PAT（甲级）2020年冬季考试自测

自测时间为2021年春季考试前一天下午(5:00 sim 8:00)。

7-1 The Closest Fibonacci Number (20 分)

The Fibonacci sequence $F_n$ is defined by $F_{n+2} = F_{n+1} + F_n$ for $n ge 0$, with $F_0$ and $F_1=1$. The closet Fibonacci number is defined as the Fibonacci number with the smallest absolute difference with the given integer $N$.

Your job is to find the closest Fibonacci number for any given $N$.

Input Specification:

Each input file contains one test case, which gives a positive integer $N (≤10^8)$.

Output Specification:

For each case, print the closest Fibonacci number. If the solution is not unique, output the smallest one.

Sample Input:

305

Sample Output:

233

Hint:

Since part of the sequence is ${ 0, 1, 1, 2, 3, 5, 8, 12, 21, 34, 55, 89, 144, 233, 377, 610, … }$, there are two solutions: $233$ and $377$, both have the smallest distance $72$ to $305$. The smaller one must be printed out.

Solution

签到题不多说了。

#include<iostream>
using namespace std;

const int INF=0x3f3f3f3f;
const int N=10010;
int f[110];
int n;
int cnt;

void init(int maxn)
{
    f[0]=0,f[1]=1;
    cnt=2;
    for(int i=2;;i++)
    {
        f[i]=f[i-1]+f[i-2];
        cnt++;
        if(f[i] > maxn) break;
    }
}

int main()
{
    cin>>n;
    init(n);

    int res=INF;
    for(int i=0;i<cnt;i++)
        if(abs(f[i]-n) < abs(res-n))
            res=f[i];
    cout<<res<<endl;
    //system("pause");
    return 0;
}

7-2 Subsequence in Substring (25 分)

A substring is a continuous part of a string. A subsequence is the part of a string that might be continuous or not but the order of the elements is maintained. For example, given the string atpaaabpabtt, pabt is a substring, while pat is a subsequence.

Now given a string $S$ and a subsequence $P$, you are supposed to find the shortest substring of $S$ that contains $P$. If such a solution is not unique, output the left most one.

Input Specification:

Each input file contains one test case which consists of two lines. The first line contains S and the second line P. S is non-empty and consists of no more than $10^4$ lower English letters. $P$ is guaranteed to be a non-empty subsequence of $S$.

Output Specification:

For each case, print the shortest substring of $S$ that contains $P$. If such a solution is not unique, output the left most one.

Sample Input:

atpaaabpabttpcat
pat

Sample Output:

pabt

Solution

写了将近(45)分钟的一个题，浪费了许多时间，先写了个二分，拿了(20)分，后来想了个双指针做法，拿了满分。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define pb push_back
using namespace std;

const int INF=0x3f3f3f3f;
const int N=10010;
string s,p;
bool vis[N];
vector<int> posl;

bool check(int st,int len)
{
    int j=0;
    for(int i=st;i<st+len;i++)
    {
        if(s[i] == p[j]) j++;
        if(j == p.size()) return true;
    }
    return false;
}

int main()
{
    cin>>s>>p;

    for(int i=0;i<s.size();i++)
        if(s[i] == p[0])
            posl.pb(i);
    posl.pb(s.size());

    int l=posl[0],k=0;
    int cnt=1;
    string ans;
    while(l<s.size())
    {
        int r=l;
        while(r < s.size() && !vis[k])
        {
            if(s[r] == p[k])
            {
                vis[k]=true;
                k++;
            }
            if(k == p.size()) break;
            r++;
        }

        //cout<<l<<' '<<r<<endl;
        if(k == p.size())
            if(ans.size() == 0 || r-l+1 < ans.size())
                ans=s.substr(l,r-l+1);
        l=posl[cnt++];
        memset(vis,0,sizeof vis);
        k=0;
    }
    cout<<ans<<endl;
    //system("pause");
    return 0;
}

7-3 File Path (25 分)

The figure shows the tree view of directories in Windows File Explorer. When a file is selected, there is a file path shown in the above navigation bar. Now given a tree view of directories, your job is to print the file path for any selected file.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer $N (≤10^3)$, which is the total number of directories and files. Then $N$ lines follow, each gives the unique $4$-digit ID of a file or a directory, starting from the unique root ID 0000. The format is that the files of depth d will have their IDs indented by d spaces. It is guaranteed that there is no conflict in this tree structure.

