if (vis[i[0]]): continue;
stack.append((i[0], st))
dp[st] = 1;
for i in g[st]:
dp[st] += dp[i[0]];
n = int(input())
for i in range(n-1):
a, b, w = map(int, sys.stdin.readline().split(' '))
g[a].append([b, w])
g[b].append([a,w])
edges.append([[a, b], w])
dfs(1);
tot = 0;
curr = 1;
div = n * (n-1) / 2;
for i in edges:
a = i[0][0];
b = i[0][1];
sa = dp[a];
sb = dp[b];
tot += min(sa, sb) * (n - min(sa, sb)) * i[1];
prod[curr] = min(sa, sb) * (n - min(sa, sb));
curr += 1;
q = int(input())
for i in range(q):
q1, q2 = map(int, sys.stdin.readline().split(' '))
tot -= prod[q1] * edges[q1-1][1];
edges[q1-1][1] = q2;
tot += prod[q1] * edges[q1-1][1];
sys.stdout.write(str(tot*3/div)+"\n")
```
No
| 2,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
New Year is coming in Tree World! In this world, as the name implies, there are n cities connected by n - 1 roads, and for any two distinct cities there always exists a path between them. The cities are numbered by integers from 1 to n, and the roads are numbered by integers from 1 to n - 1. Let's define d(u, v) as total length of roads on the path between city u and city v.
As an annual event, people in Tree World repairs exactly one road per year. As a result, the length of one road decreases. It is already known that in the i-th year, the length of the ri-th road is going to become wi, which is shorter than its length before. Assume that the current year is year 1.
Three Santas are planning to give presents annually to all the children in Tree World. In order to do that, they need some preparation, so they are going to choose three distinct cities c1, c2, c3 and make exactly one warehouse in each city. The k-th (1 ≤ k ≤ 3) Santa will take charge of the warehouse in city ck.
It is really boring for the three Santas to keep a warehouse alone. So, they decided to build an only-for-Santa network! The cost needed to build this network equals to d(c1, c2) + d(c2, c3) + d(c3, c1) dollars. Santas are too busy to find the best place, so they decided to choose c1, c2, c3 randomly uniformly over all triples of distinct numbers from 1 to n. Santas would like to know the expected value of the cost needed to build the network.
However, as mentioned, each year, the length of exactly one road decreases. So, the Santas want to calculate the expected after each length change. Help them to calculate the value.
Input
The first line contains an integer n (3 ≤ n ≤ 105) — the number of cities in Tree World.
Next n - 1 lines describe the roads. The i-th line of them (1 ≤ i ≤ n - 1) contains three space-separated integers ai, bi, li (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ li ≤ 103), denoting that the i-th road connects cities ai and bi, and the length of i-th road is li.
The next line contains an integer q (1 ≤ q ≤ 105) — the number of road length changes.
Next q lines describe the length changes. The j-th line of them (1 ≤ j ≤ q) contains two space-separated integers rj, wj (1 ≤ rj ≤ n - 1, 1 ≤ wj ≤ 103). It means that in the j-th repair, the length of the rj-th road becomes wj. It is guaranteed that wj is smaller than the current length of the rj-th road. The same road can be repaired several times.
Output
Output q numbers. For each given change, print a line containing the expected cost needed to build the network in Tree World. The answer will be considered correct if its absolute and relative error doesn't exceed 10 - 6.
Examples
Input
3
2 3 5
1 3 3
5
1 4
2 2
1 2
2 1
1 1
Output
14.0000000000
12.0000000000
8.0000000000
6.0000000000
4.0000000000
Input
6
1 5 3
5 3 2
6 1 7
1 4 4
5 2 3
5
1 2
2 1
3 5
4 1
5 2
Output
19.6000000000
18.6000000000
16.6000000000
13.6000000000
12.6000000000
Note
Consider the first sample. There are 6 triples: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1). Because n = 3, the cost needed to build the network is always d(1, 2) + d(2, 3) + d(3, 1) for all the triples. So, the expected cost equals to d(1, 2) + d(2, 3) + d(3, 1).
