山东大学（威海）程序设计竞赛2020新星赛（线上模拟赛）题解

7-1 factorial

In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.

Input:

Input consists of several lines of integer numbers. The first line contains an integer T, which is the number of cases to be tested, followed by T lines, one integer 1 ≤ n ≤ 10000000 on each line.

1≤T≤10

Output:

The output contains the number of digits in the factorial of the integers appearing in the input.

Sample Input:

2
10
20

Sample Output:

7
19

Solution:

询问一个数的阶乘结果的位数。

如果硬算，当 n 超过 12 临时变量就会爆 int，即使用 int_128 在 n 达到 100 时也顶不住，更不用说 1e7 的上限。

如果用高精度（见 7-6）运算，在这种数据量下必超时。

不难想到对于一个数 n，它的位数 dig(n) = [log10(n)] + 1。

因此 dig(n!) = [log10(1 * 2 * 3 * ... * n)] + 1 = [log10(1) + log10(2) + ... + log10(n)] + 1。  

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        double sum = 0;
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++)
            sum += log10(i);
        cout << (int)sum + 1 << endl;
    }
    return 0;
}

7-2 exam

The final exam is over. In order to give rewards, the teacher should rank the students in the class.

There are N students in the class. This semester, they have learned K courses.

The student number of the i-th student is Xi, and his score of the j-th course is expressed as Aij.

The ranking rules are as follows:

If the scores of two students are different, the first course with different scores will be the basis for ranking, and the one with higher scores will come first.

If the scores of the two students are identical, the one with the smaller student number will come first.

Input:

The first line contains N, K (N≤1000,K≤10).

In the following N lines, the i-th line contains K+1 integers X​i​​ A​i1​​ A​i2​​ A​i3​​ … A​ik​​. (X​i​​<100000,A​ij​​<100000)

Guarantee student number is different.

Output:

A line contains n integers, which are the student numbers from the first to the last, separated by spaces.

There should be a space after the last number

Sample Input:

4 3
1 1 2 3
2 1 3 3
3 2 2 2
4 2 2 3

Sample Output:

4 3 2 1 

Solution:

细心读题并观察样例，排序要求类似字符串字典序。

#include <bits/stdc++.h>
using namespace std;

int n, k;
using p = pair<vector<int>, int>;

bool cmp(const p& a, const p& b)
{
    for (int i = 0; i < k; i++)
        if (a.first[i] != b.first[i])
            return a.first[i] > b.first[i];
    return a.second < b.second;
}

int main()
{
    cin >> n >> k;
    vector<p> vec;
    for (int i = 0; i < n; i++)
    {
        int id;
        cin >> id;
        vector<int> score(k);
        for (int j = 0; j < k; j++) cin >> score[j];
        vec.push_back(p(score, id));
    }
    sort(vec.begin(), vec.end(), cmp);
    for (int i = 0; i < n; i++) cout << vec[i].second << " ";
    return 0;
}

7-3 stack

Xiaoming is learning stack. Stack is a last in, first out data structure. There are only two operations: adding an element to the end of the stack and taking out the end element. Now Xiaoming asks you to help him simulate the operation of the stack.

Input:

The first line contains an integer n indicating how many operations there are. (N≤100000)

Next N lines, two integers A and B per line. A = 1 means to add B to the tail of the stack, A = 2 means to take out the tail element of B times. (B<100000)

The stack is initially empty. Ensure that elements are not removed from the empty stack.

Output:

The first line, an integer m, indicates how many elements are left in the stack.

The second line, m integers, lists the m elements from the beginning to the end of the stack, separated by spaces.

Sample Input:

5
1 2
1 4
2 1
1 3
1 5

Sample Output:

3
2 3 5 

Solution:

模拟栈的 push 和 pop 操作，再从栈底逐个输出。可用双端队列代替。

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    deque<int> q;
    while (t--)
    {
        int n, k;
        cin >> n >> k;
        if (n == 1) q.push_back(k);
        else while (k--) q.pop_back();
    }
    cout << q.size() << endl;
    while (!q.empty())
    {
        cout << q.front() << " ";
        q.pop_front();
    }
    return 0;
}

7-4 Hateful fat

LCY lost his pen when he was studying, which is really a bad situation, because his eyes have been blocked by his fat. So can you tell him the positional relationship between him and the pencil?