Then a positive integer $K (≤100)$ is given, followed by $K$ queries of IDs.

Output Specification:

For each queried ID, print in a line the corresponding path from the root to the file in the format: 0000->ID1->ID2->...->ID. If the ID is not in the tree, print Error: ID is not found. instead.

Sample Input:

14
0000
 1234
  2234
   3234
    4234
    4235
    2333
   5234
   6234
    7234
     9999
  0001
   8234
 0002
4 9999 8234 0002 6666

Sample Output:

0000->1234->2234->6234->7234->9999
0000->1234->0001->8234
0000->0002
Error: 6666 is not found.

Solution

这题总的来说还好，花了大概(30)分钟的样子，一开始觉得挺难写的，但其实算个不是很复杂的小模拟吧。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<unordered_set>
#define pb push_back
using namespace std;
typedef pair<int,int> PII;
const int INF=0x3f3f3f3f;
const int N=10010;
int dir[N];
int fa[N];
int dep[N];
unordered_set<int> S;
int n,m;

void print(int x)
{
    if(fa[x] == -1)
    {
        printf("%04d",x);
        return;
    }
    print(fa[x]);
    printf("->%04d",x);
}

int main()
{
    memset(fa,-1,sizeof fa);
    cin>>n;
    cin.ignore();

    int last=0;
    for(int i=0;i<n;i++)
    {
        string s;
        getline(cin,s);
        int k=0;
        for(int i=0;i<s.size();i++)
        {
            if(isdigit(s[i])) break;
            else k++;
        }

        int t=stoi(s.substr(k));
        S.insert(t);
        dep[t]=k;

        if(k == dep[last]+1)
            fa[t]=last;
        else if(k)
            fa[t]=dir[k-1];

        dir[k]=t;

        last=t;
    }

    cin>>m;
    while(m--)
    {
        int x;
        cin>>x;
        if(S.count(x) == 0)
            printf("Error: %04d is not found.",x);
        else
            print(x);
        cout<<endl;
    }
    //system("pause");
    return 0;
}

7-4 Chemical Equation (30 分)

A chemical equation is the symbolic representation of a chemical reaction in the form of symbols and formulae, wherein the reactant entities are given on the left-hand side and the product entities on the right-hand side. For example, $CH_4 + 2O_2 =CO_2 +2H_2O$ means that the reactants in this chemical reaction are methane and oxygen: $CH_4$ and $O_2$, and the products of this reaction are carbon dioxide and water: $CO_2$ and $H_2O$.

Given a set of reactants and products, you are supposed to tell that in which way we can obtain these products, provided that each reactant can be used only once. For the sake of simplicity, we will consider all the entities on the right-hand side of the equation as one single product.

Input Specification:

Each input file contains one test case. For each case, the first line gives an integer $N (2≤N≤20)$, followed by N distinct indices of reactants. The second line gives an integer $M (1≤M≤10)$, followed by M distinct indices of products. The index of an entity is a $2$-digit number.

Then a positive integer $K (≤50)$ is given, followed by $K$ lines of equations, in the format:

reactant_1 + reactant_2 + ... + reactant_n -> product

where all the reactants are distinct and are in increasing order of their indices.

Note: It is guaranteed that

1. one set of reactants will not produce two or more different products, i.e. situation like 01 + 02 -> 03 and 01 + 02 -> 04 is impossible;
2. a reactant cannot be its product unless it is the only one on the left-hand side, i.e. 01 -> 01 is always true (no matter the equation is given or not), but 01 + 02 -> 01 is impossible; and
3. there are never more than $5$ different ways of obtaining a product given in the equations list.

Output Specification:

For each case, print the equations that use the given reactants to obtain all the given products. Note that each reactant can be used only once.

Each equation occupies a line, in the same format as we see in the inputs. The equations must be print in the same order as the products given in the input. For each product in order, if the solution is not unique, always print the one with the smallest sequence of reactants – A sequence {${ a_1,⋯,a_m}$} is said to be smaller than another sequence {${b_1,⋯,b_n}$} if there exists $1≤i≤min(m,n)$ so that $a_j = b_j$ for all $j<i$, and $a_i<b_i$.

It is guaranteed that at least one solution exists.

