# 112 學年度商業類學生技藝競賽
**題目出處：**[點擊](https://sci.me.ntnu.edu.tw/PortalFile/ContestData/10239/627a578c-8d17-49d0-9a7b-f9f44bcdbd3c.pdf)

# pA. 二數有權重的相加–(單列)

a,b=map(int,input().split())
print(4*a+6*b)

# pB. 剪刀石頭布單行版

n1,n2=input().split()
if n2==n1:
    print(0)
elif n1=='Y':
    if n2=='O':
        print(2)
    else:
        print(1)
elif n1=='O':
    if n2=='X':
        print(2)
    else:
        print(1)
else:
    if n2=='Y':
        print(2)
    else:
        print(1)

# pC. 閏年

n=int(input())
n+=1911

if n%4==0 and n%100!=0 or n%400==0:
    print(True)
else:
    print(False)

# pD. ASCII碼

n=int(input())
if 122>=n>=97:
    print(2)
elif 90>=n>=65:
    print(1)
else:
    print(0)

# pE. 2的冪

n=int(input())
a=1
while 2**a<=n:
    a+=1
print(2**(a-1))

# pF. 質因數分解

n = int(input())
ans = []
while n != 1:
    for i in range(2, int(n**0.5)+1):
        if not(n%i):
            ans.append(i)
            n //= i
            break
    else:
        ans.append(n)
        break
print(*ans)

# pG. 因數加總

n=int(input())
ans=0
for i in range(1,n+1):
    if n%i==0:
        ans+=i
print(ans)

# pH. 數字運算(減1或除以10)

n,times=map(int,input().split())
while times:
    if n%10!=0:
        n-=1
    else:
        n//=10
    times-=1
print(n)

# pI. 數字相加

t=int(input())
for _ in range(t):
    n=int(input())
    l=[int(i) for i in input().split()]
    l,target=l[0:n-1],l[-1]
    l.sort()
    flag=False
    for i in range(len(l)):
        if l[i]>=target:
            break
        for j in range(i+1,len(l)):
            if l[i]+l[j]==target:
                flag=True
                break
            if l[i]+l[j]>target:
                break
        if flag:
            break
    if flag:
        print('YES')
    else:
        print('NO')

# pJ. 七段顯示器
貪心者法則
性價比最高就是1，1用來衝位數，當n為奇數就把多的一段拿來當7放最前面，就會是最大數。

for _ in range(int(input())):
    n=int(input())
    ans='1'*(n//2)
    if n%2:
        ans='7'+ans[1:]
    print(ans)

# pK. 字串

for _ in range(int(input())):
    input()
    n = input()
    seen = set()
    for i in range(len(n)-1):
        seen.add(n[0:i]+n[i+2::])
    print(len(seen))

# pL. 中位數

t=int(input())
ans=[]
for i in range(t):
    n=int(input())
    ans.append(n)
    ans.sort()
    D=len(ans)//2
    if len(ans)%2==0:
        print((ans[D-1]+ans[D])//2)
    else:
        print(ans[D])

# pM. 分組

for _ in range(int(input())):
    n, r = map(int, input().split())
    arr = sorted([int(i) for i in input().split()])
    ans = 0
    temp = []
    while len(arr):
        temp.append(arr.pop())
        if len(temp)*temp[-1] >= r:
            temp =[]
            ans += 1
    print(ans) 

# pN. 函數

n = int(input())
print(-(n//2+1) if n%2 else n//2)

# pO. 洗牌

n, m = map(int, input().split())
arr = input().split()
mid = n//2
for _ in range(m):
    temp = []
    for i in range(mid):
        temp.append(arr[i])
        temp.append(arr[mid+i])
    arr = temp.copy()
print(*arr)

# pP. 魔術方陣

arr=[]
for _ in range(3):
    arr.append([int(i) for i in input().split()])
a=sum(arr[0][1::])
b=(arr[1][0]+arr[1][2])
c=sum(arr[2][0:2])
S=(a+b+c)//2
arr[0][0]=S-a
arr[1][1]=S-b
arr[2][2]=S-c
for i in arr:
    print(*i)

# pQ. 數字全部相加成本
霍夫曼編碼

while int(input()):
    q = [int(i) for i in input().split()]
    ans = 0
    while len(q) > 1:
        q.sort(reverse=True)
        a, b = q.pop(), q.pop()
        ans += a+b
        q.append(a+b)
    print(ans)

# pR. 樹的直徑
### 需要多練練
先看每一條樹的直徑，再用最長的樹來去算直徑。
BFS走訪

n = int(input())
l = [[] for _ in range(n+1)]
for _ in range(n-1):
    a, b = map(int, input().split())
    l[a].append(b)
    l[b].append(a)
q = [1]
seen = [-1]*(n+1)
seen[1] = 0
head = 0
while head < len(q):
    node = q[head]; head += 1
    for i in l[node]:
        if seen[i] == -1:
            seen[i] = seen[node]+1
            q.append(i)
i = seen.index(max(seen))
seen = [-1]*(n+1)
seen[i] = 0
q = [i]
head = 0
while head < len(q):
    node = q[head]; head += 1
    for i in l[node]:
        if seen[i] == -1:
            seen[i] = seen[node]+1
            q.append(i)
print(max(seen))

# pS. 迴圈
### 需要多練練
BFS

n,m=map(int,input().split())
l=[[] for i in range(n+1)]
for _ in range(m):
    a,b=map(int,input().split())
    l[a].append(b)
    l[b].append(a)
#print(l)

seen=set()
ans=0
for i in range(1,n+1):
    if i not in seen:
        que=[i];head=0
        arr=[]
        while head<len(que): #BFS
            node=que[head];head+=1
            for v in l[node]:
                if v not in seen:
                    arr.append(v)
                    que.append(v)
                    seen.add(v)
        if arr and all(len(l[num]) == 2 for num in arr): #如果arr裡有東西,而且每個點都是有兩條邊的
            ans += 1
print(ans)

補充第22行：因為arr有可能沒東西，所以先判斷all是否為空值，再去做all運算。