For given three points p0,p1,p2,

print " COUNTER_CLOCKWISE” if p0,p1,p2 make a counterclockwise turn (1)

print " CLOCKWISE” if p0,p1,p2 make a clockwise turn (2),

print " ONLINE_BACK” if p2 is on a line p2,p0,p1 in this order (3),

print " ONLINE_FRONT “if p2 is on a line p0,p1,p2 in this order (4),

print " ON_SEGMENT” if p2 is on a segment p0p1(5).

Input:

In the first line, integer coordinates of p0 and p1 are given.

The next line gives an integer q, which represents the number of queries.

Then, q queries are given for integer coordinates of p2.

Output:

For each query, print the above mentioned status.

Sample Input:

0 0 2 0
3
-1 1
-1 -1
2 0

Sample Output:

COUNTER_CLOCKWISE
CLOCKWISE
ON_SEGMENT

Solution:

基础计算几何，判断两个向量之间的关系。可利用点积和叉积性质分类讨论解决。

#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> p;

int dot(const p& a, const p& b) { return a.first * b.first + a.second * b.second; }

int cross(const p& a, const p& b) { return a.first * b.second - a.second * b.first; }

p vec(const p& x, const p& y) { return p(y.first - x.first, y.second - x.second); }

int main()
{
    int x0, y0, x1, y1, q;
    cin >> x0 >> y0 >> x1 >> y1 >> q;
    p p0(x0, y0);
    p p1(x1, y1);
    while (q--)
    {
        int x2, y2;
        cin >> x2 >> y2;
        p p2(x2, y2);
        p a = vec(p0, p1);
        p b = vec(p0, p2);
        int cro = cross(a, b);
        if (cro > 0) cout << "COUNTER_CLOCKWISE" << endl;
        else if (cro < 0) cout << "CLOCKWISE" << endl;
        else
        {
            int dt = dot(a, b);
            if (dt < 0) cout << "ONLINE_BACK" << endl;
            else if ((double)b.first / a.first > 1.0) cout << "ONLINE_FRONT" << endl;
            else cout << "ON_SEGMENT" << endl;
        }
    }
    return 0;
}

7-5 prime

Xiaoming is learning prime number, which is a integer greater than 1 and can only be divided by 1 and itself. Xiao Ming has learned how to judge whether a number is prime. Now, he wants to know how many prime numbers are in [1, N].

Input:

An integer N.(N<=10000000)

Output:

An integer represents how many prime numbers are in [1, N].

Sample Input:

10

Sample Output:

4

Solution:

询问 1e7 上限范围内的素数个数。

单独判断一个数 n 是否为素数复杂度为 O(√n) ，此题数据量下若逐个判断再计数显然会超时。

数论初步：素数筛法

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e7 + 10;
int n, cnt, prime[maxn];
bool vis[maxn];

int main()
{
    cin >> n;
    for (int i = 2; i <= n; i++) // 欧拉筛 O(n)
    {
        if (!vis[i]) prime[cnt++] = i;
        for (int j = 0; j < cnt; j++)
        {
            if (i * prime[j] > n) break;
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
        }
    }
    cout << cnt;
    return 0;
}

7-6 lcy‘s weight

During the holiday, LCY has been eating snacks, but sometimes he also exercises, so his weight has been changing. At the same time, LCY is an excellent college student, so he records the change of his weight the day before every day. But n days later, he found that he could not know how heavy he was now, but he could remember his weight on the first day as M. So LCY found smart you. I hope you can help him calculate his weight now, so can you help him?

Input:

The first line gives two integers, N and M, representing the days lcy experienced and the weight of the first day.(n≤10​4​​,m≤10​10000​​)

Then n lines followed.

Each line contains two integers x and y.(y≤10​10000​​)

When x equals 1, it means that LCY's current weight is y heavier than the previous day.

When x equals 2, it means that LCY's current weight is y lighter than the previous day.

When x equals 3, it means that LCY's current weight is y times heavier than the previous day.

Ensure LCY's weight is always less than 10​10000​​

Output:

Output a single line represents the weight after n days of LCY

Sample Input:

4 70
1 3
3 10
2 100
3 10000000000

Sample Output:

6300000000000

1e10000 上限的数值运算当然要用大数。

所谓高精度就是用字符串模拟加减乘除等数值运算。Java 有内置的大数类而 C++ 没有，所以 C++ 选手需要常备模板。

#include <bits/stdc++.h>
using namespace std;

