第十三届蓝桥杯 C/C++ 大学B组 题解

第十三届蓝桥杯 C/C++ 大学B组 题解

A

进制计算简单模拟

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
signed main() {
    int a[] = {2, 0, 2, 2};
    int sum = 0;
    for (int i = 0; i < 4; i++) {
        sum = sum * 9 + a[i];
    }
    cout << sum << endl;
    return 0;
}
// 答案1478

B

遍历2022的每一天，转成字符串拼接，然后判断

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;

bool check(string s) {
    for (int i = 1; i < s.size() - 1; i++) {
        if (s[i] - s[i - 1] == 1 && s[i + 1] - s[i - 1] == 1) return true;
    }
    return false;
}

signed main() {
    int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int ans = 0;
    for (int i = 1; i <= 12; i++) {
        for (int j = 1; j <= month[i]; j++) {
            string y = "2022";
            if (i < 10)
                y = y + "0" + to_string(i);
            else {
                y = y + to_string(i);
            }
            if (j < 10)
                y = y + "0" + to_string(j);
            else {
                y = y + to_string(j);
            }
            if (check(y)) ans += 1;
            // cout << y << endl;
        }
    }
    cout << ans << endl;
    return 0;
}

C

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
signed main() {
    ll a, b, n;
    cin >> a >> b >> n;
    ll week = n / (a * 5 + b * 2);
    ll dy = n % (a * 5 + b * 2);
    ll ans = week * 7;
    if (dy <= 5 * a) {
        ans = ans + (dy + a - 1) / (a);
    } else {
        ans += 5;
        dy -= 5 * a;
        ans = ans + (dy + b - 1) / b;
    }
    cout << ans << endl;
    return 0;
}

D

找规律

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
// 暴力找出规律
vector<int> solve(int n) {
    // int n;
    // cin >> n;
    vector<int> a(n + 1, 0), ans(n + 1, 0);
    int op = 1;
    while (op <= 100) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                a[j] += 1;
            }
            // cout << "第" << op << "天傍晚"
            //      << "\n";
            for (int j = 1; j <= n; j++) {
                // cout << a[j] << " ";
                ans[j] = max(ans[j], a[j]);
            }
            // cout << "\n";
            a[i] = 0;
            op += 1;
        }
        for (int i = n - 1; i > 1; i--) {
            for (int j = 1; j <= n; j++) {
                a[j] += 1;
            }
            // cout << "第" << op << "天傍晚"
            //      << "\n";
            for (int j = 1; j <= n; j++) {
                // cout << a[j] << " ";
                ans[j] = max(ans[j], a[j]);
            }
            a[i] = 0;
            // cout << "\n";
            op += 1;
        }
    }
    // cout << n << endl;
    // for (int i = 1; i <= n; i++) {
    //     cout << ans[i] << " ";
    // }
    return ans;
}

void work() {
    int n;
    cin >> n;
    if (n == 1) {
        cout << 1 << endl;
        return;
    }
    vector<int> vec(n + 1, 0);
    int num = (n - 1) * 2;
    for (int i = 1; i <= n / 2; i++) {
        vec[i] = vec[n - i + 1] = num;
        num -= 2;
    }
    if (n % 2 == 1) {
        vec[n / 2 + 1] = n - 1;
    }
    for (int i = 1; i <= n; i++) {
        cout << vec[i] << " ";
    }
    cout << endl;
}

signed main() {
    // for (int i = 1; i <= 100; i++) {
    //     solve(i);
    //     cout << endl;
    // }
    work();
    return 0;
}
/*
0 0 0 第一天早上
1 1 1 第一天晚上
0 1 1 第二天早上
1 2 2 第二天晚上
1 0 2 第三天早上
2 1 3 第三天晚上
2 1 0 第四天早上
3 2 1 第四天晚上
3 0 1 第五天早上
4 1 2 第五天晚上


*/

E

主要就是看懂题意和取模的问题。

321 对应八进制、十进制、二进制

计算过程为： 3 ∗ 10 ∗ 2 + 2 ∗ 2 + 1 = 65 3*10*2+2*2+1 = 65 3∗10∗2+2∗2+1=65

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int long long
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
int X, n, m;
int a[N], b[N];

int get_bit(int a, int b, int c) {
    return (a > b ? a : b) > c ? (a > b ? a : b) : c;
}

signed main() {
    cin >> X;
    cin >> n;
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    for (int i = n; i >= 1; i--) cin >> a[i];

    cin >> m;
    for (int i = m; i >= 1; i--) cin >> b[i];

    ll ans = 0;

    for (int i = n; i > 1; i--) {
        ans = ((ans + a[i] - b[i]) * get_bit(a[i - 1] + 1, b[i - 1] + 1, 2)) % mod;
    }

    ans += a[1] - b[1];

    cout << ans << endl;
    return 0;
}
/*
321 65
3*10*2+2*2+1 = 65
*/

F

通过枚举上下边界，和前缀和，就转成了一维数组求子段和小于等于k的问题

经典双指针解决

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const int mod = 1e9 + 7;

ll a[510][510];

signed main() {
    ll n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j];
            a[i][j] += a[i - 1][j];
        }
    }

    ll ans = 0;

    // 通过枚举上下边界，和前缀和，就转成了一维数组求子段和小于等于k的问题 
    // 经典双指针解决

    for (int i = 1; i <= n; i++) { //枚举上边届
        for (int j = i; j <= n; j++) { // 枚举下边界
            int l = 1;
            ll sum = 0;
            for (int r = 1; r <= m; r++) {
                sum += a[j][r] - a[i - 1][r];
                while (sum > k) {
                    sum -= a[j][l] - a[i - 1][l];
                    l++;
                }
                ans += r - l + 1;
            }
        }
    }

    cout << ans << endl;

    return 0;
}

G

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define int long long
const int N = 1e5 + 7;
const int mod = 1e9 + 7;
signed main() {
    int n;
    cin >> n;
    vector<int> dp(n + 4, 0);
    dp[1] = 1;
    dp[2] = 2;
    dp[3] = 5;
    for (int i = 4; i <= n; i++) {
        dp[i] = (dp[i - 1] * 2 % mod + dp[i - 3] % mod) % mod;
    }
    cout << dp[n] << endl;

    return 0;
}