C++

$\color{gold}{— > 蓝桥杯辅导课题解}$

思路：

$根据题意，不用显示出年月日，只需要显示出时分秒即可，毫秒也不用显示，直接舍去$

#include <iostream>
#include <climits>  // LONG_LONG_MAX头文件

using namespace std;

typedef long long ll; // 由于最大为10的18次方，用long long 来存
//关于long long 
// cout << LONG_LONG_MAX << endl; // long long 的最大值为：922 3372 0368 5477 5807 （10的19次方）

ll n;

int main () {
    cin >> n;
    n /= 1000; // 从毫秒转换成秒数

    n %= 24 * 60 * 60; // 把经过了多少天的秒数去掉，剩下的就是一天之内的秒数
    printf ("%02d:", n / 60 / 60); // 输出小时数
    n %= 60 * 60;
    printf ("%02d:", n / 60); // 输出分钟数
    n %= 60;
    printf ("%02d", n); // 输出秒数

    return 0;
}

蓝桥杯12届（2021）C++ B组 填空题：

A.空间

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

int main() {
    cout << 256 * 1024 * 1024 / 4;
    return 0;
}

答案：67108864

B.卡片
法一：

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

int s[15];

bool check(int x) {
    while (x) {
        int t = x % 10;
        x /= 10;
        if (-- s[t] < 0) return false;
    }
    return true;
}

int main() {
    for (int i = 0; i < 10; i ++) s[i] = 2021;

    for (int i = 1; ; i ++ ) {
        if (!check(i)) {
            cout << i - 1;
            return 0;
        }
    }
    return 0;
}

法二：

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

int s[15];

int main() {
    for (int i = 0; i < 10; i ++) s[i] = 2021;

    int i;
    for (i = 1; ; i ++ ) {
        int j = i;
        while (j) {
            if (s[j % 10] > 0) s[j % 10] --;
            else break;
            j /= 10;
        }
        if (j) break;
    }
    cout << i - 1;
    return 0;
}

答案：3181

C.直线
法一：

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

const int N = 200000;

int n;
struct Line {
    double k, b;
    bool operator< (const Line& t) const {
        if (k != t.k) return k < t.k;
        return b < t.b;
    }
}l[N];

int main()
{
    for (int x1 = 0; x1 < 20; x1 ++ )
        for (int y1 = 0; y1 < 21; y1 ++ )
            for (int x2 = 0; x2 < 20; x2 ++ )
                for (int y2 = 0; y2 < 21; y2 ++ )
                    if (x1 != x2) {
                        double k = (double)(y2 - y1) / (x2 - x1);
                        double b = y1 - k * x1;
                        l[n ++ ] = {k, b};
                    }

    sort(l, l + n);
    int res = 1;
    for (int i = 1; i < n; i ++ )
        if (fabs(l[i].k - l[i - 1].k) > 1e-8 || fabs(l[i].b - l[i - 1].b) > 1e-8)
            res ++ ;
    cout << res + 20 << endl;

    return 0;
}

法二：

#include<bits/stdc++.h>
using namespace std;
struct Point {
    double x, y;
}p[25 * 25];
map<pair<double, double>, int> line;

int main() {
    int cnt = 0;
    int row = 21, col = 20; // row :行 col : 列 
    for (int i = 0; i < row; i ++) 
        for (int j = 0; j < col; j ++)
            p[cnt].x = i, p[cnt ++].y = j;

    int linenum = row + col; // 定义直线条数    
    for (int i = 0; i < cnt; i ++) {
        for (int j = 0; j < cnt; j++) {
            if (p[i].x == p[j].x || p[i].y == p[j].y)
            continue;

            double k = (p[j].y - p[i].y) / (p[j].x - p[i].x);
            double b = (p[i].x * p[j].y - p[j].x * p[i].y) / (p[i].x - p[j].x);

            if (line[{k, b}] == 0) {
                line[{k, b}] = 1;
                linenum ++;
            }
        }
    }     

    cout << linenum;
    return 0;
}

答案：40257

D.货物摆放
法一：

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const int N = 1e5 + 10;

ll n = 2021041820210418;
ll factor[N];
ll len, ans;
int main() {
    for (ll i = 1; i * i <= n; i ++) {
        if (n % i == 0) {
            factor[len ++] = i;
            if (i != n / i) factor[len ++] = n / i;
        }
    }

    for (int a = 0; a < len; a ++) {
        for (int b = 0; b < len; b ++) {
            if (factor[a] * factor[b] > n) continue; 
            for (int c = 0; c < len; c ++) {
                if (factor[a] * factor[b] * factor[c] == n) 
                ans ++;
            }
        }
    }
    cout << ans;
    return 0;
}

法二：

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

typedef long long ll;

int main()
{
    ll n = 2021041820210418;
    vector<ll> d;
    for (ll i = 1; i * i <= n; i ++ )
        if (n % i == 0)
        {
            d.push_back(i);
            if (n / i != i) d.push_back(n / i);
        }

    int res = 0;
    for (auto a: d)
        for (auto b: d)
            for (auto c: d)
                if (a * b * c == n)
                    res ++ ;
    cout << res << endl;

    return 0;
}

答案：2430

E.路径

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

const int N = 2200, M = N * 50;

int n;
int h[N], e[M], w[M], ne[M], idx;
int q[N], dist[N];
bool st[N];

int gcd(int a, int b)  // 欧几里得算法
{
    return b ? gcd(b, a % b) : a;
}

void add(int a, int b, int c)  // 添加一条边a->b，边权为c
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

void spfa() { // 求1号点到n号点的最短路距离
    int hh = 0, tt = 0;
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    q[tt ++ ] = 1;
    st[1] = true;

    while (hh != tt) {
        int t = q[hh ++ ];
        if (hh == N) hh = 0;
        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            if (dist[j] > dist[t] + w[i]) {
                dist[j] = dist[t] + w[i];
                if (!st[j]) {    // 如果队列中已存在j，则不需要将j重复插入
                    q[tt ++ ] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
}

int main() {
    n = 2021;
    memset(h, -1, sizeof h);
    for (int i = 1; i <= n; i ++ )
        for (int j = max(1, i - 21); j <= min(n, i + 21); j ++ ) {
            int d = gcd(i, j);
            add(i, j, i * j / d);
        }

    spfa();
    printf("%d\n", dist[n]);
    return 0;
}

答案：10266837