#define MAXLEN 10010
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4

class BigNum
{
private:
    int a[MAXLEN];
    int len;
public:
    BigNum() { len = 1; memset(a,0,sizeof(a)); }
    BigNum(const int);
    BigNum(const char*);
    BigNum(const BigNum&);
    BigNum& operator =(const BigNum &);
    friend istream& operator >>(istream&, BigNum&);
    friend ostream& operator <<(ostream&, BigNum&);
    BigNum operator +(const BigNum&) const;
    BigNum operator -(const BigNum&) const;
    BigNum operator *(const BigNum&) const;
    BigNum operator /(const int&) const;
    BigNum operator ^(const int&) const;
    int operator %(const int&) const;
    bool operator >(const BigNum& T)const;
    bool operator >(const int& t)const;
    void print();
};

BigNum::BigNum(const int b)
{
    int c, d = b;
    len = 0;
    memset(a,0,sizeof(a));
    while (d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}

BigNum::BigNum(const char* s)
{
    int t, k, index, L, i;
    memset(a,0,sizeof(a));
    L = (int)strlen(s);
    len = L / DLEN;
    if (L % DLEN) len++;
    index = 0;
    for (i = L - 1; i >= 0; i -= DLEN)
    {
        t = 0;
        k = i - DLEN + 1;
        if (k < 0) k = 0;
        for (int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++] = t;
    }
}

BigNum::BigNum(const BigNum& T) : len(T.len)
{
    int i;
    memset(a, 0, sizeof(a));
    for (i = 0; i < len; i++) a[i] = T.a[i];
}

BigNum& BigNum::operator =(const BigNum& n)
{
    int i;
    len = n.len;
    memset(a, 0, sizeof(a));
    for (i = 0; i < len; i++) a[i]=n.a[i];
    return *this;
}

istream& operator >>(istream& in, BigNum& b)
{
    char ch[MAXSIZE * 4];
    int i = -1;
    in >> ch;
    int L = (int)strlen(ch);
    int count = 0, sum = 0;
    for (i = L - 1; i >= 0;)
    {
        sum = 0;
        int t = 1;
        for (int j = 0; j < 4 && i >= 0; j++, i--, t *= 10)
            sum += (ch[i] - '0') * t;
        b.a[count] = sum;
        count++;
    }
    b.len = count++;
    return in;
}

ostream& operator <<(ostream& out, BigNum& b)
{
    int i;
    cout << b.a[b.len - 1];
    for (i = b.len - 2; i >= 0; i--)
        printf("%04d", b.a[i]);
    return out;
}
    
BigNum BigNum::operator +(const BigNum& T) const
{
    BigNum t(*this);
    int i, big;
    big = T.len > len ? T.len : len;
    for (i = 0; i < big; i++)
    {
        t.a[i] += T.a[i];
        if (t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -= MAXN + 1;
        }
    }
    if (t.a[big] != 0) t.len = big + 1;
    else t.len = big;
    return t;
}

BigNum BigNum::operator -(const BigNum& T) const
{
    int i, j, big;
    bool flag;
    BigNum t1, t2;
    if (*this > T)
    {
        t1 = *this;
        t2 = T;
        flag = 0;
    }
    else
    {
        t1 = T;
        t2 = *this;
        flag = 1;
    }
    big = t1.len;
    for (i = 0; i < big; i++)
    {
        if (t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while (t1.a[j] == 0) j++;
            t1.a[j--]--;
            while (j > i) t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while (t1.a[t1.len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if (flag) t1.a[big - 1] = 0 - t1.a[big - 1];
    return t1;
}

BigNum BigNum::operator *(const BigNum& T)const
{
    BigNum ret;
    int i, j = 0, up;
    int temp, temp1;
    for (i = 0; i < len; i++)
    {
        up=0;
        for (j = 0; j < T.len; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if (temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if (up != 0) ret.a[i + j] = up;
    }
    ret.len = i + j;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
    return ret;
}

BigNum BigNum::operator /(const int& b)const
{
    BigNum ret;
    int i, down = 0;
    for (i = len - 1; i >= 0; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
    return ret;
}

int BigNum::operator %(const int& b) const
{
    int i, d = 0;
    for (i = len - 1; i >= 0; i--)
        d=((d*(MAXN+1))%b+a[i])%b;
    return d;
}