Submitted Solution:
```
n = int(input())
g, l, e = [[] for x in range(n+1)], [0] * n, [0] * n
for i in range(1,n):
x, y, e[i] = map(int,input().split())
g[x] += [[y, i]]
g[y] += [[x, i]]
print(g)
def dfs(i,p):
d = 1
for z in g[i]:
if z[0] != p:
u = dfs(z[0],i)
d += u
l[z[1]] = u * (n-u)
return d
dfs(1,0)
s = 0
for i in range(1,n):
s += l[i] * e[i]
m = int(input())
while m > 0:
m -= 1
r, w = map(int,input().split())
s += (w - e[r]) * l[r]
e[r] = w
print("%.17f"%float(6*s/n/(n-1)))
```
No
| 2,868 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
ans = 0
b = 2 ** (n + 1) - 3
while n != 0:
n -= 1
#print(n)
p = 2 ** (n + 1) - 3
while b != p:
ans += abs(arr[b] - arr[b - 1])
arr[b // 2 - 1] += max(arr[b], arr[b - 1])
#print(arr)
b -= 2
print(ans)
```
| 2,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
def main():
n, l, res = 2 ** (int(input()) + 1) - 2, [0, 0], 0
l.extend(map(int, input().split()))
while n:
a, b = l[n], l[n + 1]
if a < b:
l[n // 2] += b
res += b - a
else:
l[n // 2] += a
res += a - b
n -= 2
print(res)
if __name__ == '__main__':
main()
```
| 2,870 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
# fin = open("input.txt")
# n = int(fin.readline())
# A = [0] + list(map(int, fin.readline().split()))
n = int(input())
A = [0] + list(map(int, input().split()))
C = 0
for i in range(2 ** n - 2, -1, -1):
C += abs(A[i * 2 + 1] - A[i * 2 + 2])
A[i] += max(A[i * 2 + 1], A[i * 2 + 2])
print(C)
```
| 2,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
s = 0
n = int(input())
k = (1 << (n + 1)) - 1
a = [0, 0] + list(map(int, input().split()))
for i in range(k, 1, -2):
u, v = a[i], a[i - 1]
if u > v: u, v = v, u
s += v - u
a[i >> 1] += v
print(s)
```
| 2,872 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
#!python3
n = int(input())
a = input().split()
a = [int(i) for i in a]
def solve(n, a, added):
last = a[-2**n:]
new = []
for i in range(0, 2**n-1, 2):
#print(last[i])
x = last[i]
y = last[i+1]
new.append(max(x,y))
added = added + abs(x-y)
a = a[:-2**n]
if a==[]:
a = [0]
for i in range(1, 2**(n-1)+1):
a[-i] = a[-i] + new[-i]
n = n-1
if n==0:
print(added)
else:
solve(n, a, added)
solve(n, a, 0)
```
| 2,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
n = 2**(int(input())+1)-1;
a = [0,0] + list(map(int,input().split()))
r = 0
while n>1:
a[n//2] += max(a[n], a[n-1])
r += abs(a[n]-a[n-1])
n -= 2
print(r)
```
| 2,874 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
l = [0]+l
top = 1
lim = 2**(n+1)-1
count = 0
def solver(top):
global count
left = top*2+1
right = 2+ top*2
if(left>lim or right>lim):
#print(top,"ret ",l[top])
return l[top]
else:
ll = solver(left)
rr = solver(right)
if(ll!=rr):
count+= max(ll,rr)-min(ll,rr)
#print("change ",max(ll,rr)-min(ll,rr))
#print("ret" , max(ll,rr)+l[top])
return max(ll,rr)+l[top]
solver(0)
print(count)
```
| 2,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Tags: dfs and similar, greedy, implementation
Correct Solution:
```
from math import floor
def main():
n = int(input())
a = list(map(int, input().split()))
streets = []
for i in range(2**n, 2**(n+1)):
#print('---')
idx = i
#print(idx)
if idx > 1:
#print('Cost: %d' % a[idx-2])
res = a[idx-2]
while idx > 0:
idx = int(floor(idx/2))
if idx > 1:
#print(idx)
#print('Cost: %d' % a[idx-2])
res += a[idx-2]
#print('res: %d' % res)
streets.append(res)
res = 0
#print(streets)
while len(streets) > 2:
new_streets = []
for i in range(0, len(streets), 2):
#print('i: %d' % i)
res += abs(streets[i]-streets[i+1])
new_streets.append(max(streets[i], streets[i+1]))
#print(new_streets, cur_diff)
streets = new_streets
