text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
a, b = list(map(int, input().split()))
while True:
a += 1
sta = str(a)
m = ""
n = 0
for i in range(len(sta)):
if sta[i] == '4' or sta[i] == '7':
m += sta[i]
if len(m) > 0:
n = int(m)
if n == b:
print(a)
break
```
| 94,600 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
a,b=map(str,input().split())
def mask(x):
st=''
for i in str(x):
if i=='7' or i=='4':
st+=i
return(st)
for i in range(int(a)+1,177778):
if mask(i)!=b:
pass
else:
print(i)
quit()
```
| 94,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
a,b=map(str,input().split())
def check(x):
st=''
for i in str(x):
if i=='7' or i=='4':
st+=i
return(st)
for i in range(int(a)+1,177778):
if check(i)!=b:
pass
else:
print(i)
quit()
```
| 94,602 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
a=input()
list1 = a.split(' ')
num1=list1[0]
num2=list1[1]
def mask(num):
num=str(num)
str1=""
for i in range(0,len(num)):
if (ord(num[i])-48)==7 or (ord(num[i])-48)==4:
str1=str1+num[i]
return str1
for i in range(int(num1)+1,1000000):
if mask(i)==num2:
print(i)
quit()
```
| 94,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
target,lucky=map(int,input().split())
if lucky>target:
print(lucky)
else:
checkarr=list(map(int,str(lucky)))
while True:
target+=1
i=0
check=False
tmp=list(map(int,str(target)))
for val in tmp:
if val==checkarr[i]:
i+=1
if i==len(checkarr):
check=True
break
if check:
tmpstr=[]
for val in tmp:
if val==7:
tmpstr.append(7)
elif val==4:
tmpstr.append(4)
if tmpstr==checkarr:
print(target)
break
```
| 94,604 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
import math
def mask(x):
tmp,res=0,0
while(x):
if(x%10==4 or x%10==7):
tmp=tmp*10+(x%10)
x//=10
while(tmp):
res=res*10+(tmp%10)
tmp//=10
return res
try:
t=1
while(t):
t-=1
a,b=map(int,input().split())
a+=1
while(mask(a)!=b):
a+=1
print(a)
except:
pass
```
| 94,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
s = input()
a = int(s.split()[0])
b = int(s.split()[1])
def getMask(num):
s = str(num)
l = [c for c in s if c=='4' or c=='7']
if len(l) == 0:
return 0
return int(''.join(l))
ans = a + 1
while not getMask(ans) == b:
ans = ans + 1
print(ans)
```
| 94,606 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Tags: brute force, implementation
Correct Solution:
```
def mask(x,k):
t=""
for i in x:
if i in "47":t+=i
return t!=k
a,b=map(int,input().split())
z=a
while mask(str(a),str(b)) or a<=z:a+=1
print(a)
```
| 94,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
def f(x):
s="0"
for c in str(x):
if c=='7' or c=='4':
s+=c
return int(s)
a,b=map(int,input().split())
x=a+1
while f(x)!=b:
x+=1
print(x)
```
Yes
| 94,608 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
a, b = input().split()
for i in range(max(int(a) + 1, int(b)), 999999):
if ''.join(c for c in str(i) if c in ('4,7')) == b:
print(i);exit()
```
Yes
| 94,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
def mask(x):
s = 0
n = str(x)
for i in range(len(n)):
if n[i] == '7' or n[i] == '4':
s = s * 10 + int(n[i])
return s
a , b = (input().split())
a= int(a)
a= a + 1
b= int(b)
while mask(a) != b:
a = a + 1
print(a)
```
Yes
| 94,610 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
a = [int(x) for x in input().split()]
b=str(a[0]+1)
c=str(a[1])
def getmask(q):
s=""
for i in q:
if i in ["4","7"]:
s+=i
return s
while getmask(b)!=c :
b=str(int(b)+1)
print(b)
```
Yes
| 94,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
a = [int(x) for x in input().split()]
b=str(a[0])
c=str(a[1])
while c not in b:
b=str(int(b)+1)
print(b)
```
No
| 94,612 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
def f(a, b):
j = 0
for i in b:
j = a.find(i, j)
if j < 0: return False
return True
a, b = input().split()
if len(a) < len(b): ans = b
else:
u, v = len(a) - 1, len(b) - 1
for i in range(len(b)):
x, y = u - i, v - i
if a[x:] > b[y:]:
t = str(int(a[:x]) + 1)
if f(t, b[:y]):
ans = t + b[y:]
break
elif f(a[:x], b[:y]):
ans = a[:x] + b[y:]
break
print(ans)
```
No
| 94,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
#from dust i have come dust i will be
import sys
a,b=map(int,input().split())
sa=str(a)
sb=str(b)
if len(sa)==len(sb):
if b>a:
print(b)
else:
sys.stdout.write("1"+sb)
else:
if b>a:
print(b)
else:
#extract the same len of b from a
sc=""
l=len(sa)-len(sb)
for i in range(l,len(sa)):
sc+=sa[i]
c=int(sc)
#if c<b then write first l chars
if c<b:
for i in range(l):
sys.stdout.write(sa[i])
sys.stdout.write(sb)
# otherwise from l-1 to 0 try increasing 1
else:
sc=""
for i in range(l):
sc+=sa[i]
c=int(sc)
c+=1
sys.stdout.write(str(c)+sb)
```
No
| 94,614 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a mask of a positive integer n the number that is obtained after successive writing of all lucky digits of number n from the left to the right. For example, the mask of number 72174994 is number 7744, the mask of 7 is 7, the mask of 9999047 is 47. Obviously, mask of any number is always a lucky number.
Petya has two numbers — an arbitrary integer a and a lucky number b. Help him find the minimum number c (c > a) such that the mask of number c equals b.
Input
The only line contains two integers a and b (1 ≤ a, b ≤ 105). It is guaranteed that number b is lucky.
Output
In the only line print a single number — the number c that is sought by Petya.
Examples
Input
1 7
Output
7
Input
100 47
Output
147
Submitted Solution:
```
'''
Created on Feb 11, 2015
@author: mohamed265
'''
def fun (a, b):
#print(a ,b)
lenA = len(a)
o = 0
for i in range(lenA):
if b[o] == a[i]:
o += 1
if o == len(b):
return False
return True
# for i in range(o , l
temp = input().split()
a = int(temp[0]) +1
b = temp[1]
while fun(str(a), b):
a +=1
print(a)
```
No
| 94,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Tags: combinatorics, dfs and similar, graphs, math, shortest paths, trees
Correct Solution:
```
import sys, os
from collections import defaultdict, deque
if os.environ['USERNAME']=='kissz':
inp=open('in2.txt','r').readline
def debug(*args):
print(*args,file=sys.stderr)
else:
inp=sys.stdin.readline
def debug(*args):
pass
# SCRIPT STARTS HERE
def solve():
n,m=map(int,inp().split())
G=defaultdict(list)
for i in range(m):
u,v=map(int,inp().split())
G[u-1]+=[v-1]
G[v-1]+=[u-1]
combs=[[-1]*n for _ in range(n)]
def bfs(i):
Q=deque([i])
D=[-1]*n
D[i]=0
while Q:
node=Q.pop()
for neighbor in G[node]:
if D[neighbor]==-1:
Q.appendleft(neighbor)
D[neighbor]=D[node]+1
return D
dists=[bfs(i) for i in range(n)]
for i in range(n):
for j in range(i+1):
node=j
failed=False
while node!=i:
c=0
next_node=-1
for neighbor in G[node]:
if dists[i][neighbor]<dists[i][node]:
c+=1
next_node=neighbor
if c==1:
node=next_node
else:
failed=True
break
call=1
if failed:
combs[i][j]=combs[j][i]=0
else:
for k in range(n):
if dists[i][k]+dists[j][k]>dists[i][j]:
c=0
for l in G[k]:
if dists[i][l]+1==dists[i][k] and dists[j][l]+1==dists[j][k]:
c+=1
call=(call*c)% 998244353
combs[i][j]=combs[j][i]=call
for i in range(n):
print(*combs[i])
solve()
```
| 94,616 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Tags: combinatorics, dfs and similar, graphs, math, shortest paths, trees
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
MOD = 998244353
adjList = []
dist = []
ans = []
for _ in range(n):
adjList.append([])
for _ in range(m):
a,b = map(int,input().split())
adjList[a - 1].append(b - 1)
adjList[b - 1].append(a - 1)
for i in range(n):
curDist = [-1] * n
curDist[i] = 0
bfsQueue = [i]
bfsIndex = 0
while len(bfsQueue) > bfsIndex:
c = bfsQueue[bfsIndex]
bfsIndex += 1
for elem in adjList[c]:
if curDist[elem] == -1:
curDist[elem] = curDist[c] + 1
bfsQueue.append(elem)
dist.append(curDist)
for _ in range(n):
ans.append([0] * n)
for i in range(n):
for j in range(i, n):
# Find shortest path between i and j
isCovered = [False] * n
isCovered[i] = True
isCovered[j] = True
jCpy = j
flag = True
while jCpy != i:
cnt = 0
cntIndex = -1
for elem in adjList[jCpy]:
if dist[i][elem] == dist[i][jCpy] - 1:
cnt += 1
cntIndex = elem
if cnt >= 2:
flag = False
break
else:
isCovered[cntIndex] = True
jCpy = cntIndex
if not flag:
ans[i][j] = 0
ans[j][i] = 0
else:
curAns = 1
for k in range(n):
if isCovered[k]:
continue
cnt = 0
for elem in adjList[k]:
if dist[i][elem] == dist[i][k] - 1 and dist[j][elem] == dist[j][k] - 1:
cnt += 1
curAns *= cnt
curAns %= MOD
ans[i][j] = curAns
ans[j][i] = curAns
for elem in ans:
print(" ".join(map(str,elem)))
# 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")
# endregion
if __name__ == "__main__":
main()
```
| 94,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Tags: combinatorics, dfs and similar, graphs, math, shortest paths, trees
Correct Solution:
```
import io
import os
from collections import deque
def solve(N, M, edges):
graph = [[] for i in range(N)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
def bfs(source):
q = deque([source])
dist = [-1] * N
dist[source] = 0
while q:
node = q.popleft()
d = dist[node]
for nbr in graph[node]:
if dist[nbr] == -1:
q.append(nbr)
dist[nbr] = d + 1
return dist
dists = [bfs(u) for u in range(N)]
MOD = 998244353
ans = [[0 for i in range(N)] for j in range(N)]
for x in range(N):
for y in range(x, N):
onPath = 0
for i in range(N):
if dists[x][i] + dists[y][i] == dists[x][y]:
onPath += 1
numVertices = dists[x][y] + 1 # including x and y
ways = 1
if onPath > numVertices:
ways = 0
else:
for i in range(N):
if dists[x][i] + dists[y][i] != dists[x][y]:
count = 0
for j in graph[i]:
if (
dists[x][i] == dists[x][j] + 1
and dists[y][i] == dists[y][j] + 1
):
count += 1
ways = (ways * count) % MOD
if not ways:
break
ans[x][y] = ways
ans[y][x] = ways