BigNum BigNum::operator ^(const int& n) const
{
    BigNum t, ret(1);
    int i;
    if (n < 0) exit(-1);
    if (n == 0) return 1;
    if (n == 1) return *this;
    int m = n;
    while (m > 1)
    {
        t = *this;
        for (i = 1; (i << 1) <= m; i <<= 1) t = t * t;
        m -= i;
        ret = ret * t;
        if (m == 1) ret = ret * (*this);
    }
    return ret;
}

bool BigNum::operator >(const BigNum& T) const
{
    int ln;
    if (len > T.len) return true;
    else if (len == T.len)
    {
        ln = len - 1;
        while (a[ln] == T.a[ln] && ln >= 0) ln--;
        if (ln >= 0 && a[ln] > T.a[ln]) return true;
        else return false;
    }
    else return false;
}

bool BigNum::operator >(const int& t) const
{
    BigNum b(t);
    return *this > b;
}

void BigNum::print()
{
    int i;
    printf("%d", a[len - 1]);
    for (i = len - 2; i >= 0; i--)
        printf("%04d", a[i]);
    printf("\n");
}
    
int main()
{
    int n;
    BigNum m;
    cin >> n >> m;
    while (n--)
    {
        int x;
        BigNum y;
        cin >> x >> y;
        if (x == 1) m = m + y;
        else if (x == 2) m = m - y;
        else m = m * y;
    }
    cout << m;
    return 0;
}

7-7 Prepare for CET-6

In order to prepare the CET-6 exam, LCY is reciting English words recently. Because he is already very clever, he can only recite n words to get full marks. He is going to memorize K words every day. But he found that if the length of the longest prefix shared by all the strings in that day is a​i​​, then he could get a​i​​ laziness value. The lazy LCY must hope that the lazier the better, so what is the maximum laziness value he can get?

For example:

The group {RAINBOW, RANK, RANDOM, RANK} has a laziness value of 2 (the longest prefix is 'RA').

The group {FIRE, FIREBALL, FIREFIGHTER} has a laziness value of 4 (the longest prefix is 'FIRE').

The group {ALLOCATION, PLATE, WORKOUT, BUNDLING} has a laziness value of 0 (the longest prefix is '').

Input:

The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the two integers N and K. Then, N lines follow, each containing one of Pip's strings.

1≤T≤100

2≤N≤10​5​​

2≤K≤N

Each of Pip's strings contain at least one character.

Each string consists only of letters from A to Z.

K divides N.

The total number of characters in Pip's strings across all test cases is at most .

Output:

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum sum of scores possible.

Sample Input:

2
2 2
KICK
START
8 2
G
G
GO
GO
GOO
GOO
GOOO
GOOO

Sample Output:

Case #1: 0
Case #2: 10

Solution:

利用 Trie 进行字符串前缀计数，树上每位的值除以 k 求和。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e6 + 7;
int cnt, val[maxn], trie[maxn][26];

void insert(const string& s)
{
    int u = 0;
    int len = (int)s.length();
    for (int i = 0; i < len; i++)
    {
        int id = s[i] - 'A';
        if (!trie[u][id]) trie[u][id] = ++cnt;
        val[trie[u][id]]++;
        u = trie[u][id];
    }
}

int main()
{
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        cnt = 0;
        memset(val, 0, sizeof(val));
        memset(trie, 0, sizeof(trie));
        int n, k;
        cin >> n >> k;
        for (int j = 0; j < n; j++)
        {
            string s;
            cin >> s;
            insert(s);
        }
        int res = 0;
        for (int j = 1; j <= cnt; j++) res += val[j] / k;
        cout << "Case #" << i << ": " << res << endl;
    }
    return 0;
}

7-8 Computer Games

LCY is playing games with his computer at home. Each playing card has two attributes: A​i​​ and B​i​​, LCY and his computer take turns selecting playing cards,with LCY going first. In each turn, a player can select one card, as long as that playing card either has an A​i​​ greater than each of the all soldiers selected so far, or has a B​i​​ greater than each of the all cards selected so far.

To be precise:

let A​i​​ and B​i​​ be the values for the i-th cards, for i from 1 to N, and let S be the set of cards that have been selected so far. Then a player can select card x if and only if at least one of the following is true:

A​x​​>A​s​​ for all s in S

B​x​​>B​s​​ for all s in S

If no selection can be made, then the selection process ends and the player with more playing cards win. LCY wants to select more cards and win the game. If both players play optimally to accomplish their goals, can LCY succeed?

Input:

The first line of each case contains a positive integer N, the number of cards. N more lines follow; the i-th of these line contains two integers Ai and Bi, indicating the values of the i-th cards.(N≤4000,0≤A​i​​,B​i​​≤10000)

Output:

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is YES or NO, indicating whether LCY can guarantee that he selects more cards than computer, even if computer plays optimally to prevent this.