print(res+abs(streets[0]-streets[1]))
if __name__ == '__main__':
main()
```
| 2,876 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# author: firolunis
# version: 0.1
n = int(input())
park = input().split(' ')
park = [int(i) for i in park]
park.insert(0, 0)
lights = [0 for i in range(2 ** (n + 1) - 1)]
res = 0
for k in range(n, 0, -1):
for i, j in tuple(enumerate(park))[(2 ** k) - 1:2 ** (k + 1) - 1:2]:
res += abs(j + lights[i] - park[i + 1] - lights[i + 1])
lights[i // 2] = max(j + lights[i], park[i + 1] + lights[i + 1])
print(res)
```
Yes
| 2,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
n = int(input())
park = [0] * (2 ** (n + 1))
park[2: 2 ** (n + 1)] = [int(i) for i in input().split()]
cnt = 0
for i in range(2 ** n - 1, 0, -1):
cnt += abs(park[i * 2 + 1] - park[i * 2])
park[i] += max(park[i * 2 + 1], park[i * 2])
print(cnt)
```
Yes
| 2,878 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
# print ("Enter n")
n = int(input())
alen = 2**(n+1)
a = [0 for i in range(alen)]
# print ("Enter all values on the same line")
inp = input().split()
for i in range(len(inp)):
a[i+2] = int(inp[i])
answer = 0
while (n > 0):
index = 2**n
for i in range(index, index*2, 2):
left = a[i]
right = a[i+1]
diff = abs(left-right)
bigone = max(left, right)
answer += diff
a[i//2] += bigone
n = n - 1
print (answer)
```
Yes
| 2,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
#!/usr/bin/python3
n = 2**(int(input())+1)-1
d = input().split(' ')
for i in range(len(d)):
d[i] = int(d[i])
p = 0
for i in range(len(d)-1, 0, -2):
p += abs(d[i]-d[i-1])
d[i//2-1] += max(d[i], d[i-1])
print(p)
```
Yes
| 2,880 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
# fin = open("input.txt")
# n = int(fin.readline())
# A = [0] + list(map(int, fin.readline().split()))
n = int(input())
A = list(map(int, input().split()))
C = 0
for i in range(n - 1, -1, -1):
C += abs(A[i * 2] - A[i * 2 + 1])
A[i] += max(A[i * 2], A[i * 2 + 1])
print(C)
```
No
| 2,881 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
import math
a = int(input())
b = list(map(int, input().split()))
co = 2
sum1 = 0
sum2 = 0
j = co
for i in b:
if j == 0:
co = co * 2
j = co
if j <= co // 2:
sum2 += i
else:
sum1 += i
j -= 1
print(int(math.fabs(sum2-sum1)))
```
No
| 2,882 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
import sys, os
import fileinput
n = int(input()) + 1
a = [int(x) for x in input().split()]
all_count = 2 ** n - 1
b = [0] * all_count
counter = 0
for i in range(n, 1, -1):
lcnt = 2 ** (i - 1)
first = 2 ** (i - 1) - 2
#print(lcnt, first)
level = a[first:first + lcnt]
# print(level)
for j in range(0, lcnt, 2):
index = first + 2 + j
if i == n:
# print(i)
# print("-" * 10)
diff = abs(level[j] - level[j + 1])
# print(j, level[j], level[j+1], diff)
counter += diff
# print(index//2 - 1, level[j] + level[j + 1] + diff)
b[index//2 - 1] += level[j] + level[j + 1] + diff
# print("=" * 10)
else:
# print(i)
# print("*" * 10)
# print(index)
# print(b)
# print(level)
diff = abs(abs(b[index - 1]//2 + level[j]) - abs(b[index]//2 + level[j + 1]))
counter += diff
b[index//2 - 1] += b[index-1] + b[index] + level[j] + level[j + 1] + diff
# print("+" * 10)
print(counter)
# print(b)
```
No
| 2,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
<image>
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square <image>. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
<image> To light the path roads in the evening, the park keeper installed street lights along each road. The road that leads from square i to square <image> has ai lights.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads.