return "\n".join(" ".join(map(str, row)) for row in ans)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = 1
for tc in range(1, TC + 1):
N, M = [int(x) for x in input().split()]
edges = [[int(x) - 1 for x in input().split()] for i in range(M)] # 0 indexed
ans = solve(N, M, edges)
print(ans)
```
| 94,618 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):#排他的論理和の階乗
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n):
self.BIT=[0]*(n+1)
self.num=n
def query(self,idx):
res_sum = 0
while idx > 0:
res_sum += self.BIT[idx]
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
while idx <= self.num:
self.BIT[idx] += x
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class Matrix():
mod=10**9+7
def set_mod(m):
Matrix.mod=m
def __init__(self,L):
self.row=len(L)
self.column=len(L[0])
self._matrix=L
for i in range(self.row):
for j in range(self.column):
self._matrix[i][j]%=Matrix.mod
def __getitem__(self,item):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
return self._matrix[i][j]
def __setitem__(self,item,val):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
self._matrix[i][j]=val
def __add__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]+other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __sub__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]-other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __mul__(self,other):
if type(other)!=int:
if self.column!=other.row:
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(other.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(other.column):
temp=0
for k in range(self.column):
temp+=self._matrix[i][k]*other._matrix[k][j]
res[i][j]=temp%Matrix.mod
return Matrix(res)
else:
n=other
res=[[(n*self._matrix[i][j])%Matrix.mod for j in range(self.column)] for i in range(self.row)]
return Matrix(res)
def __pow__(self,m):
if self.column!=self.row:
raise MatrixPowError("the size of row must be the same as that of column")
n=self.row
res=Matrix([[int(i==j) for i in range(n)] for j in range(n)])
while m:
if m%2==1:
res=res*self
self=self*self
m//=2
return res
def __str__(self):
res=[]
for i in range(self.row):
for j in range(self.column):
res.append(str(self._matrix[i][j]))
res.append(" ")
res.append("\n")
res=res[:len(res)-1]
return "".join(res)
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
mod = 998244353
N = 2*10**3
g1 = [1]*(N+1)
g2 = [1]*(N+1)
inverse = [1]*(N+1)
for i in range( 2, N + 1 ):
g1[i]=( ( g1[i-1] * i ) % mod )
inverse[i]=( ( -inverse[mod % i] * (mod//i) ) % mod )
g2[i]=( (g2[i-1] * inverse[i]) % mod )
inverse[0]=0
mod = 998244353
n,m = mi()
edge = [[] for i in range(n)]
Edge = []
for _ in range(m):
a,b = mi()
edge[a-1].append(b-1)
edge[b-1].append(a-1)
Edge.append((a-1,b-1))
dist = [[10**17 for j in range(n)] for i in range(n)]
for s in range(n):
deq = deque([s])
dist[s][s] = 0
while deq:
v = deq.popleft()
for nv in edge[v]:
if dist[s][nv]==10**17:
dist[s][nv] = dist[s][v] + 1
deq.append(nv)
ans = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
for j in range(n):
tmp = [0 for i in range(n)]
tmp_edge = [[] for i in range(n)]
deg = [0 for i in range(n)]
for k in range(m):
a,b = Edge[k]
if dist[i][a]==dist[i][b]+1 and dist[j][a]==dist[j][b]+1:
tmp[a] += 1
elif dist[i][a]+1==dist[i][b] and dist[j][a]+1==dist[j][b]:
tmp[b] += 1
elif dist[i][a]==dist[i][b]+1 and dist[j][a]+1==dist[j][b]:
tmp_edge[b].append(a)
deg[a] += 1
elif dist[i][a]+1==dist[i][b] and dist[j][a]==dist[j][b]+1:
tmp_edge[a].append(b)
deg[b] += 1
deq = deque([i])
res = []
while deq:
v = deq.popleft()
res.append(v)
for nv in tmp_edge[v]:
deg[nv] -= 1
if not deg[nv]:
deq.append(nv)
check = [v for v in res if tmp[v]==0]
ttt = len(check)
dp = [0 for i in range(n)]
dp[i] = 1
for k in range(1,ttt):
if dist[i][check[k]]==dist[i][check[k-1]]:
dp[i] = 0
pos = 0
L = 0
for v in res:
if check[pos] == v:
for k in range(L,v):
dp[k] = 0
L = v
pos += 1
else:
dp[v] *= inverse[tmp[v]]
dp[v] %= mod
for nv in tmp_edge[v]:
dp[nv] += dp[v]
dp[nv] %= mod
res = dp[j]
for k in range(n):
if tmp[k]!=0:
res *= tmp[k]
res %= mod
ans[i][j] = res
for i in range(n):
print(*ans[i])
```
No
| 94,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Submitted Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):#排他的論理和の階乗
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n):
self.BIT=[0]*(n+1)
self.num=n
def query(self,idx):
res_sum = 0
while idx > 0:
res_sum += self.BIT[idx]
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
while idx <= self.num:
self.BIT[idx] += x
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class Matrix():
mod=10**9+7
def set_mod(m):
Matrix.mod=m
def __init__(self,L):
self.row=len(L)
self.column=len(L[0])
self._matrix=L
for i in range(self.row):
for j in range(self.column):
self._matrix[i][j]%=Matrix.mod
def __getitem__(self,item):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
return self._matrix[i][j]
def __setitem__(self,item,val):
if type(item)==int:
raise IndexError("you must specific row and column")
elif len(item)!=2:
raise IndexError("you must specific row and column")
i,j=item
self._matrix[i][j]=val
def __add__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]+other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __sub__(self,other):
if (self.row,self.column)!=(other.row,other.column):
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(self.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(self.column):
res[i][j]=self._matrix[i][j]-other._matrix[i][j]
res[i][j]%=Matrix.mod
return Matrix(res)
def __mul__(self,other):
if type(other)!=int:
if self.column!=other.row:
raise SizeError("sizes of matrixes are different")
res=[[0 for j in range(other.column)] for i in range(self.row)]
for i in range(self.row):
for j in range(other.column):
temp=0
for k in range(self.column):
temp+=self._matrix[i][k]*other._matrix[k][j]
res[i][j]=temp%Matrix.mod
return Matrix(res)
else:
n=other
res=[[(n*self._matrix[i][j])%Matrix.mod for j in range(self.column)] for i in range(self.row)]
return Matrix(res)
def __pow__(self,m):
if self.column!=self.row:
raise MatrixPowError("the size of row must be the same as that of column")
n=self.row
res=Matrix([[int(i==j) for i in range(n)] for j in range(n)])
while m:
if m%2==1:
res=res*self
self=self*self
m//=2
return res
def __str__(self):
res=[]
for i in range(self.row):
for j in range(self.column):
res.append(str(self._matrix[i][j]))
res.append(" ")
res.append("\n")
res=res[:len(res)-1]
return "".join(res)
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import log,gcd
input = lambda :sys.stdin.buffer.readline()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
mod = 998244353
N = 2*10**3
g1 = [1]*(N+1)
g2 = [1]*(N+1)
inverse = [1]*(N+1)
for i in range( 2, N + 1 ):
g1[i]=( ( g1[i-1] * i ) % mod )
inverse[i]=( ( -inverse[mod % i] * (mod//i) ) % mod )
g2[i]=( (g2[i-1] * inverse[i]) % mod )
inverse[0]=0
mod = 998244353
n,m = mi()
edge = [[] for i in range(n)]
Edge = []
for _ in range(m):
a,b = mi()
edge[a-1].append(b-1)
edge[b-1].append(a-1)
Edge.append((a-1,b-1))
dist = [[10**17 for j in range(n)] for i in range(n)]
for s in range(n):
deq = deque([s])
dist[s][s] = 0
while deq:
v = deq.popleft()
for nv in edge[v]:
if dist[s][nv]==10**17:
dist[s][nv] = dist[s][v] + 1
deq.append(nv)
node_dist = [[j for j in range(n)] for i in range(n)]
for i in range(n):
node_dist[i].sort(key=lambda v:dist[i][v])
ans = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
di = dist[i]
for j in range(i,n):
dj = dist[j]
tmp = [0 for i in range(n)]
tmp_edge = [[] for i in range(n)]
deg = [0 for i in range(n)]
for k in range(m):
a,b = Edge[k]
if di[a]==di[b]+1:
if dj[a]==dj[b]+1:
tmp[a] += 1
elif dj[a]+1==dj[b]:
tmp_edge[b].append(a)
deg[a] += 1
elif di[a]+1==di[b]:
if dj[a]+1==dj[b]:
tmp[b] += 1
elif dj[a]==dj[b]+1:
tmp_edge[a].append(b)
deg[b] += 1
deq = deque([i])
res = []
while deq:
v = deq.popleft()
res.append(v)
for nv in tmp_edge[v]:
deg[nv] -= 1
if not deg[nv]:
deq.append(nv)
res = [v for v in node_dist[i] if not tmp[v]]
ttt = len(res)
dp = [0 for i in range(n)]
dp[i] = 1
for k in range(1,ttt):
if di[res[k]]==di[res[k-1]]:
dp[i] = 0
for v in res:
if tmp[v]!=0:
dp[v] *= inverse[tmp[v]]
dp[v] %= mod
for nv in tmp_edge[v]:
dp[nv] += dp[v]
dp[nv] %= mod
res = dp[j]
for k in range(n):
if tmp[k]!=0:
res *= tmp[k]
res %= mod
ans[i][j] = ans[j][i] = res % mod
for i in range(n):
print(*ans[i])
```
No
| 94,620 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Submitted Solution:
```
import sys
from sys import stdin
from collections import deque
tt = 1
for loop in range(tt):
n,m = map(int,stdin.readline().split())
lis = [ [] for i in range(n) ]
for i in range(m):
a,b = map(int,stdin.readline().split())
a -= 1
b -= 1
lis[a].append(b)
lis[b].append(a)
able = [True] * n
ans = [[0] * n for i in range(n)]
mod = 998244353
for st in range(n):
if able[st] and len(lis[st]) == 1:
nums = [0] * n
d = [float("inf")] * n
d[st] = 0
q = deque([st])
while q:
v = q.popleft()
for nex in lis[v]:
if d[nex] >= d[v] + 1:
nums[nex] += 1
if d[nex] > d[v] + 1:
d[nex] = d[v] + 1
q.append(nex)
ed = 1
for i in range(n):
if nums[i] > 1:
ed *= nums[i]
ed %= mod
#print (nums,ed)
for i in range(n):
for j in range(n):
if nums[i] <= 1 and nums[j] <= 1:
ans[i][j] = ed
ans[j][i] = ed
for i in range(n):
if nums[i] <= 1:
able[i] = False
for ii in ans:
print (*ii)
```
No
| 94,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define a spanning tree of a graph to be a BFS tree rooted at vertex s if and only if for every node t the shortest distance between s and t in the graph is equal to the shortest distance between s and t in the spanning tree.
