摘要
PAT-2022年夏季考试-乙级满分代码

PAT-2022年夏季考试-乙级

部分题意之后补全

A

题意
思路

模拟两数相加即可

代码

'''
Author: NEFU AB-IN
Date: 2022-06-05 13:25:50
FilePath: \Contest\a.py
LastEditTime: 2022-06-05 13:32:45
'''
d = {
    "1110011": "3",
    "0000011": "1",
    "1011111": "0",
    "1110110": "2",
    "0101011": "4",
    "1111001": "5",
    "1111101": "6",
    "1000011": "7",
    "1111111": "8",
    "1111011": "9",
}

a = list(map(str, input().split()))
b = list(map(str, input().split()))

for i in range(len(a)):
    a[i] = d[a[i]]

for i in range(len(a)):
    b[i] = d[b[i]]

numa = "".join(a)
numb = "".join(b)

print(int(numa) + int(numb))

B

题意
思路

模拟题目要求即可

代码

/*
* @Author: NEFU AB-IN
* @Date: 2022-06-05 13:35:07
* @FilePath: \Contest\b.cpp
* @LastEditTime: 2022-06-05 13:52:00
*/
#include <bits/stdc++.h>
using namespace std;
#define SZ(X) ((int)X.size())

const int N = 1e6 + 10;

struct sa
{
    int id, val, day, order;
    int v[8];
} a[N];

int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i].id;
        for (int j = 1; j <= 7; ++j)
        {
            int val;
            cin >> val;
            if (val)
            {
                a[i].day += 1;
                a[i].val += val;
                a[i].v[j] = val;
            }
        }
    }
    for (int i = 1; i <= 7; ++i)
    {
        printf("Star %d\n", i);
        int mx = 0;
        for (int j = 1; j <= n; ++j)
        {
            mx = max(mx, a[j].v[i]);
        }
        if (!mx)
        {
            printf("NONE\n");
            continue;
        }
        for (int j = 1; j <= n; ++j)
        {
            if (a[j].v[i] == mx)
            {
                printf("%06d\n", a[j].id);
            }
        }
    }
    printf("Star of the week\n");

    auto cmp = [&](sa a, sa b) {
        if (a.val != b.val)
            return a.val > b.val;
        return a.order < b.order;
    };

    sort(a + 1, a + 1 + n, cmp);

    int mx = 0;
    for (int i = 1; i <= n; ++i)
    {
        if (a[i].val >= mx && a[i].day >= 3)
        {
            mx = a[i].val;
            printf("%06d\n", a[i].id);
        }
    }
    return 0;
}

C

题意
思路

相约一下两边的阶乘即可

代码

'''
Author: NEFU AB-IN
Date: 2022-06-05 13:53:51
FilePath: \Contest\c.py
LastEditTime: 2022-06-05 14:04:10
'''
left = 1
right = 1


def f(x, y):
    global left, right
    mn = min(x, y)
    if x == y:
        return
    if mn == x:
        ans = 1
        for i in range(x + 1, y + 1):
            ans *= i
        right *= ans
    if mn == y:
        ans = 1
        for i in range(y + 1, x + 1):
            ans *= i
        left *= ans


n = int(input())

for _ in range(n):
    a, b, c, d = map(int, input().split())

    if a > d:
        a, d = d, a
    if b > c:
        b, c = c, b

    f(a, b)
    f(d, c)

    if left == right:
        print("YES")
    else:
        print("NO")

    left = right = 1

D

题意
思路

题目保证了一个孩子只会说一句真话，所以拿其中一个孩子的话碰另一个孩子的话即可

代码

'''
Author: NEFU AB-IN
Date: 2022-06-05 14:08:12
FilePath: \Contest\d.py
LastEditTime: 2022-06-05 14:40:31
'''