Input
The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and <image>. All numbers ai are positive integers, not exceeding 100.
Output
Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe.
Examples
Input
2
1 2 3 4 5 6
Output
5
Note
Picture for the sample test. Green color denotes the additional street lights.
<image>
Submitted Solution:
```
import math
a = input()
b = list(map(int, input().split()))
co = 2
sum1 = 0
sum2 = 0
j = co
if b == [1, 2, 3, 3, 2, 2]:
print(0)
exit(0)
for i in b:
if j == 0:
co = co * 2
j = co
if j <= co // 2:
sum2 += i
else:
sum1 += i
j -= 1
print(int(math.fabs(sum2-sum1)))
```
No
| 2,884 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
# -*- coding: utf-8 -*-
# @Time : 2021/1/15 5:25 下午
# @Author : qu
# @Email : quzhenqing@zju.edu.cn
# @File : C. GukiZ hates Boxes.py
from sys import stdin
#
n, m = map(int, stdin.buffer.readline().split())
a = list(map(int, stdin.buffer.readline().split()))
max_not_zero = 0
for index, value in enumerate(a):
if value > 0:
max_not_zero = index
total = sum(a)
def remove(t):
boxes = 0
s = m
for i in range(max_not_zero+1):
boxes += a[i]
# 够一个人用的了
while boxes + (i + 1) >= t:
boxes -= (t - i - 1)
s -= 1
if s < 0:
return False
if s == 0:
if boxes > 0:
return False
return True
def binary_search(left, right):
mid = int((left + right) // 2)
if right - left <= 1 and remove(left):
return left
if remove(mid):
return binary_search(left, mid)
else:
return binary_search(mid + 1, right)
print(binary_search(2, total + n))
```
| 2,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
n, m = map(int, input().split())
# a = [0] * n
# s = 0
a = list(map(int, input().split()))
s = sum(a)
l = 2
r = s + n
while (l < r):
z = l + r >> 1
b = a.copy()
p = n - 1
for i in range(m):
while (p >= 0 and b[p] == 0):
p -= 1
t = z - p - 1
if (t <= 0):
break
while (p >= 0 and b[p] <= t):
t -= b[p]
p -= 1 # or do it before?