Given a graph, we define f(x,y) to be the number of spanning trees of that graph that are BFS trees rooted at vertices x and y at the same time.
You are given an undirected connected graph with n vertices and m edges. Calculate f(i,j) for all i, j by modulo 998 244 353.
Input
The first line contains two integers n, m (1 ≤ n ≤ 400, 0 ≤ m ≤ 600) — the number of vertices and the number of edges in the graph.
The i-th of the next m lines contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i < b_i), representing an edge connecting a_i and b_i.
It is guaranteed that all edges are distinct and the graph is connected.
Output
Print n lines, each consisting of n integers.
The integer printed in the row i and the column j should be f(i,j) mod 998 244 353.
Examples
Input
4 4
1 2
2 3
3 4
1 4
Output
2 1 0 1
1 2 1 0
0 1 2 1
1 0 1 2
Input
8 9
1 2
1 3
1 4
2 7
3 5
3 6
4 8
2 3
3 4
Output
1 0 0 0 0 0 0 0
0 2 0 0 0 0 2 0
0 0 1 0 1 1 0 0
0 0 0 2 0 0 0 2
0 0 1 0 1 1 0 0
0 0 1 0 1 1 0 0
0 2 0 0 0 0 2 0
0 0 0 2 0 0 0 2
Note
The following picture describes the first example.
<image>
The tree with red edges is a BFS tree rooted at both 1 and 2.
<image>
Similarly, the BFS tree for other adjacent pairs of vertices can be generated in this way.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
MOD = 998244353
adjList = []
dist = []
ans = []
for _ in range(n):
adjList.append([])
for _ in range(m):
a,b = map(int,input().split())
adjList[a - 1].append(b - 1)
adjList[b - 1].append(a - 1)
for i in range(n):
curDist = [-1] * n
curDist[i] = 0
bfsQueue = [i]
bfsIndex = 0
while len(bfsQueue) > bfsIndex:
c = bfsQueue[bfsIndex]
bfsIndex += 1
for elem in adjList[c]:
if curDist[elem] == -1:
curDist[elem] = curDist[c] + 1
bfsQueue.append(elem)
dist.append(curDist)
for i in range(n):
ans.append([])
for j in range(n):
# Find shortest path between i and j
isCovered = [False] * n
isCovered[i] = True
isCovered[j] = True
jCpy = j
flag = True
while jCpy != i:
cnt = 0
cntIndex = -1
for elem in adjList[jCpy]:
if dist[i][elem] == dist[i][jCpy] - 1:
cnt += 1
cntIndex = elem
if cnt >= 2:
flag = False
break
else:
isCovered[cntIndex] = True
jCpy = cntIndex
if not flag:
ans[i].append(0)
else:
curAns = 1
for k in range(n):
if isCovered[k]:
continue
cnt = 0
for elem in adjList[k]:
if dist[i][elem] == dist[i][k] - 1 and dist[j][elem] == dist[j][k] - 1:
cnt += 1
curAns = (curAns * cnt) % MOD
if not curAns:
break
ans[i].append(curAns)
for elem in ans:
print(" ".join(map(str,elem)))
# 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")
# endregion
if __name__ == "__main__":
main()
```
No
| 94,622 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
# cook your dish here
#jai_shree_raam
#jai_bajrang_bali
#this function is taken from GeekForGeeks
import math
from collections import defaultdict as dfc
from collections import Counter
from math import gcd
def SOE(n):
prime=[True for i in range(n+1)]
p=2
while (p * p <= n):
if(prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p+=1
l=[]
for p in range(2, n+1):
if prime[p]:
l+=[p]
return l
def i1(): return int(input())#single integer
def i2(): return map(int,input().split())#two integers
def i3(): return list(map(int,input().split()))#list of integers
def i4(): return input()#string input
def i5(): return list(str(i1()))#list of characters of a string
def kbit(a, k):
if ((a>>(k-1)) and 1):
return True
else:
return False
for i in range(i1()):
r,b,d=i2()
k1=abs(r-b)
if(k1==0):
print("YES")
else:
k2=min(r,b)
c=k1//k2
if(k1%k2!=0):
c=c+1
if(c>d):
print("NO")
else:
print("YES")
```
| 94,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a, b, d = map(int, input().split())
x, diff = min(a, b), abs(a - b)
if diff / x > d:
print("NO")
else:
print("YES")
```
| 94,624 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
from sys import stdin, stdout
for testcase in range(int(stdin.readline())):
r, b, d = list(map(int, stdin.readline().split()))
if d==0: print(["NO", "YES"][r==b])
else: print(["NO", "YES"][ abs(r-b)/d <= min(r, b) ])
```
| 94,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a,b,c=map(int, input().split())
if b>a:
a,b=b,a
if b*(c+1)>=a:
print("YES")
else:
print("NO")
```
| 94,626 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
t=int(input())
for _ in range(t):
r,b,d=map(int,input().split())
if r==b:
print("YES")
elif r<b:
if b<=(1+d)*r:
print("YES")
else:
print("NO")
elif r>b:
if r<=(1+d)*b:
print("YES")
else:
print("NO")
else:
print("NO")
```
| 94,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
for _ in range(int(input())):
r, b, d = map(int, input().split())
if r > b:
r, b = b, r
if b <= r * (d+1) and b and r:
print('YES')
else:
print('NO')
```
| 94,628 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
r,b,d = [int(i) for i in input().strip().split()]
check = True
if d==0:
if r!=b:
print("NO")
check = False
else:
max_packets = min(r,b)
if (r+b-max_packets)>(d+1)*max_packets:
print("NO")
check = False
if check:
print("YES")
```
| 94,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
r, b, d = map(int, input().split())
x = min(r, b) * d
print('YES' if abs(r - b) <= x else 'NO')
```
| 94,630 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
import math
from sys import stdin,stdout
from operator import itemgetter
t=int(input())
for i in range(t):
n,m,d=map(int,stdin.readline().split())
flag=0
if(d==0):
if(n!=m):
flag=1
else:
mi=min(n,m)
k=mi*(d+1)
if(max(m,n)>k):
flag=1
if(flag==1):
print("NO")
else:
print("YES")
```
Yes
| 94,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
import sys,os,io
from sys import stdin,stdout
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
# input = stdin.readline
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
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 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
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
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for _ in range(int(input())):
a,b,c = map(int,input().split())
if(a<b):
if(b> a*(c+1)):
print("NO")
else:
print("YES")
else:
if(a> b*(c+1)):
print("NO")
else:
print("YES")
# li = list(map(int,input().split()))
```
Yes
| 94,632 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
for _ in range(int(input())):
r,b,d=map(int,input().split())
z,y=min(r,b),max(r,b)
diff=y-z
zz=diff//z+(diff%z!=0)
if d<zz:
print("NO")
else:
print("YES")
```
Yes
| 94,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
import math
def getint():
return [int(i) for i in input().split()]
def getstr():
return [str(i) for i in input().split()]
#--------------------------------------------------------------------------
def solve():
a,b,k=getint()
a_=min(a,b)*(1+k)
b_=max(a,b)
if a_>=b_:
print("YES")
else:
print("NO")
#--------------------------------------------------------------------------
for _ in range(int(input())):
solve()
```
Yes
| 94,634 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
from math import ceil
for tc in range(int(input())):
r,b,d = map(int,input().split())
mini = min(r,b)
maxi = max(r,b)
ans = ceil(maxi/mini)
if abs(ans-mini) <= d:
print('YES')
else:
print('NO')
```
No
| 94,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
for _ in range(int(input())):
r,b,d=map(int,input().split())
if d==0:
if r!=b:
print('No')
else:
print('Yes')
elif min(r,b)==1:
if abs(r-b)<=d:
print('Yes')
else:
print("No")
else:
print("Yes")
```
No
| 94,636 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
try:
n=int(input())
for i in range(n):
r,b,d=map(int,input().split())
if(r==b and d==0):
print("YES")
else:
print("NO")
if(r>=1):
if(b>=1):
if((r-b)<=d):
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
except:
pass
```
No
| 94,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have r red and b blue beans. You'd like to distribute them among several (maybe, one) packets in such a way that each packet:
* has at least one red bean (or the number of red beans r_i ≥ 1);
* has at least one blue bean (or the number of blue beans b_i ≥ 1);
* the number of red and blue beans should differ in no more than d (or |r_i - b_i| ≤ d)
Can you distribute all beans?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first and only line of each test case contains three integers r, b, and d (1 ≤ r, b ≤ 10^9; 0 ≤ d ≤ 10^9) — the number of red and blue beans and the maximum absolute difference in each packet.
Output
For each test case, if you can distribute all beans, print YES. Otherwise, print NO.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
4
1 1 0
2 7 3
6 1 4
5 4 0
Output
YES
YES
NO
NO
Note
In the first test case, you can form one packet with 1 red and 1 blue bean. The absolute difference |1 - 1| = 0 ≤ d.
In the second test case, you can form two packets: 1 red and 4 blue beans in the first packet and 1 red and 3 blue beans in the second one.
In the third test case, since b = 1, you can form only one packet with 6 red and 1 blue beans. The absolute difference |6 - 1| = 5 > d.
In the fourth test case, since d = 0 so each packet should contain the same number of red and blue beans, but r ≠ b.