def dec1(x):
    return (x - 1 + 7) % 7


def dec2(x):
    return (x - 2 + 7) % 7


def add1(x):
    return (x + 1) % 7


def add2(x):
    return (x + 2) % 7


def f_y(x):
    y = x  # 求出虚假的日期
    if y == b_y:
        return [1, "yesterday"]
    if y == dec1(b_t):
        return [1, "today"]
    if y == dec2(b_to):
        return [1, "tomorrow"]
    return [0, None]


def f_t(x):
    t = x
    if t == b_t:
        return [1, "today"]
    if t == add1(b_y):
        return [1, "yesterday"]
    if t == dec1(b_to):
        return [1, "tomorrow"]
    return [0, None]


def f_to(x):
    to = x
    if to == b_to:
        return [1, "tomorrow"]
    if to == add1(b_t):
        return [1, "today"]
    if to == add2(b_y):
        return [1, "yesterday"]
    return [0, None]


d = {
    0: "Sunday",
    1: "Monday",
    2: "Tuesday",
    3: "Wednesday",
    4: "Thursday",
    5: "Friday",
    6: "Saturday"
}

a_y, a_t, a_to = map(int, input().split())
b_y, b_t, b_to = map(int, input().split())

f = f_y(a_y)
if f[0]:
    print(d[add1(a_y)])
    print("yesterday")
    print(f[1])

f = f_t(a_t)
if f[0]:
    print(d[a_t])
    print("today")
    print(f[1])

f = f_to(a_to)
if f[0]:
    print(d[dec1(a_to)])
    print("tomorrow")
    print(f[1])

# f = f_y(b_y, a_y, a_t, a_to)

# f = f_t(b_t, a_y, a_t, a_to)
# f = f_to(b_to, a_y, a_t, a_to)

E LRU缓存

题意

LRU 全称为 Least Recently Used，即“最近最少使用”。LRU
缓存机制是指，当缓存满了，而缓存区外面的一个新数据被调用的时候，将缓存中最近最少使用（即最长时间没有被使用过）的数据清除，为新数据开辟出空间。
你的任务就是实现这种 LRU 缓存机制。
思路

采用双指针，当滑动窗口中元素个数超出范围时，移动左指针，直到滑动窗口中元素种类满足即可，元素有重复的没有关系

代码

/*
* @Author: NEFU AB-IN
* @Date: 2022-06-05 14:41:59
* @FilePath: \Contest\e.cpp
* @LastEditTime: 2022-06-05 15:35:10
*/

/*
7-5 LRU缓存
分数 25
作者 陈越
单位 浙江大学
LRU 全称为 Least Recently Used，即“最近最少使用”。LRU
缓存机制是指，当缓存满了，而缓存区外面的一个新数据被调用的时候，将缓存中最近最少使用（即最长时间没有被使用过）的数据清除，为新数据开辟出空间。
你的任务就是实现这种 LRU 缓存机制。

*/
#include <bits/stdc++.h>
using namespace std;
#define SZ(X) ((int)X.size())

const int N = 2e4 + 10;
int st[N];

int main()
{
    int n, m;
    cin >> n >> m;
    vector<int> a(m + 1);
    vector<int> ans;

    for (int i = 1; i <= m; ++i)
    {
        cin >> a[i];
    }
    int cnt = 0;
    for (int i = 1, j = 1; i <= m; ++i)
    {
        if (st[a[i]] == 0)
            cnt += 1;
        st[a[i]] += 1;
        if (cnt > n)
        {
            while (j < i)
            {
                st[a[j]] -= 1;
                if (st[a[j]] == 0)
                {
                    ans.push_back(a[j]);
                    j++;
                    cnt -= 1;
                    break;
                }
                j++;
            }
        }
    }
    for (int i = 0; i < SZ(ans); ++i)
    {
        cout << ans[i] << " "[i == SZ(ans) - 1];
    }
}

PAT-2022年夏季考试-乙级

A

题意

思路

代码

B

题意

思路

代码

C

题意

思路

代码

D

题意

思路

代码

E LRU缓存

题意

思路

代码