if (p >= 0):
b[p] -= t
if (p < 0):
r = z
else:
l = z + 1
print(r)
```
| 2,886 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
from sys import stdin
import copy
def check(id,b,m,t):
ans = 0
i = 0
while i < len(b):
ans += b[i]
#可以用掉一个人了
while ans + id[i] >= t:
ans -= (t - id[i])
m -= 1
if m <= 0:
if m == 0 and ans == 0 and i + 1 == len(b):
return True
return False
i = i + 1
return True
n, m = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
#右边界
id = [i + 1 for i, e in enumerate(a) if e != 0]
b = [a[i - 1] for i in id]
R = sum(b) + id[-1]
#左边界
L = 0
while L <= R:
mid = (L + R) // 2
if L == R:
break
if check(id,b,m,mid):
R = mid
else:
L = mid + 1
print(R)
```
| 2,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
def read_data():
n, m = map(int, input().split())
A = list(map(int, input().split()))
while A and not A[-1]:
del A[-1]
return len(A), m, A
def solve(n, m, A):
total = sum(A)
upper = n + (total + m - 1) // m
lower = n
while lower + 1 < upper:
mid = (lower + upper) // 2
if is_enough(mid, n, m, A):
upper = mid
else:
lower = mid
return lower + 1
def is_enough(t, n, m, A):
pool = 0
for i in range(n-1, -1, -1):
a = A[i]
delta = t - i - 1
b = (a - pool + delta - 1) // delta
m -= b
if m < 0:
return False
pool += b * delta - a
return True
if __name__ == '__main__':
n, m, A = read_data()
print(solve(n, m, A))
```
| 2,888 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
N, M = map(int, input().split())
books_list = list(map(int, input().split()))
while books_list[-1] == 0:
books_list.pop()
books_list.insert(0, 0)
def check(Time):
piles = books_list[:]
last_pile_no = len(piles) - 1
for i in range(M): #student
i_time = Time - last_pile_no
while True:
if i_time >= piles[last_pile_no]:
i_time -= piles[last_pile_no]
last_pile_no -= 1
if last_pile_no == 0:
return True
else:
piles[last_pile_no] -= i_time
break
return False
l = 0
r = int(sum(books_list)/M) + len(books_list) + 1
while r-l > 1:
mid = int((l+r)/2)
if check(mid):
r = mid
else:
l = mid
print(r)
```
| 2,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
f = lambda: map(int, input().split())
n, m = f()
h = list(f())[::-1]
def g(t, m):
t -= n
d = 0
for u in h:
t += 1
if u > d:
if t < 1: return 1
u -= d
d = -u % t
m -= (u + d) // t
if m < 0: return 1
else:
d -= u
return 0
a, b = 0, int(11e13)
while b - a > 1:
c = (b + a) // 2
if g(c, m):
a = c
else:
b = c
print(b + 1)
```
| 2,890 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
def c(t):
s,r,p,b=m,0,n,0
while 1:
while b==0:
if p==0: return 1
p-=1
b=a[p]
if r==0:
if s==0: return 0
r=t-p-1
s-=1
d=min(b,r)
b-=d
r-=d
l,h=0,n+sum(a)+9
while h-l>1:
md=(l+h)//2
if c(md):
h=md
else:
l=md
print(h)
```
| 2,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Tags: binary search, greedy
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue May 28 20:03:18 2019
@author: fsshakkhor
"""
N,M = map(int,input().split())
ara = list(map(int,input().split()))
while ara[-1] == 0:
ara.pop()
ara.insert(0,0)
def check(Time):
piles = ara[:]
last_pile_no = len(piles) - 1
for i in range(M):
i_time = Time - last_pile_no
while True:
if i_time >= piles[last_pile_no]:
i_time -= piles[last_pile_no]
last_pile_no -= 1
if last_pile_no == 0:
return True
else:
piles[last_pile_no] -= i_time
break
return False
lo = 0
hi = int(sum(ara)/M) + len(ara) + 1
ans = 0
while lo <= hi:
mid = int((lo+hi)/2)
if check(mid):
ans = mid
hi = mid - 1
else:
lo = mid + 1
print(ans)
```
| 2,892 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
from sys import stdin
import collections
import copy
def check(t):
global m, end
stu = m
time = 0
for i in range(end):
# print(time,stu)
time += a[i]
while(time+i+1>=t):
time -= t-i-1
stu -=1
if stu<0:
return 0
if stu ==0:
return time<=0
return 1
# def check(t):
# global end
# flag = end
# for i in range(m):
# time = t-flag
# while(time > 0):
# # print(time, flag)
# if time>=a[flag-1]:
# a[flag-1]=0
# flag -=1
# time -= a[flag-1]
# else:
# a[flag-1] -= time
# time =0
# if flag <1:
# return True
# if(flag<1):
# return True
# else:
# return False
n, m = list(map(int, stdin.readline().split()))
a = list(map(int, stdin.readline().split()))
l=0
r=0
for i in range(len(a)):
r += a[i]
if a[i]>0:
l = i+1
r += l
end = l
while(l<=r):
mid = (l+r)//2
if(check(mid)):
ans = mid
r = mid-1
else:
# print(l)
l = mid+1
print(ans)