Submitted Solution:
```
t=int(input())
while(t>0):
r,b,d=[int(x) for x in input().split()]
if d==0:
if r==b:
print("YES")
else:
print("NO")
elif r<=(b+d):
print("YES")
else:
print("NO")
t=t-1
```
No
| 94,638 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if sum(a) != sum(b):
print(-1)
else:
s = 0
for i in range(n):
s += abs(a[i] - b[i])
print(s // 2)
for i in range(n):
if a[i] < b[i]:
while a[i] != b[i]:
for j in range(i + 1, n):
if a[i] == b[i]:
break
if a[j] >= 0 and a[j] > b[j]:
print(j+1, i+1)
a[j] -= 1
a[i] += 1
else:
while a[i] != b[i]:
for j in range(i + 1, n):
if a[i] == b[i]:
break
if a[j] < b[j]:
print(i+1, j+1)
a[j] += 1
a[i] -= 1
```
| 94,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
pos=[]
# f,g=0,0
# for i in range(n):
# if arr1[i]<arr2[i]:
# pos.append((i,arr2[i]-arr1[i]))
# elif arr1[i]>arr2[i]:
# neg.append((i,arr1[i]-arr2[i]))
neg=[]
if sum(arr1)!=sum(arr2):
print(-1)
else:
for i in range(n):
if arr2[i]-arr1[i]>0:
for j in range(arr2[i]-arr1[i]):
pos.append(i+1)
elif arr2[i]-arr1[i]<0:
for j in range(arr1[i]-arr2[i]):
neg.append(i+1)
print(len(pos))
for i in range(len(pos)):
print(neg[i],pos[i])
```
| 94,640 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
def sex(a, b):
if sum(a) != sum(b):
print(-1)
return
if len(a) != len(b):
print(-1)
return
result = ''
plus = []
minus = []
for i in range(len(a)):
if a[i] < b[i]:
for _ in range(b[i] - a[i]):
plus.append(i)
else:
for _ in range(a[i] - b[i]):
minus.append(i)
m = len(plus)
print(m)
for i in range(m):
print(minus[i] + 1, plus[i] + 1)
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
sex(a, b)
```
| 94,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
for idfghjk in range(int(input())):
n=(int(input()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
ai=[]
aj=[]
for i in range(n):
if a[i]>b[i]:
for j in range(a[i]-b[i]):
ai.append(i)
if a[i]<b[i]:
for j in range(b[i]-a[i]):
aj.append(i)
if len(ai)==len(aj):
print(len(ai))
for i in range(len(ai)):
print(ai[i]+1,aj[i]+1)
else:
print(-1)
```
| 94,642 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
a=int(input())
for i in range(a):
s=int(input())
z=list(map(int,input().split()))
q=list(map(int,input().split()))
large=[]
small=[]
for i in range(len(z)):
if(z[i]>q[i]):
large.append(i)
elif(z[i]<q[i]):
small.append(i)
i=0
j=0
ans=[]
flag=0
for i in range(len(small)):
t=small[i]
while(j<len(large) and z[t]!=q[t]):
m=large[j]
if(z[m]==q[m]):
j+=1
continue;
ans.append([m+1,t+1])
z[t]+=1
z[m]-=1
if(z[t]!=q[t]):
flag=1
break;
for i in range(len(z)):
if(z[i]!=q[i]):
flag=1
break;
if(flag==1):
print(-1)
continue;
print(len(ans))
for i in range(len(ans)):
print(ans[i][0],ans[i][1])
```
| 94,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
t = int(input())
for i in range(t):
l = int(input())
a = list( map(int,input().split()) )
b = list( map(int,input().split()) )
if a==b:
print(0)
continue
r=[]
rc = 0
for j in range(l):
if a[j]>b[j]:
for k in range(j+1,l):
if a[k]<b[k]:
tmp = min([abs(a[j]-b[j]),abs(a[k]-b[k])])
r.append([str(j+1)+" "+str(k+1),tmp])
a[j]-=tmp
a[k]+=tmp
rc+=tmp
if a[j]==0:
break
elif a[j]<b[j]:
for k in range(j+1,l):
if a[k]>b[k]:
tmp = min([abs(a[j]-b[j]),abs(a[k]-b[k])])
r.append([str(k+1)+" "+str(j+1),tmp])
a[j]+=tmp
a[k]-=tmp
rc+=tmp
if a[j]==0:
break
if a!=b:
print(-1)
else:
print(rc)
for j in r:
for k in range(j[1]):
print(j[0])
```
| 94,644 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
import sys
LI=lambda:list(map(int,sys.stdin.readline().split()))
MI=lambda:map(int,sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())
for _ in range(II()):
n, a, b=II(), LI(), LI()
v=[b[i]-a[i] for i in range(n)]
if sum(v):
print(-1)
else:
p, n=[], []
for i,x in enumerate(v):
if x>0:
for j in range(x):p.append(i+1)
else:
for j in range(-x):n.append(i+1)
print(len(n))
while p or n:
print(n.pop(), p.pop())
```
| 94,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Tags: brute force, greedy
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a = list(map(int,input().strip().split()))
b = list(map(int,input().strip().split()))
c=0
if len(a)==1:
if a[0]==b[0]:print(0)
else:print(-1)
else:
x=[]
y=[]
for i in range(n):
if a[i]>b[i]:
x+=[i+1]*(a[i]-b[i])
c+=(a[i]-b[i])
elif b[i]>a[i]:
y+=[i+1]*(b[i]-a[i])
c+=(a[i]-b[i])
if c==0:
print(len(x))
for j in range(len(x)):
print(x[j],y[j])
else:print(-1)
```
| 94,646 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
R=lambda:map(int,input().split())
t,=R()
while t:
t-=1;R();a=[[],[]];s=i=0
for x,y in zip(R(),R()):i+=1;d=y-x;s+=d;a[d>0]+=[i]*abs(d)
print(*s and[-1]or[len(a[0])]+[f'{i} {j}'for i,j in zip(*a)])
```
Yes
| 94,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
def ans(a,b,n):
if(sum(a)!=sum(b)):
print(-1)
return
d = {}
count = 0
for i in range(len(a)):
#we count adds
if(a[i]<b[i]):
count+=(b[i]-a[i])
if(a[i]!=b[i]):
d[i+1] = b[i]-a[i]
print(count)
d1 = {}
d2 = {}
for i in d:
if d[i]<0:
d2[i] = d[i]
else:
d1[i] = d[i]
l1 = list(d1.keys())#add
l2 = list(d2.keys())#subtract
i = 0
j = 0
while(i<len(l1) and j<len(l2)):
x = d1[l1[i]]
y = d2[l2[j]]
print(str(l2[j])+' '+str(l1[i]))
d1[l1[i]]-=1
d2[l2[j]]+=1
if d1[l1[i]]==0:
i+=1
if d2[l2[j]]==0:
j+=1
return
m = int(input())
for i in range(m):
n = int(input())
arr = input().split()
a = []
for i in arr:
a.append(int(i))
arr = input().split()
b = []
for i in arr:
b.append(int(i))
ans(a,b,n)
```
Yes
| 94,648 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
import sys
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=[]
d=[]
if sum(a)!=sum(b):
print(-1)
continue
for i in range(n):
if a[i]>b[i]:
for j in range(a[i]-b[i]):
c.append(i+1)
if a[i]<b[i]:
for j in range(b[i]-a[i]):
d.append(i+1)
l=len(c)
print(l)
for i in range(l):
print(c[i],d[i])
```
Yes
| 94,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
from sys import stdin
for _ in range(int(stdin.readline())):
n=int(stdin.readline())
a=list(map(int,stdin.readline().split()))
b=list(map(int,stdin.readline().split()))
ans=0
if sum(a)!=sum(b):
print(-1)
else:
ans=0
arr=[]
for i in range(n):
j=i+1
while j<n and a[i]!=b[i]:
if (a[i]>b[i] and a[j]>=b[j]) or (a[i]<b[i] and a[j]<=b[j]):
j+=1
continue
while a[j]!=b[j] and a[i]!=b[i]:
if a[i]>b[i]:
a[i]-=1
a[j]+=1
arr.append([i+1,j+1])
else:
a[i]+=1
a[j]-=1
arr.append([j+1,i+1])
#print(i,j)
#print(a)
ans+=1
j+=1
print(ans)
#print(a)
for a in arr:
print(*a)
```
Yes
| 94,650 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
n=int(input())
for _ in range(n):
p=int(input())
d=[]
a=list(map(int,input().split()))
b=list(map(int,input().split()))
if a==b:
print(0)
continue
for i in range(len(a)):
if a[i]==b[i]:
d.append(i)
j=0
w=0
i = len(a) - 1
po=[]
while w==0:
while i in d:
i -= 1
while j in d:
j += 1
a[i],a[j]=a[j],a[i]
po.append([i+1,j+1])
if a[i]==b[i]:
d.append(i)
if a[j]==b[j]:
d.append(j)
if a==b:
for k in po:
print(k[0],k[-1])
break
if i==j:
print(-1)
w=90
break
```
No
| 94,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if sum(a) != sum(b):
print(-1)
continue
ac = a.copy()
bc = b.copy()
swaps = []
for i in range(n-1, -1, -1):
#print(a, i)
for j in range(i-1, -1, -1):
if a[i] < b[i]:
if a[j] >= 1:
add = b[i]-a[i]
if a[j] >= add:
a[i] += add
a[j] -= add
for k in range(add):
swaps.append([i+1,j+1])
else:
tmp = a[j]
a[i] += a[j]
a[j] = 0
for k in range(tmp):
swaps.append([i+1,j+1])
if a==b:
print(len(swaps))
for swap in swaps:
print(*sorted(swap))
continue
swaps = []
for i in range(n-1, -1, -1):
#print(bc, i)
for j in range(i-1, -1, -1):
if ac[i] > bc[i]:
if bc[j] >= 1:
add = ac[i]-bc[i]
if bc[j] >= add:
bc[i] += add
bc[j] -= add
for k in range(add):
swaps.append([i+1,j+1])
else:
tmp = bc[j]
bc[i] += bc[j]
bc[j] = 0
for k in range(tmp):
swaps.append([i+1,j+1])
if ac==bc:
print(len(swaps))
for swap in swaps:
print(*sorted(swap))
continue
else:
print(-1)
```
No
| 94,652 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
li=[]
if sum(a)==sum(b):
li1=[]
for i in range(n):
li.append(a[i]-b[i])
for i in range(n):
for j in range(i+1,n):
if li[i]>0 and li[j]<0:
y=min(li[i],abs(li[j]))
li[i]-=y
li[j]+=y
for k in range(y):
li1.append([i+1,j+1])
for j in range(i+1,n):
if li[i]<0 and li[j]>0:
y=min(abs(li[i]),li[j])
li[i]+=y
li[j]-=y
for k in range(y):
li1.append([i+1,j+1])
print(len(li1))
for i in range(len(li1)):
li1[i].sort()
print(*li1[i])
else:
print(-1)
```
No
| 94,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon and Cirno are playing an interesting game with arrays. Cirno has prepared two arrays a and b, both consist of n non-negative integers. AquaMoon can perform the following operation an arbitrary number of times (possibly zero):
* She chooses two indices i and j (1 ≤ i, j ≤ n), then decreases the i-th element of array a by 1, and increases the j-th element of array a by 1. The resulting values at i-th and j-th index of array a are a_i - 1 and a_j + 1, respectively. Each element of array a must be non-negative after each operation. If i = j this operation doesn't change the array a.
AquaMoon wants to make some operations to make arrays a and b equal. Two arrays a and b are considered equal if and only if a_i = b_i for all 1 ≤ i ≤ n.
Help AquaMoon to find a sequence of operations that will solve her problem or find, that it is impossible to make arrays a and b equal.
Please note, that you don't have to minimize the number of operations.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains a single integer n (1 ≤ n ≤ 100).
The second line of each test case contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 100). The sum of all a_i does not exceed 100.
The third line of each test case contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 100). The sum of all b_i does not exceed 100.
Output
For each test case print "-1" on the only line if it is impossible to make two arrays equal with some sequence of operations.
