第十四届蓝桥杯模拟赛（第二期）

写在前面 

1. 包含本次模拟赛的10道题题解
2. 能过样例，应该可以AC
3. 若有错误，欢迎评论区指出
4. 有疑问可私信我哈

 

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从2023开始暴力枚举每一个数，直到找到正确答案 

start = 2022

def check(num) :
    t = str(bin(num))
    if t[-6:] == '000000' : return True
    return False

while 1 :
    
    start += 1
    if check(start) : print(start) ; break
    
    

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使用Python自带的datetime库统计一共多少天 

import datetime

start = datetime.date(1949,10,1) 
end = datetime.date(2022,1,1) 
diff = datetime.timedelta(1)

ans = 0

while start < end :
    ans += 1
    start += diff 
    
print(ans)

 

从10开始枚举直到找到正确答案

start = 10 

def check(num) :
    t = int(str(num),16)
    
    if t % num == 0 : print(num,t) ; return True
    return False

while 1 :
    if check(start) : print(start) ; break
    start += 1

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动态规划，dp[i][j]代表走到第i行，第j列的最大值是多少，它是由左边的前一个元素dp[i][j-1]和上边的dp[i-1][j] 两者之间的最大值+原数组第i行，第j列的值相加 更新而来的。

s = "174094882455171152761423221685761892795431233411387427793198\
650286024865090061389344606618496378829135984076361542097372\
601657541200146071777733599818266038012509478351201640618984\
143988087783837107349651099683484992553337438088068198972282\
890781586124258626539246182119762952003918195325258677229419\
698255491250839396799769357665825441616335532825361862146291\
503649293440596342887581257444442930778730382520372975343211\
325351222640703400531067500454956482168314849207060705673849\
265774579830223671554026061117300483012903885770893074783710\
083450145620356667677191627276513995926532444279237315785832\
411595106453089134746365281031552217482363035280722591085079\
053410485925413958279617719034175332412908745680774313630190\
429314820559328748143552689295945058801322270313370955837837\
939182801848609300876356583948397645861551964542532682663945\
625356614462682551015176002433628234343684739800880514363921\
982340231989891351425389287014819359798014755509282450440511\
590838726938103384801541373585690893606978941566666714061214\
952341523168827712604946036245881214982452998386986623826275\
782780208928205527678781609589000725521486468983551558405472\
149903035076783644195574734088152324666290493119955560594634\
905391288186024902215444250421277955403412298227858394469856\
607272647132163832860126054679347881638761723785858733108109\
249157334220127702410373959720286708183036202841837581704881\
367895556630088230650972282944827258473951902831431040790814\
079538232104075905120989173307660289899942087873076421916033\
622143260549608274076012938515668898707915863945382394851328\
164677964192631597026176253407553188801750590935427267220117\
591817866992665840378311257621611574856498432538327068011953\
631534031790352912617015229051836886166704989498756486878095\
690013558017746707412183571476823027885971347137127534455141"

Map = [[0] * 60 for i in range(30)]

for i in range(30) : 
    for j in range(60) :
        Map[i][j] = int(s[i*60+j])

dp = [[0] * 60 for i in range(30)]

for i in range(30) :
    for j in range(60) : 
        if i == 0 and j == 0 : dp[i][j] = Map[0][0]
        elif j == 0 : dp[i][j] = dp[i-1][j] + Map[i][j]
        else : dp[i][j] = max(dp[i-1][j],dp[i][j-1]) + Map[i][j]
        
print(dp[29][59])

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将小于2022的质数都找出来（质数筛），之后枚举质数列表的每一项prime，看2022-prime是否在列表中

def find(num) :
    
    prime = []
    flag = [True] * (num+1)
    
    for i in range(2,num+1) :
        if flag[i] : prime.append(i)
        j = 0 
        while prime[j] <= num // i : 
            flag[prime[j] * i] = False
            if i % prime[j] == 0 : break
            j += 1
            
    return prime 

prime = find(2022)
ans = 0 

for i in prime :
    if 2022 - i in prime : ans += 1
    
print(ans >> 1)

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签到题 略

t,c,s = map(int,input().split())
print(s*t//c - t)

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开一个集合s，读入每一个单词，若该单词不在集合中，则输出并加入集合中。若在集合中则跳过。 

N = int(input())
s = set()
for _ in range(N) : 
    num = input()
    if num not in s : 
        print(num) 
        s.add(num)
        
        

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将原字符串颠倒，看从第一个字符开始到第几个字符，以这两个字符为边界的字符串是回文串（至少为1，找最大长度） ，则剩下的字符（不属于回文串的，颠倒后+原字符串 就是答案，有点绕，看不懂可以私信我）

s = input()
leng = len(s)

res = 0 
s = s[::-1]

def check(s) :
    return s == s[::-1]
    
if check(s) : res = leng 
else :
    for i in range(1,leng) : 
        if check(s[:i]) : res = i
        
t = leng - res 
s = s[::-1]
tmp = s[:t]
print(s + tmp[::-1])

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枚举每一个不在边界上的点，看以它为中心，能贡献几个X图形，相加起来得到的和就是答案

n,m = map(int,input().split())
Map = []
for i in range(n) : Map.append(list(input()))

def bound(x,y) : 
    
    return 0 <= x < n and 0 <= y < m 

def deep_check(t,a,b,c,d) :

    return t == a and t == b and t == c and t == d

def check(x,y) : 
    
    l = 0 
    while 1 : 
        l += 1 
        ax = x + (-1) * l ; ay = y + (-1) * l
        bx = x + (-1) * l ; by = y + 1 * l
        cx = x + 1 * l ; cy = y + 1 * l 
        dx = x + 1 * l ; dy = y + (-1) * l
        if bound(ax,ay) and bound(bx,by) and bound(cx,cy) and bound(dx,dy) : pass
        else : return l - 1
        if not deep_check(Map[x][y],Map[ax][ay],Map[bx][by],Map[cx][cy],Map[dx][dy]): return l - 1    
        
ans = 0 
for i in range(1,n-1) : 
    for j in range(1,m-1) :
        ans += check(i,j) 

print(ans) 

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归并排序求逆序对的模版题，只不过这里不是求逆序对的数量了，答案等于每一对逆序对中较大的那个数（即在左边的数）的和（仔细思考下为什么呢？） 如果没有了解过归并排序，建议先学一下哈

N = int(input())
lst = list(map(int,input().split()))

if N < 2 : print(0) ; exit()
ans = 0 

def mergeSort(nums,l,r,tmp) : 
    if l >= r : return 
    mid = l + r >> 1 
    mergeSort(nums,l,mid,tmp)
    mergeSort(nums,mid+1,r,tmp)
    merge(nums,l,mid,r,tmp)
    
def merge(nums,l,mid,r,tmp) :
    global ans 
    
    i, j = l, mid + 1
    while i <= mid and j <= r : 
        if nums[i] <= nums[j] : 
            tmp.append(nums[i])
            i += 1
        else : 
            ans += nums[i] 
            tmp.append(nums[j])
            j += 1
    while i <= mid : tmp.append(nums[i]) ; i += 1 
    while j <= r : tmp.append(nums[j]) ; j += 1
    
    for k in range(len(tmp)) : nums[l+k] = tmp[k]
    tmp.clear()
    
mergeSort(lst,0,N-1,[])
print(ans)

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