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Those who have come to refer to this article. Sorry. A, B, C, D understood A - Registration

s==t[0:-1]#(I mean[:-1]But I feel good...)

B - Easy Linear Programming If you pull the number in order, it's ok Dirty source code.

def j():
    a,b,c,k=map(int,input().split())
    o=0
    x=k
    if a!=0:
        o+=a
        k-=a
        if k==0:
            return a
        elif k<0:
            return x
    if b!=0:
        k-=b
        if k<=0:
            return a
    if c!=0:
        k-=c
        o-=c
        if k==0:
            return o
        elif k<0:
            return o-k
print(j())

C - Skill Up I skipped it. Not good at all searches. By the next time, learn.

Postscript It seems that it can be done by bit search. So learn I tried it. As a result, it was a one-shot clear. The bit search method turned out to be convenient.

n,m,x=map(int,input().split())
item=[list(map(int,input().split()))for i in range(n)]#Gold, algorithm up number
bag=[]#How to buy a book that meets the conditions
for i in range(2**n):#Examine all
    temp=[]#Books to buy
    job=[0]*m#Comprehension
    for s in range(n):#Examine all digits
        if (i>>s)&1:#If you buy
            temp.append(s)#Add books to buy
            for y in range(m):
                job[y]+=item[s][y+1]#Add comprehension
    for s in range(len(job)):
        if job[s]<x:#If the comprehension is less than X, break
            break
    else:#If all comprehension is X or higher
        money=0
        for s in temp:#Calculate the total amount
            money+=item[s][0]
        bag.append(money)
#If there is not more than one combination to buy-Output 1
print(min(bag) if len(bag)>=1 else -1)

D - Teleporter If you turn K times for, it will only be TLE, so you can go if there is a surplus at the beginning and end of the loop. I thought, but I couldn't go. Someone tell me. By the way, an example of a problem sentence is made.


n,k=map(int,input().split())
a=[int(i)-1 for i in input().split()]
ed=[False]*n
go=0
c=0
cc=0
while not ed[go]:#End if you have already passed
    print(a[go]+1)
    ed[go]=True
    go=a[go]
    c+=1
ed=[False]*n
print()
while not ed[go]:#The second loop can be different 6,5,2,5,3,In case of 2, 6,2,5,3,2,5,3,2,5,...Becomes
    print(a[go]+1)
    ed[go]=True
    go=a[go]
    cc+=1#Save as another variable
print()
for i in range((k-c)%cc):
    print(a[go]+1)
    go=a[go]
print()
print(go+1)

Postscript I could do it by putting together while, saving the value without turning print with for, and adding the output when the loop was never looped.

n,k=map(int,input().split())
a=[int(i)-1 for i in input().split()]
p=0
visited=[0]*n
first=[]
roop=[]
while visited[p]!=2:#Stop if you go twice
    if visited[p]==0:#I went for the first time
        first.append(p)
    else:#Passed once
        roop.append(p)
    visited[p]+=1#Make a mark
    p=a[p]#Teleport
if len(first)>k:#If it never loops
    print(first[k]+1)#First kth
else:
    print(roop[(k-(len(first)-len(roop)))%len(roop)]+1)#Otherwise, subtract the first loop and the second loop from k, and the surplus of the number of times of the second loop

Finally

E and F couldn't reach. It was my third challenge, but it was difficult. I thought I had to learn more algorithms and be able to think logically. By the way, the new Rating is 68.

AtCoder Beginner Contest 167 Memorandum

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Finally