Otherwise, print an integer m (0 ≤ m ≤ 100) in the first line — the number of operations. Then print m lines, each line consists of two integers i and j — the indices you choose for the operation.
It can be proven that if it is possible to make two arrays equal with some sequence of operations, there exists a sequence with m ≤ 100.
If there are multiple possible solutions, you can print any.
Example
Input
4
4
1 2 3 4
3 1 2 4
2
1 3
2 1
1
0
0
5
4 3 2 1 0
0 1 2 3 4
Output
2
2 1
3 1
-1
0
6
1 4
1 4
1 5
1 5
2 5
2 5
Note
In the first example, we do the following operations:
* i = 2, j = 1: [1, 2, 3, 4] → [2, 1, 3, 4];
* i = 3, j = 1: [2, 1, 3, 4] → [3, 1, 2, 4];
In the second example, it's impossible to make two arrays equal.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(c) for c in input().split()]
b = [int(d) for d in input().split()]
x1 = sum(a)
x2 = sum(b)
if x1!=x2:
print(-1)
elif a==b:
print(0)
else:
for i in range(n):
for j in range(n):
if a[i] > b[i] and a[j] < b[j]:
while a[i]!=b[i] and a[j]!=b[j]:
a[i]-=1
a[j]+=1
print(i+1, j+1)
elif a[i] < b[i] and a[j] > b[j]:
while a[j]>0 and a[i]!=b[i] and a[j]!=b[j]:
a[i]+=1
a[j]-=1
print(j+1, i+1)
```
No
| 94,654 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Tags: dfs and similar, shortest paths
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
import math
import bisect
mod = 998244353
# for _ in range(int(input())):
from collections import Counter
# sys.setrecursionlimit(10**6)
# dp=[[-1 for i in range(n+5)]for j in range(cap+5)]
# arr= list(map(int, input().split()))
# n,l= map(int, input().split())
# arr= list(map(int, input().split()))
# for _ in range(int(input())):
# n=int(input())
# for _ in range(int(input())):
import bisect
from heapq import *
from collections import defaultdict,deque
n,m=map(int, input().split())
g=[[] for i in range(n+m)]
for i in range(n):
s=input()
for j,x in enumerate(s):
if x=="#":
g[i].append(n+j)
g[n+j].append(i)
q=deque([0])
dis=[10**9]*(n+m)
dis[0]=0
while q:
node=q.popleft()
for i in g[node]:
if dis[i]>dis[node]+1:
dis[i]=dis[node]+1
q.append(i)
print(-1 if dis[n-1]==10**9 else dis[n-1])
# ls=list(map(int, input().split()))
# d=defaultdict(list)
#for _ in range(int(input())):
#import math
#print(math.factorial(20)//200)
# n=int(input())
# n,k= map(int, input().split())
# arr=list(map(int, input().split()))
```
| 94,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Tags: dfs and similar, shortest paths
Correct Solution:
```
from collections import deque
from collections import defaultdict
n, m = map(int, input().split())
INF = 10 ** 9
visited=defaultdict(int)
g = [[] for _ in range(n + m)] # 0..(n-1) -- rows, n..n+m-1 -- columns
for i in range(n):
s=input()
for j,c in enumerate(s):
j_v=n+j
if(c=="#"):
g[i].append(j_v)
g[j_v].append(i)
q = deque([0])
total_col=0
visited[0]=0
while q:
u=q.popleft()
visited[u] = visited[u] + 1
for v in g[u]:
if(v==n-1):
visited[n-1]=visited[u]
q.clear()
break
if(visited[v]==0):
visited[v]=visited[u]
q.append(v)
print(-1 if (visited[n-1]==0) else visited[n-1])
```
| 94,656 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Tags: dfs and similar, shortest paths
Correct Solution:
```
n, m = map(int, input().split())
t = [input() for i in range(n)]
p = list(zip(*t))
u, v = [True] * n, [True] * m
x, u[n - 1], d = [n - 1], False, 0
while x:
y = []
for i in x:
for j, k in enumerate(t[i]):
if k == '#' and v[j]:
v[j] = False
y.append(j)
x = []
for i in y:
for j, k in enumerate(p[i]):
if k == '#' and u[j]:
u[j] = False
x.append(j)
d += 1
if 0 in x:
print(2 * d)
break
else: print(-1)
```
| 94,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Tags: dfs and similar, shortest paths
Correct Solution:
```
from collections import deque
n, m = map(int, input().split())
INF = 10 ** 9
g = [[] for _ in range(n + m)] # 0..(n-1) -- rows, n..n+m-1 -- columns
for i in range(n):
s = input()
for j, c in enumerate(s):
j_v = n + j
if c == '#':
g[i].append(j_v)
g[j_v].append(i)
q = deque([0])
dist = [INF for _ in range(n + m)]
dist[0] = 0
while q:
u = q.popleft()
for v in g[u]:
n_dist_v = dist[u] + 1
if n_dist_v < dist[v]:
dist[v] = n_dist_v
q.append(v)
print(-1 if dist[n - 1] == INF else dist[n - 1])
```
| 94,658 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Submitted Solution:
```
from collections import deque
inf = 1000000000000
tmp = list(map(int, input().split(' ')[:2]))
n = tmp[0]
m = tmp[1]
graph = [[]for _ in range(n)]
for i in range(n):
tmp = input()
graph[i] = tmp
used = [False]*(n + m)
finalGraph = [[] for _ in range(n + m)]
for i in range(n):
for j in range(m):
#print(graph[i][j])
if graph[i][j] == '#':
#print(i, " ", j + n)
finalGraph[i].append(j + n)
finalGraph[n + j].append(i)
#print(finalGraph, " final graph")
d = deque()
d.append(n - 1)
weight = []
for _ in range(n + m):
weight.append(inf)
weight[n - 1] = 0
while d:
cur = d.pop()
for check in finalGraph[cur]:
if not used[check]:
used[check] = True
weight[check] = weight[cur] + 1
d.append(check)
if weight[0] == inf:
print(-1)
else :
print(weight[0])
```
No
| 94,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Submitted Solution:
```
n, m = map(int, input().split())
first = set()
firstStatus = 0
lastStatus = 0
curr_intersection = set()
last = set()
row = []
intersections=[]
i = 0
intersectPoint = 0
for elem in (input()):
if (elem == "#"):
first.add(i)
firstStatus = 1
i += 1
intersections.append(set())
row.append(first.copy())
for a in range(n - 1):
i = 0
lastStatus = 0
last.clear()
for elem in (input()):
if (elem == "#"):
last.add(i)
i=i+1
j=0
if(last!=set()):
lastStatus=0
row.append(last.copy())
for i in range(len(row) - 1):
curr_intersection = row[len(row) - i - 2].intersection(row[len(row) - 1])
if (curr_intersection == set()):
j = j + 1
lastStatus = 0
row.pop()
else:
lastStatus = 1
break
intersections.append(curr_intersection.copy())
if(intersections[len(intersections) - j - 2] == set() and len(intersections)!=2):
lastStatus=0
if(lastStatus==1):
if (intersections[len(intersections)-j-2].intersection(intersections[len(intersections)-1]) == set() ):
intersectPoint = intersectPoint + 2-j
if (firstStatus == 0 or lastStatus == 0):
print(-1)
else:
print(intersectPoint)
```
No
| 94,660 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Submitted Solution:
```
n, m = map(int, input().split())
first = set()
firstStatus = 0
lastStatus = 0
curr_intersection = set()
last = set()
row = []
intersections=[]
i = 0
intersectPoint = 0
for elem in (input()):
if (elem == "#"):
first.add(i)
firstStatus = 1
i += 1
intersections.append(set())
row.append(first.copy())
for a in range(n - 1):
i = 0
lastStatus = 0
last.clear()
for elem in (input()):
if (elem == "#"):
last.add(i)
i=i+1
j=0
if(last!=set()):
lastStatus=0
row.append(last.copy())
for i in range(len(row) - 1):
curr_intersection = row[len(row) - i - 2].intersection(row[len(row) - 1])
if (curr_intersection == set()):
j = j + 1
lastStatus = 0
else:
lastStatus = 1
break
intersections.append(curr_intersection.copy())
if(lastStatus==1):
if (intersections[len(intersections)-j-2].intersection(intersections[len(intersections)-1]) == set()):
intersectPoint = intersectPoint + 2-j
if (firstStatus == 0 or lastStatus == 0):
print(-1)
else:
print(intersectPoint)
```
No
| 94,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"The Chamber of Secrets has been opened again" — this news has spread all around Hogwarts and some of the students have been petrified due to seeing the basilisk. Dumbledore got fired and now Harry is trying to enter the Chamber of Secrets. These aren't good news for Lord Voldemort. The problem is, he doesn't want anybody to be able to enter the chamber. The Dark Lord is going to be busy sucking life out of Ginny.
The Chamber of Secrets is an n × m rectangular grid in which some of the cells are columns. A light ray (and a basilisk's gaze) passes through the columns without changing its direction. But with some spell we can make a column magic to reflect the light ray (or the gaze) in all four directions when it receives the ray. This is shown in the figure below.
<image> The left light ray passes through a regular column, and the right ray — through the magic column.
The basilisk is located at the right side of the lower right cell of the grid and is looking to the left (in the direction of the lower left cell). According to the legend, anyone who meets a basilisk's gaze directly dies immediately. But if someone meets a basilisk's gaze through a column, this person will get petrified. We know that the door to the Chamber is located on the left side of the upper left corner of the grid and anyone who wants to enter will look in the direction of its movement (in the direction of the upper right cell) from that position.
<image> This figure illustrates the first sample test.
Given the dimensions of the chamber and the location of regular columns, Lord Voldemort has asked you to find the minimum number of columns that we need to make magic so that anyone who wants to enter the chamber would be petrified or just declare that it's impossible to secure the chamber.
Input
The first line of the input contains two integer numbers n and m (2 ≤ n, m ≤ 1000). Each of the next n lines contains m characters. Each character is either "." or "#" and represents one cell of the Chamber grid. It's "." if the corresponding cell is empty and "#" if it's a regular column.
Output
Print the minimum number of columns to make magic or -1 if it's impossible to do.
Examples
Input
3 3
.#.
...
.#.
Output
2
Input
4 3
##.
...
.#.
.#.
Output
2
Note
The figure above shows the first sample test. In the first sample we should make both columns magic. The dragon figure represents the basilisk and the binoculars represent the person who will enter the Chamber of secrets. The black star shows the place where the person will be petrified. Yellow lines represent basilisk gaze moving through columns.
