2024 蓝桥杯模拟赛 1

P8761 [蓝桥杯 2021 国 BC] 大写

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;


i32 main() {
    string s;
    cin >> s;
    for( auto & i : s ){
        if( i >= 'a' and i <= 'z' ) cout << char( i - 'a' + 'A');
        else cout << i;
    }
    return 0;
}

P8195 [传智杯 #4 决赛] 小智的疑惑

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;


i32 main() {
    string s;
    cin >> s;
    int res = 0;
    for( int i = 0 ; i + 8 <= s.size() ; i ++ ){
        if( s.substr( i , 8 ) == "chuanzhi" ) res ++;
    }
    cout << res << "\n";
    return 0;
}

P8651 [蓝桥杯 2017 省 B] 日期问题

注意去重

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

int t[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

i32 main() {
    int a, b, c;
    cin >> a, cin.ignore();
    cin >> b, cin.ignore();
    cin >> c, cin.ignore();
    int y, m, d;
    set<array<int, 3>> res;

    if (a <= 59) y = a + 2000;
    else y = a + 1900;
    m = b, d = c;
    if ( m >= 1 and m <= 12) {
        if (m != 2) {
            if (d >= 1 and d <= t[m]) res.insert({y, m, d});
        } else {
            if ((y % 4 == 0 and y % 100 != 0) or (y % 400) == 0) {
                if (d >= 1 and d <= 29) res.insert({y, m, d});
            } else if (d >= 1 and d <= 28)res.insert({y, m, d});
        }
    }

    if (c <= 59) y = c + 2000;
    else y = c + 1900;
    m = a, d = b;
    if (m >= 1 and m <= 12) {
        if (m != 2) {
            if (d >= 1 and d <= t[m]) res.insert({y, m, d});
        } else {
            if ((y % 4 == 0 and y % 100 != 0) or (y % 400) == 0) {
                if (d >= 1 and d <= 29) res.insert({y, m, d});
            } else if (d >= 1 and d <= 28)res.insert({y, m, d});
        }
    }
    if (c <= 59) y = c + 2000;
    else y = c + 1900;
    m = b, d = a;
    if (m >= 1 and m <= 12) {
        if (m != 2) {
            if (d >= 1 and d <= t[m]) res.insert({y, m, d});
        } else {
            if ((y % 4 == 0 and y % 100 != 0) or (y % 400) == 0) {
                if (d >= 1 and d <= 29) res.insert({y, m, d});
            } else if (d >= 1 and d <= 28)res.insert({y, m, d});
        }
    }
    for (auto [x, y, z]: res)
        printf("%d-%02d-%02d\n", x, y, z);
    return 0;
}

P9240 [蓝桥杯 2023 省 B] 冶炼金属

两次二分

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

i32 main() {
    int n;
    cin >> n;
    vector<pii> a(n);
    for (auto &[x, y]: a) cin >> x >> y;
    int l = 1, r = inf, res = -1;
    for (int mid, flag; l <= r;) {
        mid = (l + r) / 2, flag = 1;
        for (const auto &[x, y]: a) {
            if (x / mid <= y) continue;
            flag = 0;
            break;
        }
        if (flag) res = mid, r = mid - 1;
        else l = mid + 1;
    }
    cout << res << " ";
    l = 1, r = inf, res = -1;
    for (int mid, flag; l <= r;) {
        mid = (l + r) / 2, flag = 1;
        for (const auto &[x, y]: a) {
            if (x / mid >= y) continue;
            flag = 0;
            break;
        }
        if (flag) res = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout << res << "\n";
    return 0;
}

P8627 [蓝桥杯 2015 省 A] 饮料换购

模拟

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    int n, res = 0;
    cin >> n;
    for (int a = n, b = 0; a > 0 or b >= 3; a = b / 3, b %= 3)
        res += a, b += a, a = 0;
        cout << res << "\n";
    return 0;
}

P8647 [蓝桥杯 2017 省 AB] 分巧克力

二分

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    int n, m;
    cin >> n >> m;
    vector<pii> a(n);
    for (auto &[x, y]: a) cin >> x >> y;
    int l = 1, r = 1e5, res = -1;
    for (int mid, cnt; l <= r;) {
        mid = (l + r) / 2, cnt = 0;
        for (const auto &[x, y]: a)
            cnt += (x / mid) * (y / mid);
        if (cnt >= m) res = mid, l = mid + 1;
        else r = mid - 1;
    }
    cout << res << "\n";

    return 0;
}

P8637 [蓝桥杯 2016 省 B] 交换瓶子

直接从小到大，每次判断瓶子在不在位置上，如果不在就进行一次交换

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++)
        cin >> a[i], a[i]--, b[a[i]] = i;
    int res = 0;
    for (int i = 0, x, y, p, q; i < n; i++) {
        if (b[i] == i) continue;
        res++;
        x = i, y = a[i], p = b[x], q = b[y];
        swap(a[p], a[q]), swap(b[x], b[y]);
    }
    cout << res << "\n";
    return 0;
}

P8625 [蓝桥杯 2015 省 B] 生命之树

其实就是求树上的最大连通块，所以直接进行树形dp，枚举每个叶子选不选即可

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    int n, res = 0;
    cin >> n;
    vector<int> f(n);
    for (auto &i: f) cin >> i;
    vector<vi> e(n);
    for (int i = 1, u, v; i < n; i++)
        cin >> u >> v, --u, --v, e[u].push_back(v), e[v].push_back(u);
    auto dfs = [&res, &f, e](auto &&self, int x, int fa) -> void {
        for (const auto &y: e[x]) {
            if (y == fa)continue;
            self(self, y, x);
            f[x] += max(0ll, f[y]);
        }
        res = max(res, f[x]);
    };

    dfs(dfs, 0, -1);
    cout << res << "\n";

    return 0;
}

P8638 [蓝桥杯 2016 省 A] 密码脱落

把原序列反序之后求一下lcs就行

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    string a, b;
    cin >> a, b = a, reverse(b.begin(), b.end());
    int n = a.size();
    a = " " + a, b = " " + b;
    auto f = vector(n + 1, vi(n + 1));
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            if (a[i] == b[j]) f[i][j] = f[i - 1][j - 1] + 1;
            else f[i][j] = max(f[i - 1][j], f[i][j - 1]);
    cout << n - f[n][n] << "\n";

    return 0;
}

P8774 [蓝桥杯 2022 省 A] 爬树的甲壳虫

几何概型，求出每一步所需要爬的次数即可

#include<bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;

const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}

i32 main() {
    int n, res = 0;
    cin >> n;
    for (int x, y; n; n--)
        cin >> x >> y, res = (res + 1) * y % mod * inv(y - x) % mod;
    cout << res << "\n";
    return 0;
}