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：$3\ast 10\ast 2+2\ast 2+1=65$

#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;
}