Submitted Solution:
```
from collections import deque
inf = 1000000000000
tmp = list(map(int, input().split(' ')[:2]))
n = tmp[0]
m = tmp[1]
graph = [[]for _ in range(n)]
for i in range(n):
tmp = input()
graph[i] = tmp
used = [False]*(n + m)
finalGraph = [[] for _ in range(n + m)]
for i in range(n):
for j in range(m):
#print(graph[i][j])
if graph[i][j] == '#':
#print(i, " ", j + n)
finalGraph[i].append(j + n)
finalGraph[n + j].append(i)
#print(finalGraph, " final graph")
d = deque()
d.append(n - 1)
weight = []
for _ in range(n + m):
weight.append(inf)
#print(weight)
weight[n - 1] = 0
end = False
while d and not end:
cur = d.pop()
if not used[cur]:
used[cur] = True
#print(cur, " cur ")
for check in finalGraph[cur]:
if not used[check]:
#print(check, " check")
#used[check] = True
if check == 0:
end = True
weight[check] = weight[cur] + 1
d.append(check)
if weight[0] == inf:
print(-1)
else :
print(weight[0])
```
No
| 94,662 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
BA = h[1][2] - BB
AB = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
```
| 94,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Tags: constructive algorithms, greedy, math, matrices
Correct Solution:
```
def get_input():
a, b, d = map(int, input().split())
c, e = map(int, input().split())
f = int(input())
return [a, b, c, d, e, f]
def check_condition(a, b, c, d, e, f):
condition1 = (a + b + c) % 2 == 0
condition2 = (d + e + a) % 2 == 0
condition3 = (e + f + c) % 2 == 0
condition4 = (d + f + b) % 2 == 0
condition = condition1 and condition2 and condition3 and condition4
return condition
def find_t(a, b, c, d, e, f):
t_min1 = round((d + f - a - c) / 2)
t_min2 = round((e + f - a - b) / 2)
t_min3 = round((d + e - b - c) / 2)
t_min4 = 0
t_min = max(t_min1, t_min2, t_min3, t_min4)
t_max1 = round((d + e - a) / 2)
t_max2 = round((e + f - c) / 2)
t_max3 = round((d + f - b) / 2)
t_max = min(t_max1, t_max2, t_max3)
if t_min <= t_max:
return t_min
else:
return -1
def find_all(a, b, c, d, e, f, t):
x1 = round((a + c - d - f) / 2 + t)
x2 = round((d + f - b) / 2 - t)
y1 = round((a + b - e - f) / 2 + t)
y2 = round((e + f - c) / 2 - t)
z1 = round((b + c - d - e) / 2 + t)
z2 = round((d + e - a) / 2 - t)
return [x1, x2, y1, y2, z1, z2]
def generate_string(x1, x2, y1, y2, z1, z2, t):
n = x1 + x2 + y1 + y2 + z1 + z2 + t
s1 = ''.join(['a'] * n)
s2 = ''.join(['a'] * (z1 + z2 + t)) + ''.join(['b'] * (x1 + x2 + y1 + y2))
s3 = ''.join(['a'] * t) + ''.join(['b'] * (y1 + y2 + z1 + z2)) + ''.join(['a'] * (x1 + x2))
s4 = ''.join(['b'] * (t + z2)) + ''.join(['a'] * (z1 + y2)) + ''.join(['b'] * (y1 + x2)) + ''.join(['a'] * x1)
return [s1, s2, s3, s4]
def __main__():
fail_output = "-1"
a, b, c, d, e, f = map(int, get_input())
if not(check_condition(a, b, c, d, e, f)):
print(fail_output)
return False
t = find_t(a, b, c, d, e, f)
if t < 0:
print(fail_output)
return False
x1, x2, y1, y2, z1, z2 = map(int, find_all(a, b, c, d, e, f, t))
s1, s2, s3, s4 = map(str, generate_string(x1, x2, y1, y2, z1, z2, t))
print(str(x1 + x2 + y1 + y2 + z1 + z2 + t) + '\n')
print(s1 + '\n')
print(s2 + '\n')
print(s3 + '\n')
print(s4 + '\n')
__main__()
# Made By Mostafa_Khaled
```
| 94,664 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += Now
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
```
No
| 94,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB
Now34 = AB + NowB + BB
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB - ABB)] + ["b" for j in range(BBB)] + ["a" for j in range(BB - BBB)] + ["b" for j in range(BAB)] + ["a" for j in range(BA - BAB)] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
```
No
| 94,666 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Hamming distance between strings a and b of equal length (denoted by h(a, b)) is equal to the number of distinct integers i (1 ≤ i ≤ |a|), such that ai ≠ bi, where ai is the i-th symbol of string a, bi is the i-th symbol of string b. For example, the Hamming distance between strings "aba" and "bba" equals 1, they have different first symbols. For strings "bbba" and "aaab" the Hamming distance equals 4.
John Doe had a paper on which four strings of equal length s1, s2, s3 and s4 were written. Each string si consisted only of lowercase letters "a" and "b". John found the Hamming distances between all pairs of strings he had. Then he lost the paper with the strings but he didn't lose the Hamming distances between all pairs.
Help John restore the strings; find some four strings s'1, s'2, s'3, s'4 of equal length that consist only of lowercase letters "a" and "b", such that the pairwise Hamming distances between them are the same as between John's strings. More formally, set s'i must satisfy the condition <image>.
To make the strings easier to put down on a piece of paper, you should choose among all suitable sets of strings the one that has strings of minimum length.
Input
The first line contains space-separated integers h(s1, s2), h(s1, s3), h(s1, s4). The second line contains space-separated integers h(s2, s3) and h(s2, s4). The third line contains the single integer h(s3, s4).
All given integers h(si, sj) are non-negative and do not exceed 105. It is guaranteed that at least one number h(si, sj) is positive.
Output
Print -1 if there's no suitable set of strings.
Otherwise print on the first line number len — the length of each string. On the i-th of the next four lines print string s'i. If there are multiple sets with the minimum length of the strings, print any of them.
Examples
Input
4 4 4
4 4
4
Output
6
aaaabb
aabbaa
bbaaaa
bbbbbb
Submitted Solution:
```
h = [[0 in range(10)] for j in range(10)]
for i in range(1, 4):
h[i] = [0 for j in range(i + 1)] + list(map(int, input().split()))
#for i in range(1, 4):
# print(" ".join(map(str, h[i][1:5])))
if (h[1][2] + h[1][3] < h[2][3] or (h[1][2] + h[1][3] - h[2][3]) % 2 == 1):
print("-1")
exit(0)
BB = (h[1][2] + h[1][3] - h[2][3]) // 2
AB = h[1][2] - BB
BA = h[1][3] - BB
NowB = h[1][4]
NowLen = BB + AB + BA
Now24 = BA + NowB + BB;
Now34 = AB + NowB + BB;
BAB = 0
ABB = 0
BBB = 0
Dif = (BA - AB) - (h[2][4] - h[3][4])
if (abs(Dif) % 2 == 1):
print("-1")
exit(0)
if Dif < 0:
ABB += abs(Dif) // 2
Now34 -= ABB * 2
#Now24 += ABB
if (AB < ABB or NowB < ABB):
print("-1")
exit(0)
NowB -= ABB
else:
BAB += Dif // 2
Now24 -= BAB * 2
#Now34 += BAB
if (BA < BAB or NowB < BAB):
print("-1")
exit(0)
NowB -= BAB
if (Now24 < h[2][4] or (Now24 - h[2][4]) % 2 == 1):
print("-1")
exit(0)
#print(Now34 - h[3][4])
for i in range(BB + 1):
if (i > NowB):
break
Now = i * 2
if (Now > Now24 - h[2][4]):
break
if min([(NowB - i) // 2, BA - BAB, AB - ABB]) * 2 >= Now24 - h[2][4] - Now:
#print(i, Now24, h[2][4], NowB)
BBB += i
BAB += (Now24 - h[2][4] - Now) // 2
ABB += (Now24 - h[2][4] - Now) // 2
NowB -= i + (Now24 - h[2][4] - Now)
print(NowLen + NowB)
print("".join(["a" for j in range(NowLen)] + ["a" for j in range(NowB)]))
print("".join(["a" for j in range(AB)] + ["b" for j in range(BB + BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(AB + BB)] + ["a" for j in range(BA)] + ["a" for j in range(NowB)]))
print("".join(["b" for j in range(ABB)] + ["a" for j in range(AB)[ABB:]] + ["b" for j in range(BBB)] + ["a" for j in range(BB)[BBB:]] + ["b" for j in range(BAB)] + ["a" for j in range(BA)[BAB:]] + ["b" for j in range(NowB)]))
exit(0)
print("-1")
```
No
| 94,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
input()
s=list(input())
c = 0
for i in range(len(s)-1):
if s[i] == s[i+1]:
c=c+1
print(c)
```
| 94,668 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n = input()
s = input()
a = 0
p = 'C'
for letter in s:
if letter == p:
a +=1
else:
p = letter
print(a)
```
| 94,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
s = input()
ptr1 = 0
ptr2 = 1
cnt = 0
while ptr1 < n and ptr2 < n:
while ptr2 < n and s[ptr1] == s[ptr2]:
ptr2 += 1
cnt += 1
ptr1 = ptr2
ptr2 += 1
print(cnt)
```
| 94,670 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n=int(input())
k=input()
count=0
for i in range(len(k)-1):
if k[i]==k[i+1]:
count=count+1
print(count)
```
| 94,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
l = int(input())
inp = list(input())
p = inp[0]
c = ''
n = 0
for i in range(l):
if(i==0):
continue
c=inp[i]
if(c==p):
n+=1
else:
p=c
print(n)
```
| 94,672 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n=int(input())
f=str(input())
count=0
for i in range(len(f)-1):
if(f[i]==f[i+1]):
count+=1
print(count)
```
| 94,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
s = str(input())
current = "P"
result = 0
for i in s:
if i != current:
current = i
else:
result+=1
print(result)
```
| 94,674 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Tags: implementation
Correct Solution:
```
n = int(input())
s = [i for i in input()]
count = 0
for i in range(0,n-1):
if s[i] == s[i+1]:
count += 1
print(count)
```
| 94,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
u=int(input())
k=input()
c=0
for i in range(len(k)-1):
if(k[i]==k[i+1]):
c+=1
print(c)
```
Yes
| 94,676 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
n=int(input())
s=input()
cnt=0
for i in range(n-1):
if s[i]==s[i+1]:
cnt+=1
print(cnt)
```
Yes
| 94,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
n = int(input())
m = input()
s = 0
for i in range(n-1):
if m[i] == m[i+1]:
s += 1
print(s)
```
Yes
| 94,678 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
n = input()
ent = input()
lista = []
for e in ent:
lista.append(e)
cont = 0
for i in range(len(lista)-1, 0, -1):
if lista[i] == lista[i-1]:
cont+= 1
print(cont)
```
Yes
| 94,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
n=int(input())
m=input()
a=[]
r,b,g=0,0,0
for i in m:
if(i=="R"):
r=r+1
b=0
g=0
if(r>1):
a.append(r)
elif(i=="B"):
b=b+1
r=0
g=0
if(b>1):
a.append(b)
else:
g=g+1
r=0
g=0
if(g>1):
a.append(g)
if(len(a)>0):
print(max(a)-1)
else:
print("0")
```
No
| 94,680 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
n = int(input())
st = input()
count = 0
for i in range (1,n):
if st[i-1] == st[i]:
count +=i
print (count )
```
No
| 94,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
num = int(input())
string = input()
i = 1
while i < len(string):
print(string)
if string[i] == string[i - 1]:
if i != len(string) - 1:
string = string[:i] + string[i + 1:]
else:
string = string[:i]
else:
i += 1
print(num - len(string))
```
No
| 94,682 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input
The first line contains integer n (1 ≤ n ≤ 50) — the number of stones on the table.