```
Yes
| 2,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
# The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109)
# where ai represents the number of boxes on i-th pile.
# The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
NUM_OF_PILES, NUM_OF_STUDENTS = map(int, input().split())
# a = [0] * n
# s = 0
boxes = list(map(int, input().split()))
s = sum(boxes)
# It's guaranteed that at least one pile of is non-empty. So walk to it and remove a box
lower = 2
# That's a case of only one student working
upper = s + NUM_OF_PILES
while (lower < upper):
# z = lower + upper >> 1
guess = (lower + upper) // 2
b = boxes.copy()
pile = NUM_OF_PILES - 1
for i in range(NUM_OF_STUDENTS):
# walk to a nonempty pile
while (pile >= 0 and b[pile] == 0):
pile -= 1
time = guess - pile - 1
if (time <= 0):
break
# remove all boxes from some piles
while (pile >= 0 and b[pile] <= time):
time -= b[pile]
pile -= 1 # or do it before?
# remove some boxes in the end
if (pile >= 0):
b[pile] -= time
if (pile < 0):
upper = guess
else:
lower = guess + 1
print(upper)
```
Yes
| 2,894 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
# The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
# NUM_OF_PILES, NUM_OF_STUDENTS = map(int, input().split())
NUM_OF_PILES, NUM_OF_STUDENTS = map(int, input().split())
a = list(map(int, input().split()))
def test(guess):
students_free, time, pile, b = NUM_OF_STUDENTS, 0, NUM_OF_PILES, 0
while True:
while b == 0:
if pile == 0:
return True
pile -= 1
b = a[pile]
if time == 0:
if students_free == 0:
return False
time = guess - pile - 1
students_free -= 1
d = min(b, time)
b -= d
time -= d
l, h = 0, NUM_OF_PILES + sum(a) + 9
while h - l > 1:
md = (l + h) // 2
if test(md):
h = md
else:
l = md
print(h)
```
Yes
| 2,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
###############################
# https://codeforces.com/problemset/problem/551/C
# 2021/01/13
# WenhuZhang
################################
from sys import stdin
import collections
import copy
# def check(t):
# global n,m,a, end
# aa = a.copy()
# flag = 0
# for i in range(m):
# time = t - (flag+1)
# while(time>0):
# # print(i, time,aa)
# # print(t,flag, time,aa)
# if flag >= end or aa[end-1] ==0:
# return True
# while(aa[flag]==0):
# flag+=1
# time-=1
# if time ==0:
# break
# if aa[flag]>time:
# aa[flag] -= time
# time=0
# else:
# # print("?",time,aa[flag])
# time -= aa[flag]
# aa[flag] =0
# if flag >= end or aa[end-1] ==0:
# return True
# return False
def check(t):
global m, end
stu = m
time = 0
for i in range(end):
# print(time,stu)
time += a[i]
while(time+i+1>=t):
time -= t-i-1
stu -=1
if stu<0:
return 0
if stu ==0:
return time<=0
return 1
n, m = list(map(int, stdin.readline().split()))
a = list(map(int, stdin.readline().split()))
l=0
r=0
for i in range(len(a)):
r += a[i]
if a[i]>0:
l = i+1
r += l
end = l
while(l<=r):
mid = (l+r)//2
# print(l,r,mid)
if(check(mid)):
ans = mid
r = mid-1
else:
# print(l)
l = mid+1
print(ans)
```
Yes
| 2,896 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
N,M=list(map(int,input().split()))
p=list(map(int,input().split()))
total=sum(p)
if(total>M):
print(3*total-2*M)
elif (total<M):
print(N+1)
elif (total==M):
print(2*total)
```
No
| 2,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
n, m = map(int, input().split())
boxes = list(map(int, input().split()))
cache = [0 for i in range(n + 1)]
if boxes.count(0) == n:
print(0)
exit(0)
seconds = 1
cache[0] = m
while True:
if boxes.count(0) == n:
break
seconds += 1
stus = cache.copy()
for i in range(n):
# print(stus[i], boxes[i], sep = "==", end = " ")
if stus[i] > boxes[i]:
cache[i + 1] += stus[i] - boxes[i]
cache[i] = boxes[i]
boxes[i] = 0
else:
boxes[i] -= stus[i]
print(seconds)
```
No
| 2,898 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Professor GukiZ is concerned about making his way to school, because massive piles of boxes are blocking his way.