Sample Input:

3
3
10 2
1 10
10 3
3
10 1
10 10
1 10
3
10 2
1 10
4 9

Sample Output:

Case #1: NO
Case #2: YES
Case #3: YES

Solution:

这种博弈的精髓在于模仿。查找最内层是否成双。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<int, int> p;

bool solve(const vector<p>& v)
{
    for (p last(1e4 + 10, 1e4 + 10), now;; last = now)
    {
        now = p(0, 0);
        for (int i = 0; i < (int)v.size(); i++)
        {
            if (v[i].first < last.first && v[i].second < last.second)
            {
                now.first = max(now.first, v[i].first);
                now.second = max(now.second, v[i].second);
            }
        }
        for (int i = 0; i < (int)v.size(); i++) if (v[i] == now) return true;
        if (now == p(0, 0)) return false;
    }
}

int main()
{
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        int n;
        cin >> n;
        vector<p> vec(n);
        for (int i = 0; i < n; i++) cin >> vec[i].first >> vec[i].second;
        bool flag = solve(vec);
        cout << "Case #" << i << ": " << (flag ? "YES" : "NO") << endl;
    }
    return 0;
}

7-9 sort

Xiaoming is tired of asking for help. This time he wants to test you.

He gave you N integers. Please find the number with the largest M.

Input:

The first row has two numbers N, M (0<m≤n≤1000000).

The second line contains N integers that are different and are all in the interval [- 500000,500000].

Output:

Output the number of the first M in the order of large to small.

Sample Input:

5 3 
3 -35 92 213 -644

Sample Output:

213 92 3 

Solution:

签到题

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n, m;
    cin >> n >> m;
    vector<int> v(n);
    for (int i = 0; i < n; i++) cin >> v[i];
    sort(v.begin(), v.end(), greater<int>());
    for (int i = 0; i < m; i++) cout << v[i] << " ";
    return 0;
}

7-10 lcy eats biscuits

In order to grow nine abdominal muscles, LCY began to eat all the N biscuits in his room, but LCY didn't like walking, because walking would cause his abdominal muscles to become smaller.How long does he have to run to get n biscuits. At the beginning, LCY is at (0,0)

Input:

The first line is a positive integer N.(N≤14)

Next, there are 2 real numbers in each line, representing the coordinates of the ith biscuits.(−10000≤x​i​​,y​i​​≤10000)

The distance formula between two points is√​(x​1​​−x​2​​)​2​​+(y​1​​−y​2​​)​2​​​​​

Output:

A number, representing the minimum distance to run, with 2 decimal places reserved.

Sample Input:

4
1 1
1 -1
-1 1
-1 -1

Sample Output:

7.41

Solution:

在二维坐标系下，询问从原点开始经过 n 个点的最短路程。

如果时限 1s，n 在 10 以内随便爆搜，14 以内需要剪枝，20 以上需要状态压缩。

在这里给出我的方法：首先根据坐标预处理任意两点之间的距离，然后进行状态压缩 DFS。

定义状态为到达某点时历史访问的点集，故对于 n 个点可用 2 ^ n 以内的二进制数表示任意一种状态。

开一个 dp 数组存放每个状态的最优解，需要注意的是还需要第二维：当前位置。搜索结束后遍历每个点的最终状态维护最优解（因为不确定最后来到哪个点）。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> p;
const int inf = 0x3f3f3f3f;
const int maxn = 15;
const int maxm = 2e5;
int n;
bool vis[maxn];
double dis[maxn][maxn], dp[maxm][maxn];
p pos[maxn];

double getDis(p a, p b) { return sqrt(pow(a.first - b.first, 2) + pow(a.second - b.second, 2)); }

void dfs(int u, int sta, double tot)
{
    vis[u] = true;
    if (dp[sta][u] < tot) return; // 曾经以相同的状态来过并且更优
    dp[sta][u] = tot;
    for (int v = 0; v <= n; v++)
    {
        if (vis[v]) continue;
        int sta_nxt = sta | (1 << (v - 1));
        dfs(v, sta_nxt, tot + dis[u][v]);
        vis[v] = false;
    }
}

int main()
{
    cin >> n;
    for (int i = 0; i < 1 << n; i++)
        for (int j = 0; j <= n; j++)
            dp[i][j] = inf;
    for (int i = 1; i <= n; i++)
        cin >> pos[i].first >> pos[i].second;
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
            dis[i][j] = getDis(pos[i], pos[j]);
    dfs(0, 0, 0);
    double res = inf;
    for (int i = 1; i <= n; i++) res = min(res, dp[(1 << n) - 1][i]);
    printf("%.2lf", res);
    return 0;
}