Sample Input:

8 09 05 03 04 02 01 16 10
3 08 03 04
6
03 + 09 -> 08
02 + 08 -> 04
02 + 04 -> 03
01 + 05 -> 03
01 + 09 + 16 -> 03
02 + 03 + 05 -> 08

Sample Output:

02 + 03 + 05 -> 08
01 + 09 + 16 -> 03
04 -> 04

Solution

额，写这题之前跑去吃了个晚饭，这题的实际花费时间大概不到一小时的样子。

先是读题，读完一直在想是不是啥特殊性质题，后来想PAT应该不会考那些高深的东西吧，直接暴搜了，其实暴搜想想的话效率应该还行的，毕竟如果选定一组反应物的话会剪掉很多分支。

考试结束前(10 sim 15)分钟写完加调完，交上去(28)分，有点吃惊，然后结束前(5)分钟也不知道哪个case没考虑到，就等它结束了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<unordered_set>
#define pb push_back
using namespace std;
typedef pair<int,int> PII;
const int INF=0x3f3f3f3f;
const int N=25;
int a[N],b[N];
int n,m,k;
vector<vector<int>> mp[110];
vector<vector<int>> ans;
vector<vector<int>> path;
unordered_set<int> S;
bool vis[110];

void dfs(int u)
{
    if(u == m)
    {
        if(ans.size() == 0 || path < ans) ans=path;
        return;
    }

    for(int i=0;i<mp[b[u]].size();i++)
    {
        //cout<<u<<' '<<mp[b[u]][i].size()<<endl;
        bool ok=true;
        for(int j=0;j<mp[b[u]][i].size();j++)
        {
            int t=mp[b[u]][i][j];
            //cout<<"---"<<t<<endl;
            if(vis[t])
            {
                ok=false;
                break;
            }
        }

        if(ok)
        {
            for(int j=0;j<mp[b[u]][i].size();j++)
            {
                int t=mp[b[u]][i][j];
                vis[t]=true;
            }
            path.pb(mp[b[u]][i]);
            dfs(u+1);
            path.pop_back();
            for(int j=0;j<mp[b[u]][i].size();j++)
            {
                int t=mp[b[u]][i][j];
                vis[t]=false;
            }
        }
    }
}

int main()
{
    cin>>n;
    for(int i=0;i<n;i++) cin>>a[i],S.insert(a[i]);

    cin>>m;
    for(int i=0;i<m;i++) cin>>b[i];

    cin>>k;
    cin.ignore();
    for(int i=0;i<k;i++)
    {
        string s;
        getline(cin,s);
        vector<int> reactant;
        string num;
        bool ok=true;
        for(int i=0;i<s.size()-2;i++)
        {
            if(isdigit(s[i]))
            {
                num+=s[i];
            }
            else if(num.size())
            {
                if(S.count(stoi(num)) == 0)//左侧不是反应物，那么就是不合法的方程式
                {
                    ok=false;
                    break;
                }
                reactant.pb(stoi(num));
                num.clear();
            }
        }

        int product=stoi(s.substr(s.size()-2));
        if(ok) mp[product].pb(reactant);
    }

    for(int i=0;i<m;i++)
    {
        vector<int> reactant;
        if(S.count(b[i]))
        {
            reactant.pb(b[i]);
            mp[b[i]].pb(reactant);
        }
    }

    dfs(0);

    for(int i=0;i<ans.size();i++)
    {
        for(int j=0;j<ans[i].size();j++)
        {
            int t=ans[i][j];
            if(j) printf(" + %02d",t);
            else printf("%02d",t);
        }
        printf(" -> %02d",b[i]);
        puts("");
    }

    //system("pause");
    return 0;
}

结束后发现没考虑左侧反应物为产物的情况，又发现样例中已经提示了这种情况，啊这，属实可惜。

总结

简单题不能出岔子啊，第一题大概10分钟以内解决，第二三题加起来最好不超过一个小时，剩下的两个小时留给最后一题。优先想暴力，比如上面的第二题结束后发现暴力也可以过。

string s,p;

int main()
{
    cin>>s>>p;

    string ans;
    for(int i=0;i<s.size();i++)
        if(s[i] == p[0])
        {
            string t;
            int k=0;
            for(int j=i;j<s.size();j++)
            {
                if(s[j] == p[k]) k++;
                if(k == p.size())
                {
                    t=s.substr(i,j-i+1);
                    break;
                }
            }
            if(ans.size() == 0 || t.size() && t.size() < ans.size()) ans=t;
        }

    cout<<ans<<endl;
    //system("pause");
    return 0;
}

注意要判断子串(t)是否为空串，若为空串则不是有效子串。

(update)：
春季考了(98)分，原来考前一晚的分数已经暗示了结局（草）。