蓝桥杯2019年省赛JAVA大学B组

一：组队

答案：490

二：不同的字串

答案：100`

#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
string s = "0100110001010001";
set<string> ans;
int main()
{
    for (int i = 0; i < s.size(); i++)
        for (int j = 0; j < s.size(); j++)
        {
            string t = s.substr(i, j);
            ans.insert(t);
        }
    cout << ans.size();
    return 0;
}

三：数列求值

答案：4659

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod = 10000;

int main()
{
    int a = 1;
    int b = 1;
    int c = 3;
    int d;
    for (int i = 5; i <= 20190324; i++)
    {
        d = (a + b + c) % mod;
        a = b % mod;
        b = c % mod;
        c = d;
    }
    cout << d << endl;
    return 0;
}

四：数的分解

答案 40785

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int ans;

bool check(int n)
{
    while (n)
    {
        if (n % 10 == 2 || n % 10 == 4)
            return false;
        n /= 10;
    }
    return true;
}

int main()
{
    for (int i = 1; i < 2019; i++)
        for (int j = i + 1; j < 2019; j++)
            for (int k = j + 1; k < 2019; k++)
                if (check(i) && check(j) && check(k))
                    if (i + j + k == 2019)
                        ans++;

    cout << ans << endl;
    return 0;
}

五：迷宫

思路 用bfs， 每次按照最小的字典序走，第一次走到终点对应的路径就是最小的字典序。
答案 DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
using namespace std;
const int n = 30, m = 50;
struct node
{
    int x, y, sum;
    char p[500];
};
char map[35][55];
bool vis[35][55];
char path[200];
int dx[] = {1, 0, 0, -1};
int dy[] = {0, -1, 1, 0};
char p[] = {'D', 'L', 'R', 'U'};
void bfs()
{
    queue<node> q;
    q.push({1, 1, 1});
    while (q.size())
    {
        node t = q.front();
        q.pop();
        int x = t.x;
        int y = t.y;
        int s = t.sum;
        if (vis[x][y])
            continue;
        vis[x][y] = true;
        for (int i = 0; i < 4; i++)
        {
            int fx = x + dx[i];
            int fy = y + dy[i];
            if (fx > 0 && fx <= n && fy > 0 && fy <= m && map[fx][fy] == '0' && !vis[fx][fy])
            {
                t.p[s] = p[i];
                if (fx == n && fy == m)
                {
                    for (int i = 1; i <= s; i++)
                        cout << t.p[i];
                    cout << s << endl;
                }
                node te;
                te.x = fx, te.y = fy, te.sum = s + 1;
                memcpy(te.p, t.p, sizeof t.p);
                q.push(te);
            }
        }
    }
}
int main()
{
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> map[i][j];
    bfs();
    return 0;
}

六：特别数的和

思路
直接暴力，数据最大是10000，所以运算最大也是40000，题目给了1s，所以应该是可以过的

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
int ans;
bool check(int n)
{
    while (n)
    {
        if (n % 10 == 2 || n % 10 == 0 || n % 10 == 1 || n % 10 == 9)
            return true;
        n /= 10;
    }
    return false;
}
int n;
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        if (check(i))
            ans += i;
    }
    printf("%d\n", ans);
    return 0;
}

七：外卖店的优先级

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 100010;
struct node
{
    int ts, id;

    bool operator<(const node &a) const
    {
        return a.ts > ts;
    }

} a[N];

int n, m, t, ans;
int dian[10010];
int vis[10010]; //记录上一次修改的位置
bool iscode[10010]; //记录这个id是否已经被计算

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> n >> m >> t;
    int ts, id;
    for (int i = 1; i <= m; i++)
    {
        cin >> ts >> id;
        a[i].ts = ts;
        a[i].id = id;
    }

    sort(a + 1, a + m + 1);

    for (int i = 1; i <= m; i++)
    {
        int id = a[i].id;
        int ts = a[i].ts;
        if (!vis[id])
        {
            dian[id] = 2;
            vis[id] = ts;
        }
        else if (ts == vis[id])
        {
            dian[id] += 2;
            vis[id] = ts;
        }
        else
        {

            dian[id] -= (ts - vis[id] - 1);
            dian[id] += 2;
            if (dian[id] <= 0)
            {
                vis[id] = 0;
                dian[id] = 0;
            }
            else
                vis[id] = ts;
        }

        if (ts == t && !iscode[id] && dian[id] >= 5) //特判刚好在最后一个时间达到5
        {
            iscode[id] = true;
            ans++;
        }
        if (t != ts && dian[id] - (t - ts - 1) >= 5 && !iscode[id]) //在最后一个时间以前达到某个值，可以让这个id一直减到t还保持在>=4
        {
            iscode[id] = true;
            ans++;
        }
    }

    cout << ans << endl;
    return 0;
}

八：人物相关性分析

思路：双指针算法，分别扫描Alice 和 Bob

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
string s;
int n, k, ans;

bool check(int n)
{
    if (s[n] >= 'a' && s[n] <= 'z')
        return false;
    if (s[n] >= 'A' && s[n] <= 'Z')
        return false;
    return true;
}
int main()
{
    scanf("%d", &k);
    getchar();
    getline(cin, s);
    int n = s.size();
   