The next line contains string s, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to n from left to right. Then the i-th character s equals "R", if the i-th stone is red, "G", if it's green and "B", if it's blue.
Output
Print a single integer — the answer to the problem.
Examples
Input
3
RRG
Output
1
Input
5
RRRRR
Output
4
Input
4
BRBG
Output
0
Submitted Solution:
```
_=input('')
word=input('')
ct = 0
ct += word.count('RR')
ct += word.count('BB')
ct += word.count('GG')
print(ct)
```
No
| 94,683 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
from collections import defaultdict
n, m = map(int, input().split())
a, b = map(int, input().split())
x, y = a, b
s = [0] * n
for i in range(n - 1):
u, v = map(int, input().split())
if u == a: s[i], b = abs(v - b), v
else: s[i], a = abs(u - a), u
s[n - 1] = abs(a - x) + abs(b - y)
a, b = defaultdict(list), defaultdict(list)
for i, j in enumerate(map(int, input().split()), 1): a[j].append(i)
for j in a: b[j] = a[j][: ]
q = []
for i in range(0, n, 2):
j = s[i] + s[i + 1]
if not a[j]: break
q.append(str(a[j].pop()))
if 2 * len(q) == n: print('YES\n-1 ' + ' -1 '.join(q))
else:
q = []
for i in range(0, n, 2):
j = s[i] + s[i - 1]
if not b[j]: break
q.append(str(b[j].pop()))
if 2 * len(q) == n: print('YES\n' + ' -1 '.join(q) + ' -1')
else: print('NO')
```
| 94,684 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
from collections import defaultdict
n, m = map(int, input().split())
tmp = list(tuple(map(int, input().split())) for _ in range(n))
nails = [abs(a - c) + abs(b - d) for (a, b), (c, d) in zip(tmp, tmp[2:] + tmp[:2])]
segments = defaultdict(list)
for i, s in enumerate(map(int, input().split()), 1):
segments[s].append(i)
for shift in -1, 0:
res = [-1] * n
for nailidx in range(shift, n + shift, 2):
nail = nails[nailidx]
if nail in segments and segments[nail]:
res[(nailidx + 1) % n] = segments[nail].pop()
else:
break
else:
print("YES")
print(" ".join(map(str, res)))
exit(0)
print("NO")
```
| 94,685 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
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-----------------------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
def dist(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def get_sorted_required_pruts(dists):
res = [dists[i * 2] + dists[i * 2 + 1] for i in range(len(dists) // 2)]
res = [(i, x) for i, x in enumerate(res)]
return sorted(res, key=lambda x: x[1])
def get_answer(pruts, required_pruts):
i = 0
j = 0
answer = "YES"
seq = []
while i < len(required_pruts):
if j == len(pruts):
answer = "NO"
return answer, None
if pruts[j][1] > required_pruts[i][1]:
answer = "NO"
return answer, None
if pruts[j][1] < required_pruts[i][1]:
j += 1
else:
seq.append((required_pruts[i][0], pruts[j][0] + 1))
i += 1
j += 1
return answer, [x[1] for x in sorted(seq)]
n, m = map(int,input().split())
gvozdi = [None] * n
for i in range(n):
gvozdi[i] = list(map(int,input().split()))
pruts = list(map(int,input().split()))
pruts = [(i, p) for i, p in enumerate(pruts)]
pruts = sorted(pruts, key=lambda x: x[1])
dists = [dist(gvozdi[i], gvozdi[i + 1]) for i in range(len(gvozdi) - 1)]
dists.append(dist(gvozdi[0], gvozdi[-1]))
even_required_pruts = get_sorted_required_pruts(dists)
# print(dists[-1:] + dists[:-1])
odd_required_pruts = get_sorted_required_pruts(dists[-1:] + dists[:-1])
even_answer, even_seq = get_answer(pruts, even_required_pruts)
odd_answer, odd_seq = get_answer(pruts, odd_required_pruts)
if even_answer == "NO" and odd_answer == "NO":
print("NO")
elif even_answer == "YES":
print("YES")
even_seq = [even_seq[i // 2] if i % 2 == 1 else -1 for i in range(n)]
print(" ".join(map(str, even_seq)))
else:
print("YES")
# print(odd_seq)
odd_seq = [odd_seq[i // 2] if i % 2 == 0 else -1 for i in range(n)]
print(" ".join(map(str, odd_seq)))
```
| 94,686 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
def dist(a, b):
return abs(a[0] - b[0]) + abs(a[1] - b[1])
def get_sorted_required_pruts(dists):
res = [dists[i * 2] + dists[i * 2 + 1] for i in range(len(dists) // 2)]
res = [(i, x) for i, x in enumerate(res)]
return sorted(res, key=lambda x: x[1])
def get_answer(pruts, required_pruts):
i = 0
j = 0
answer = "YES"
seq = []
while i < len(required_pruts):
if j == len(pruts):
answer = "NO"
return answer, None
if pruts[j][1] > required_pruts[i][1]:
answer = "NO"
return answer, None
if pruts[j][1] < required_pruts[i][1]:
j += 1
else:
seq.append((required_pruts[i][0], pruts[j][0] + 1))
i += 1
j += 1
return answer, [x[1] for x in sorted(seq)]
n, m = map(int,input().split())
gvozdi = [None] * n
for i in range(n):
gvozdi[i] = list(map(int,input().split()))
pruts = list(map(int,input().split()))
pruts = [(i, p) for i, p in enumerate(pruts)]
pruts = sorted(pruts, key=lambda x: x[1])
dists = [dist(gvozdi[i], gvozdi[i + 1]) for i in range(len(gvozdi) - 1)]
dists.append(dist(gvozdi[0], gvozdi[-1]))
even_required_pruts = get_sorted_required_pruts(dists)
# print(dists[-1:] + dists[:-1])
odd_required_pruts = get_sorted_required_pruts(dists[-1:] + dists[:-1])
even_answer, even_seq = get_answer(pruts, even_required_pruts)
odd_answer, odd_seq = get_answer(pruts, odd_required_pruts)
if even_answer == "NO" and odd_answer == "NO":
print("NO")
elif even_answer == "YES":
print("YES")
even_seq = [even_seq[i // 2] if i % 2 == 1 else -1 for i in range(n)]
print(" ".join(map(str, even_seq)))
else:
print("YES")
# print(odd_seq)
odd_seq = [odd_seq[i // 2] if i % 2 == 0 else -1 for i in range(n)]
print(" ".join(map(str, odd_seq)))
```
| 94,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
n,m = map(int,input().split())
s = []
for i in range(n):
a = map(int,input().split())
a = list(a)
s.append(a)
s_chet = []
for i in range(1,n-1,2): #Проход по четным гвоздям
q = abs((s[i][0]-s[i-1][0])+(s[i][1]-s[i-1][1])) + abs((s[i][0]-s[i+1][0])+(s[i][1]-s[i+1][1]))
s_chet.append((i + 1, q))
q = abs((s[-1][0]-s[-2][0])+(s[-1][1]-s[-2][1])) + abs((s[-1][0]-s[0][0])+(s[-1][1]-s[0][1]))
s_chet.append((n, q))
s_nechet = []
for i in range(2,n-1,2): #Проход по нечетным гвоздям
q = abs((s[i][0]-s[i-1][0])+(s[i][1]-s[i-1][1])) + abs((s[i][0]-s[i+1][0])+(s[i][1]-s[i+1][1]))
s_nechet.append((i + 1, q))
q = abs((s[-1][0]-s[0][0])+(s[-1][1]-s[0][1])) + abs((s[1][0]-s[0][0])+(s[1][1]-s[0][1]))
s_nechet.append((1, q))
ss = map(int, input().split())
ss = [[i + 1, ss_i] for i, ss_i in enumerate(ss)]
s_chet.sort(key=lambda x: x[1])
s_nechet.sort(key=lambda x: x[1])
ss.sort(key=lambda x: x[1])
prut_chet = [-1] * n
chet = True
j = 0
for i in range(len(s_chet)):
while j < len(ss) and ss[j][1] < s_chet[i][1]:
j += 1
if j == len(ss) or ss[j][1] > s_chet[i][1]:
chet = False
break
if s_chet[i][1] == ss[j][1]:
prut_chet[s_chet[i][0] - 1] = ss[j][0]
j += 1
continue
if chet:
print('YES')
print(" ".join(map(str,prut_chet)))
else:
prut_nechet = [-1] * n
nechet = True
j = 0
for i in range(len(s_nechet)):
while j < len(ss) and ss[j][1] < s_nechet[i][1]:
j += 1
if j == len(ss) or ss[j][1] > s_nechet[i][1]:
nechet = False
break
if s_nechet[i][1] == ss[j][1]:
prut_nechet[s_nechet[i][0] - 1] = ss[j][0]
j += 1
continue
if nechet:
print('YES')
print(" ".join(map(str, prut_nechet)))
else:
print('NO')
```
| 94,688 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
import sys
n, m = map(int, input().split())
points = [tuple(map(int, input().split())) for i in range(n)]
points.append(points.pop(0))
segments = []
x, y = points[-1]
for a, b in points:
if x == a:
segments.append(abs(y - b))
else:
segments.append(abs(x - a))
x, y = a, b
rods = list(map(int, input().split()))
rod_indices = {}
for index, rod in enumerate(rods):
rod_indices.setdefault(rod, []).append(index+1)
for offset in range(2):
target_indices = rod_indices.copy()
assignment = [-1 for i in range(n)]
for i in range(offset, n, 2):
target = segments[(i-1)%n] + segments[i]
if target not in target_indices or target_indices[target] == []:
assignment = None
break
assignment[i] = target_indices[target].pop()
if assignment != None:
print('YES')
print(' '.join(map(str, assignment)))
sys.exit()
print('NO')
```
| 94,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Tags: implementation
Correct Solution:
```
from collections import defaultdict
def main():
n, m = map(int, input().split())
tmp = list(tuple(map(int, input().split())) for _ in range(n))
nails = [abs(a - c) + abs(b - d) for (a, b), (c, d) in zip(tmp, tmp[2:] + tmp[:2])]
segments = defaultdict(list)
for i, s in enumerate(map(int, input().split()), 1):
segments[s].append(i)
for shift in -1, 0:
res = [-1] * n
for nailidx in range(shift, n + shift, 2):
nail = nails[nailidx]
if nail in segments and segments[nail]:
res[(nailidx + 1) % n] = segments[nail].pop()
else:
break
else:
print("YES")
print(" ".join(map(str, res)))
return
print("NO")
if __name__ == '__main__':
main()
```
| 94,690 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Submitted Solution:
```
n,m = map(int,input().split())
s = []
for i in range(n):