In total there are n piles of boxes, arranged in a line, from left to right, i-th pile (1 ≤ i ≤ n) containing ai boxes. Luckily, m students are willing to help GukiZ by removing all the boxes from his way. Students are working simultaneously. At time 0, all students are located left of the first pile. It takes one second for every student to move from this position to the first pile, and after that, every student must start performing sequence of two possible operations, each taking one second to complete. Possible operations are:
1. If i ≠ n, move from pile i to pile i + 1;
2. If pile located at the position of student is not empty, remove one box from it.
GukiZ's students aren't smart at all, so they need you to tell them how to remove boxes before professor comes (he is very impatient man, and doesn't want to wait). They ask you to calculate minumum time t in seconds for which they can remove all the boxes from GukiZ's way. Note that students can be positioned in any manner after t seconds, but all the boxes must be removed.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105), the number of piles of boxes and the number of GukiZ's students.
The second line contains n integers a1, a2, ... an (0 ≤ ai ≤ 109) where ai represents the number of boxes on i-th pile. It's guaranteed that at least one pile of is non-empty.
Output
In a single line, print one number, minimum time needed to remove all the boxes in seconds.
Examples
Input
2 1
1 1
Output
4
Input
3 2
1 0 2
Output
5
Input
4 100
3 4 5 4
Output
5
Note
First sample: Student will first move to the first pile (1 second), then remove box from first pile (1 second), then move to the second pile (1 second) and finally remove the box from second pile (1 second).
Second sample: One of optimal solutions is to send one student to remove a box from the first pile and a box from the third pile, and send another student to remove a box from the third pile. Overall, 5 seconds.
Third sample: With a lot of available students, send three of them to remove boxes from the first pile, four of them to remove boxes from the second pile, five of them to remove boxes from the third pile, and four of them to remove boxes from the fourth pile. Process will be over in 5 seconds, when removing the boxes from the last pile is finished.
Submitted Solution:
```
###############################
# https://codeforces.com/problemset/problem/551/C
# 2021/01/09
# WenhuZhang
################################
from sys import stdin
import collections
import copy
def check(t):
global end
flag = end
for i in range(m):
time = t-flag
while(time > 0):
# print(time, flag)
if time>=a[flag-1]:
a[flag-1]=0
flag -=1
time -= a[flag-1]
else:
a[flag-1] -= time
time =0
if flag <1:
return True
if(flag<1):
return True
else:
return False
n, m = list(map(int, stdin.readline().split()))
a = list(map(int, stdin.readline().split()))
l=0
r=0
for i in range(len(a)):
r += a[i]
if a[i]>0:
l = i+1
r += l
end = l
while(l<=r):
mid = (l+r)//2
if(check(mid)):
ans = mid
r = mid-1
else:
l = mid+1
print(ans)
```
No
| 2,899 |
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