    string a = "Alice", b = "Bob";
    int x = 5, y = 3;
    int s1, s2, e1, e2;
    bool flag1 = false, flag2 = false;
    for (int i = 0, j = 0; i + 4 < n && j + 2 < n;)
    {
        if (s[i] == 'A' && s[i + 1] == 'l' && s[i + 2] == 'i' && s[i + 3] == 'c' && s[i + 4] == 'e' && (i + 5 == n || check(i + 5)) && (i == 0 || check(i - 1)))
        {
            s1 = i, e1 = i + 4;
            flag1 = true;
        }
        if (s[j] == 'B' && s[j + 1] == 'o' && s[j + 2] == 'b' && ((j + 3 == n) || check(j + 3))  && (j == 0 || check(j - 1)))
        {
            
            s2 = j, e2 = j + 2;
            flag2 = true;
        }
        if (flag1 && flag2)
        {
            if (s1 > e2)
            {
                int dis = s1 - e2 + 1;
                if (dis <= k)
                {
                    ans++;
                    flag2 = false;
                }
                else
                    flag2 = false;
            }

            else if (s2 > e1)
            {
                int dis = s2 - e1 + 1;
                if (dis <= k)
                {
                    ans++;
                    flag1 = false;
                }
                else
                    flag1 = false;
            }
        }

        if (!flag1)
            i++;
        if (!flag2)
            j++;
    }

    cout << ans << endl;
    return 0;
}

九：后缀表达式

这道题比较有争议，我用的贪心，有m个减号，就相当于把m个负数变成正数，如果负数不足m个，就把最小的那几个数变成负数，然后相加。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 100010;
int a[N];
int n, m;
bool cmp(int a, int b)
{
    return abs(a) > abs(b);
}
int main()
{
    scanf("%d%d", &n, &m);
    int s = n + m + 1;
    for (int i = 0; i < s; i++)
    {
        scanf("%d", a + i);
    }
    sort(a, a + s, cmp);
    for (int i = 0; i < s && m > 0; i++)
        if (a[i] < 0)
            m--, a[i] = -a[i];
    if (m > 0)
    {
        for (int i = s - 1; i >= 0 && m > 0; i--)
        {
            a[i] = -a[i];
            m--;
        }
    }
    ll ans = 0;
    for (int i = 0; i < s; i++)
    {
        ans += a[i];
        cout << a[i] << " ";
    }
    cout << endl;
    printf("%lld\n", ans);
    return 0;
}

十：灵能传输

#include <algorithm>
#include <cstring>
#include <iostream>
#include <cstdio>
using namespace std;
const int N = 10010;
int a[N];
int n, t;
bool dif(int a, int b)
{
    if (a > 0 && b < 0)
        return true;
    if (a < 0 && b > 0)
        return true;
    return false;
}
bool Same(int x, int y)
{
    if (x > 0 && y > 0)
        return true;
    if (x < 0 && y < 0)
        return true;
    return false;
}
void sove(int i)
{
    a[i - 1] += a[i];
    a[i + 1] += a[i];
    a[i] -= 2 * a[i];
}
int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        bool zheng = false, fu = false;
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", a + i);
            if (a[i] > 0)
                zheng = true;
            if (a[i] < 0)
                fu = true;
        }
        if (zheng && fu)
        {
            bool flag = false;
            do
            {
                flag = false;
                for (int i = 2; i <= n - 1; i++)
                {
                    //1
                    if (a[i] > 0 && dif(a[i + 1], a[i - 1]))
                    {
                        if (a[i - 1] < 0 && abs(a[i - 1]) > a[i] + a[i + 1])
                        {
                            sove(i);
                            flag = true;
                        }
                        else if (a[i - 1] > 0 && abs(a[i + 1]) > a[i] + a[i - 1])
                        {
                            sove(i);
                            flag = true;
                        }
                    }
                    //2
                    if (a[i] < 0 && dif(a[i + 1], a[i - 1]))
                    {
                        if (a[i - 1] > 0 && a[i - 1] > abs(a[i] + a[i + 1]))
                        {
                            sove(i);
                            flag = true;
                        }
                        if (a[i - 1] < 0 && a[i + 1] > abs(a[i] + a[i - 1]))
                        {
                            sove(i);
                            flag = true;
                        }
                    }
                    //3
                    if (Same(a[i - 1], a[i + 1]) && dif(a[i], a[i - 1]) && dif(a[i], a[i + 1]))
                    {
                        if (a[i] < 0)
                        {
                            if (a[i - 1] > abs(a[i] || a[i + 1] > abs(a[i])))
                            {
                                sove(i);
                                flag = true;
                            }
                        }
                        if (a[i] > 0)
                        {
                            if (abs(a[i - 1]) > a[i] || abs(a[i + 1]) > a[i])
                            {
                                sove(i);
                                flag = true;
                            }
                        }
                    }
                }

            } while (flag);
        }
        int ans = -0x3f3f3f3f;

        for (int i = 1; i <= n; i++)
            ans = max(ans, abs(a[i]));
        printf("%d\n", ans);
    }
    return 0;
}