a = map(int,input().split())
a = list(a)
s.append(a)
s_chet = []
for i in range(1,n-1,2): #Проход по четным гвоздям
q = abs((s[i][0]-s[i-1][0])+(s[i][1]-s[i-1][1])) + abs((s[i][0]-s[i+1][0])+(s[i][1]-s[i+1][1]))
s_chet.append(q)
q1 = abs((s[-1][0]-s[-2][0])+(s[-1][1]-s[-2][1])) + abs((s[-1][0]-s[0][0])+(s[-1][1]-s[0][1]))
s_chet.append(q1)
s_ch = s_chet.copy()
s_nechet = []
for i in range(2,n-1,2): #Проход по нечетным гвоздям
qq = abs((s[i][0]-s[i-1][0])+(s[i][1]-s[i-1][1])) + abs((s[i][0]-s[i+1][0])+(s[i][1]-s[i+1][1]))
s_nechet.append(qq)
qq1 = abs((s[-1][0]-s[0][0])+(s[-1][1]-s[0][1])) + abs((s[1][0]-s[0][0])+(s[1][1]-s[0][1]))
s_nechet.append(qq1)
s_n = s_nechet.copy()
ss = map(int,input().split())
ss = list(ss)
ss1 = ss.copy()
ss2 = ss.copy()
for i in s_chet:
if i in ss1:
s_ch.remove(i)
ss1.remove(i)
if len(s_ch) == 0:
print('YES')
sss = []
for i in range(m):
sss.append(-1)
sss.append(i+1)
print(" ".join(map(str,sss)))
else:
for i in s_nechet:
if i in ss2:
s_n.remove(i)
ss2.remove(i)
if len(s_n) == 0:
print('YES')
sss = []
for i in range(m):
sss.append(i+1)
sss.append(-1)
print(" ".join(map(str,sss)))
else:
print('NO')
```
No
| 94,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Submitted Solution:
```
n,m = map(int,input().split())
x = 0
y = 0
sx = []
sy = []
for i in range(n):
A = map(int,input().split())
A = list(A)
if A[0] in sx:
sx = sx
else:
sx.append(A[0])
if A[1] in sy:
sy = sy
else:
sy.append(A[1])
q = 2 * sx[-1] + 2 * sy[-1]
ss = map(int,input().split())
ss = list(ss)
if sum(ss) >= q:
print('YES')
ss = []
for i in range(1,m+1):
ss.append(i)
ss.append(-1)
print(" ".join(map(str,ss)))
else:
print('NO')
```
No
| 94,692 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Submitted Solution:
```
import sys
n, m = map(int, input().split())
points = [tuple(map(int, input().split())) for i in range(n)]
segments = []
x, y = points[-1]
for a, b in points:
if x == a:
segments.append(abs(y - b))
else:
segments.append(abs(x - a))
x, y = a, b
rods = list(map(int, input().split()))
rod_indices = {}
for index, rod in enumerate(rods):
rod_indices.setdefault(rod, []).append(index+1)
for offset in range(2):
target_indices = rod_indices.copy()
assignment = [-1 for i in range(n)]
for i in range(offset, n, 2):
target = segments[(i-1)%n] + segments[i]
if target not in target_indices or target_indices[target] == []:
assignment = None
break
assignment[i] = target_indices[target].pop()
if assignment != None:
print('YES')
print(' '.join(map(str, assignment)))
sys.exit()
print('NO')
```
No
| 94,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.
Help Bender to solve this difficult task.
Input
The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.
Output
If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.
If there are multiple solutions, print any of them.
Examples
Input
4 2
0 0
0 2
2 2
2 0
4 4
Output
YES
1 -1 2 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3
Output
YES
1 -1 2 -1 3 -1
Input
6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3
Output
NO
Submitted Solution:
```
n,m = map(int,input().split())
s = []
for i in range(n):
a = map(int,input().split())
a = list(a)
s.append(a)
s_chet = []
for i in range(1,n-1,2): #Проход по четным гвоздям
q = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_chet.append(q)
q1 = abs(sum(s[-1])-sum(s[-2])) + abs(sum(s[-1])-sum(s[0]))
s_chet.append(q1)
s_ch = s_chet.copy()
s_nechet = []
for i in range(2,n-1,2): #Проход по нечетным гвоздям
qq = abs(sum(s[i])-sum(s[i-1])) + abs(sum(s[i])-sum(s[i+1]))
s_nechet.append(qq)
qq1 = abs(sum(s[-1])-sum(s[0])) + abs(sum(s[1])-sum(s[0]))
s_nechet.append(qq1)
s_n = s_nechet.copy()
ss = map(int,input().split())
ss = list(ss)
ss1 = ss.copy()
ss2 = ss.copy()
for i in s_chet:
if i in ss1:
s_ch.remove(i)
ss1.remove(i)
if len(s_ch) == 0:
print('YES')
sss = []
for i in range(m):
sss.append(-1)
sss.append(i+1)
print(" ".join(map(str,sss)))
else:
for i in s_nechet:
if i in ss2:
s_n.remove(i)
ss2.remove(i)
if len(s_n) == 0:
print('YES')
sss = []
for i in range(m):
sss.append(i+1)
sss.append(-1)
print(" ".join(map(str,sss)))
else:
print('NO')
```
No
| 94,694 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the i-th bottle is from brand ai, besides, you can use it to open other bottles of brand bi. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers ai, bi (1 ≤ ai, bi ≤ 1000) — the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
4
1 1
2 2
3 3
4 4
Output
4
Input
4
1 2
2 3
3 4
4 1
Output
0
Tags: brute force
Correct Solution:
```
n = int(input())
a = []
b=[]
freq={}
for i in range(n):
ai,bi = [int(x) for x in input().split()]
a.append(ai)
b.append(bi)
if bi not in freq:
freq[bi] = 0
freq[bi] +=1 #No of bottles than can be used to open bi
ans= n
for i in range(n):
#a[i]#can be used to open bottle bi
#to open ith bottle, we need to find which i has the a[i] in its b array
if a[i] not in freq or freq[a[i]] == 0:
continue
if freq[a[i]] == 1 and b[i] == a[i]:
continue
ans -=1
print(ans)
```
| 94,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the i-th bottle is from brand ai, besides, you can use it to open other bottles of brand bi. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers ai, bi (1 ≤ ai, bi ≤ 1000) — the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
4
1 1
2 2
3 3
4 4
Output
4
Input
4
1 2
2 3
3 4
4 1
Output
0
Tags: brute force
Correct Solution:
```
n=int(input())
arr=[]
rra=[]
for i in range (n):
z=list(map(int,input().split()))
rra.append(z[0])
arr.append(z[1])
count=0
for i in range(n) :
temp=arr[i]
indexx=i
arr.remove(temp)
if rra[i] in arr :
arr.insert(indexx,temp)
else :
count+=1
arr.insert(indexx,temp)
print(count)
```
| 94,696 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the i-th bottle is from brand ai, besides, you can use it to open other bottles of brand bi. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers ai, bi (1 ≤ ai, bi ≤ 1000) — the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
4
1 1
2 2
3 3
4 4
Output
4
Input
4
1 2
2 3
3 4
4 1
Output
0
Tags: brute force
Correct Solution:
```
import sys,math
from collections import deque,defaultdict
import operator as op
from functools import reduce
from itertools import permutations
#sys.setrecursionlimit(10**6)
I=sys.stdin.readline
#alpha="abcdefghijklmnopqrstuvwxyz"
"""
x_move=[-1,0,1,0,-1,1,1,-1]
y_move=[0,1,0,-1,1,1,-1,-1]
"""
def ii():
return int(I().strip())
def li():
return list(map(int,I().strip().split()))
def mi():
return map(int,I().strip().split())
"""def ncr(n, r):
r = min(r, n-r)
numer = (reduce(op.mul, range(n, n-r, -1), 1))%(10**9+7)
denom = (reduce(op.mul, range(1, r+1), 1))%(10**9+7)
return (numer // denom)%(10**9+7)"""
def ncr(n, r, p):
# initialize numerator
# and denominator
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def gcd(x, y):
while y:
x, y = y, x % y
return x
def valid(row,col,rows,cols,rcross,lcross):
return rows[row]==0 and cols[col]==0 and rcross[col+row]==0 and lcross[col-row]==0
def div(n):
if n==1:
return 1
cnt=2
for i in range(2,int(n**.5)+1):
if n%i==0:
if i!=n//i:
cnt+=2
else:
cnt+=1
return cnt
def isPrime(n):
if n<=1:
return False
elif n<=2:
return True
else:
flag=True
for i in range(2,int(n**.5)+1):
if n%i==0:
flag=False
break
return flag
def s(b):
ans=[]
while b>0:
tmp=b%10
ans.append(tmp)
b=b//10
return ans
def main():
n=ii()
d=defaultdict(int)
brnd=[]
for i in range(n):
x,y=mi()
brnd.append((x,y))
cnt=n
for i in range(n):
for j in range(n):
if i!=j:
if brnd[j][1]==brnd[i][0]:
cnt-=1
break
print(cnt)
if __name__ == '__main__':
main()
```
| 94,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the i-th bottle is from brand ai, besides, you can use it to open other bottles of brand bi. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers ai, bi (1 ≤ ai, bi ≤ 1000) — the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
4
1 1
2 2
3 3
4 4
Output
4
Input
4
1 2
2 3
3 4
4 1
Output
0
Tags: brute force
Correct Solution:
```
n = int(input())
a, b = [], []
for i in range(n):
l, m = input().split()
a.append(int(l))
b.append(int(m))
count = 0
for i in range(n):
for j in range(n):
if b[j] == a[i] and i != j:
count += 1
break
print(n - count)
```
| 94,698 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.
Sereja knows that the i-th bottle is from brand ai, besides, you can use it to open other bottles of brand bi. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.
Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers ai, bi (1 ≤ ai, bi ≤ 1000) — the description of the i-th bottle.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
4
1 1
2 2
3 3
4 4
Output
4
Input
4
1 2
2 3
3 4
4 1
Output
0
Tags: brute force
Correct Solution:
```
#A. Sereja and Bottles
n = int(input())
a = []
b = []
for _ in range(n):
x,y = map(int,input().split())
a.append(x)
b.append(y)
total = n
for i in range(n):
for j in range(n):
if i!=j and a[i]==b[j]:
total -= 1
break
print(total)
```
| 94,699 |
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