42分填空，58分编程

填空题基本都是智商题和算法基础，没啥大的难度

四道编程题也不难，基本没用到什么算法，除了第一题LRU看错题意了在50%耗了半天，其他题基本都是1A

没给数据范围，默认全部能暴力，没动什么脑子就过了

九点放代码

UPD：

第一题

模拟LRU

30%代码(50%的被覆盖了）

def solve1():
    n, m = map(int, input().split())

    q = []
    for i in range(m):
        s = input().split()
        op = s[0]
        # print(*q)
        if op == "s":
            k, v = int(s[1]), int(s[2])
            if len(q) < n:
                q.append([1, k, v])
            else:
                q.sort(reverse=True)
                q.pop()
                q.append([1, k, v])
        else:
            temp = int(s[1])
            for j in q:
                if j[1] == temp:
                    print(j[2])
                    t = j
                    q.remove(j)
                    q.append([t[0] + 1, t[1], t[2]])
                    break
            else:
                print(-1)
                # pass

第二题

模拟正则匹配中的*和.，a串是目标串，b串是模式串，匹配成功输出1，失败则输出0

def solve2():
    a = input()
    b = input()

    n = len(a)
    ok = 1
    last = ""
    i, j = 0, 0
    while i < n:
        if a[i] == b[j]:
            i += 1
            j += 1
            continue
        else:
            if b[j] == ".":
                i += 1
                j += 1
            elif b[j] == "*":
                while i < n and b[j - 1] == a[i]:
                    i += 1
                j += 1
            else:
                ok = 0

    print(ok)

第三题

给定一串数字，求此数字循环往右的过程中的最大值，例如1324循环的结果最大是4132

def solve3():
    n = input()
    mx = -1
    for i in n:
        mx = max(mx, int(i))
    pos = []
    for i in range(len(n)):
        if int(n[i]) == mx:
            pos.append(i)

    ans = []
    for i in pos:
        ans.append(int(n[i:] + n[:i]))
    print(max(ans))

第四题

给定字符串A和字符串B，求在字符串A中能找到包含字符串B的字串的最小长度（包含字符串B：即该字串含有字符串B的全部字母，不论顺序）

def solve4():
    temp = input().split(",")
    a = temp[0][1:-1]
    b = temp[1][1:-1]
    n = len(a)
    d = [0] * 29
    for i in b:
        d[ord(i) - ord('A')] += 1
    ans = []
    now = [0] * 29
    l, r = 0, 0

    def check():
        ok = 1
        for i in range(28):
            if now[i] < d[i]:
                ok = 0
                break
        return ok
    if b.find(a[l]) != -1:
        now[ord(a[l]) - ord('A')] += 1
    while r < n:
        ok = 0
        for i in range(28):
            if now[i] < d[i]:
                ok = 1
                break
        if ok:
            r += 1
            if r < n and b.find(a[r]) != -1:
                now[ord(a[r]) - ord('A')] += 1
            continue
        ans.append(a[l:r + 1])

        while check() and l < n:
            # print(l)
            if b.find(a[l]) != -1:
                now[ord(a[l]) - ord('A')] -= 1
                ans.append(a[l:r + 1])
            l += 1
    # print(ans)
    mi = 1 << 31
    for i in ans:
        mi = min(mi, len(i))
    for i in ans:
        if len(i) == mi:
            print(f'"{i}"')
            return
    print('""')