2023.3.15 30天蓝桥训练打卡题目 day 12

[szx233]

一、0的个数

#include<iostream>
using namespace std;
typedef long long ll;

int main()
{
    string s;
    cin >> s;
    int t = s.size(), p = 0;
    for(int i = t - 1; i >= 0; i -- )
        if(s[i] == '0') p ++ ;
        else break;
    cout << p;

    return 0;
}

 

二、超级质数

#include<iostream>
using namespace std;
typedef long long ll;

bool is_prime(int x)
{
    if(x == 1) return false;
    for(int i = 2; i <= x / i; i ++ )
        if(x % i == 0) return false;
    return true;
}

int main()
{
    int ma = 0;
    for(int i = 2; i <= 400; i ++ )
    {
        if(is_prime(i))
        {
            string h = to_string(i);
            bool flag = false;
            for(int j = 1; j <= h.size(); j ++ )
                for(int k = 0; k < h.size() - j + 1; k ++ )
                    if(!is_prime(stoi(h.substr(k, j))))
                        flag = true;
            if(!flag) ma = max(ma, i);
        }
    }
    cout << ma;

    return 0;
}

 

 

三、卡牌

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PII;
const int N = 2e5 + 10;
PII a[N];
ll s[N];
bool cmp(PII x, PII y)
{
    return x.first < y.first;
}
int main()
{
    ll n, m, q = 0;
    cin >> n >> m;
    for(int i = 1; i <= n; i ++ ) cin >> a[i].first;
    for(int i = 1; i <= n; i ++ ) cin >> a[i].second, q += a[i].second;
    sort(a + 1, a + n + 1 ,cmp);
    int k = 1;
    for(int i = 1; i < n; i ++ )
        if(a[i + 1].first != a[i].first)
            s[k ++ ] = i * (a[i + 1].first - a[i].first);
    s[k] = n;
    for(int i = 1; i <= k; i ++ ) s[i] += s[i - 1];
    ll p = 0;
    for(int i = 1; i <= k; i ++ )
        if(s[i] > q)
        {
            p = i;
            break;
        }

    ll mi = p;
    for(int i = 1; i <= n; i ++ )
        if(a[i].first < p && a[i].second < p - a[i].first)
            mi = min(mi, a[i].second + a[i].first);

    cout << mi;

    return 0;
}

 

四、染色时间

#include<iostream>
#include<set>
using namespace std;
typedef long long ll;
const int N = 610;
int t[N][N];
bool v[N][N];
int dx[5] = {-1, 0, 1, 0};
int dy[5] = {0, 1, 0, -1};
int n, m;

struct Node
{
    int first, second, time;
    bool operator< (const Node h) const
    {
        if(time != h.time) return time < h.time;
        else if(first != h.first) return first < h.first;
        else return second < h.second;
    }
};

void bfs()
{
    set<Node> q;
    q.insert({1, 1, t[1][1]});
    v[1][1] = 1;

    while(q.size())
    {
        auto w = *q.begin();
        q.erase(q.begin());

        for(int i = 0; i < 4; i ++ )
        {
            int x = w.first + dx[i], y = w.second + dy[i], z = w.time;
            if(x < 1 || x > n || y < 1 || y > m) continue;
            if(!v[x][y])
            {
                q.insert({x, y, z + t[x][y]});
                v[x][y] = 1;
            }
        }

        if(q.size() == 1) cout << (*q.begin()).time;
    }
}

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= m; j ++ )
            cin >> t[i][j];
    bfs();

    return 0;
}