第十三届蓝桥杯 C/C++ 大学B组 题解
Posted 万伏小太阳
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了第十三届蓝桥杯 C/C++ 大学B组 题解相关的知识,希望对你有一定的参考价值。
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;
2022年4月第十三届蓝桥杯C/C++程序设计A组(省赛)试题及题解
目录
试题A:裁纸刀
答案为 n ∗ m − 1 + 4 n*m-1+4 n∗m−1+4
443
试题B:灭鼠先锋
LLLV
试题C:求和
预计得分100%
思路:维护一个前缀 s u m sum sum即可。
总时间复杂度 O ( n ) O(n) O(n)
参考代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a[N];
void solve()
int n;
scanf("%d", &n);
long long ans = 0, sum = 0;
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
ans += sum * a[i];
sum += a[i];
cout << ans << endl;
int main()
solve();
return 0;
试题 D: 选数异或
预计得分100%
对于每个位置 i i i ,设 y = a [ i ] y = a[i] y=a[i] ^ x x x,找到最近的一个 y y y的下标 i d x idx idx,记作 b [ i ] b[i] b[i]。再用线段树维护 b b b数组的区间最大值 m a x x maxx maxx,如果 m a x x > = L maxx >= L maxx>=L,那么为 y e s yes yes。
总时间复杂度 O ( n l o g n ) O(nlogn) O(nlogn)
参考代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int a[N], b[N];
struct node
int l, r, val, maxx;
tr[N * 4];
void pushup(int k) tr[k].maxx = max(tr[k * 2].maxx, tr[k * 2 + 1].maxx);
void build(int k, int l, int r)
tr[k].l = l;
tr[k].r = r;
if (tr[k].l == tr[k].r)
tr[k].val = tr[k].maxx = b[l];
return;
int mid = l + r >> 1;
build(k * 2, l, mid);
build(k * 2 + 1, mid + 1, r);
pushup(k);
int query(int k, int l, int r)
if (tr[k].l == l && tr[k].r == r)
return tr[k].maxx;
int mid = tr[k].l + tr[k].r >> 1;
if (r <= mid)
return query(k * 2, l, r);
else if (l > mid)
return query(k * 2 + 1, l, r);
else
return max(query(k * 2, l, mid), query(k * 2 + 1, mid + 1, r));
void solve()
int n, m, x;
scanf("%d %d %d", &n, &m, &x);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
map<int, int> last;
for (int i = 1; i <= n; i++)
int y = a[i] ^ x;
if (!last.count(y))
b[i] = -1;
else
b[i] = last[y];
last[a[i]] = i;
build(1, 1, n);
while (m--)
int l, r;
scanf("%d %d", &l, &r);
int maxx_pos = query(1, l, r);
puts(maxx_pos >= l ? "yes" : "no");
int main()
solve();
return 0;
试题 E: 爬树的甲壳虫
暂时不会。
试题 F: 青蛙过河
2022.4.25 UPDATE:少了一个语句,已修正
预计得分100%
二分经典题改编题,一眼二分,check有点难写。
check思路:
可以令
d
p
[
i
]
dp[i]
dp[i]表示跳到第
i
i
i个石头的最大次数。
把最后
m
i
d
mid
mid个石头的
d
p
[
i
]
dp[i]
dp[i]加起来,
s
u
m
>
=
2
∗
x
sum>= 2*x
sum>=2∗x即合法。
总时间复杂度 O ( n l o g n ) O(nlogn) O(nlogn)
参考代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int a[N];
long long dp[N];
// dp[i]表示最多能跳到位置i dp[i]次
int n, x;
bool check(int mid)
long long sum = 0;
for (int i = 1; i <= mid; i++) //前mid个都可以
dp[i] = a[i];
int l = 1;
for (int i = mid + 1; i <= n; i++)
while (i - l > mid)//控制跳跃距离。
++l;
dp[i] = 0;
while (dp[i] < a[i] && l < i)
if (dp[l] + dp[i] <= a[i])
dp[i] += dp[l];
dp[l] = 0;
++l;
else
dp[l] -= a[i] - dp[i];
dp[i] = a[i];
long long ans = 0;
int L = n - mid + 1;
for (int i = L; i <= n; i++)
ans += dp[i];
return ans >= 2 * x;
void solve()
scanf("%d %d", &n, &x);
--n;
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int l = 1, r = n, ans = n + 1;
while (l <= r)
int mid = l + r >> 1;
if (check(mid))
ans = mid;
r = mid - 1;
else
l = mid + 1;
printf("%d\\n", ans);
int main()
solve();
return 0;
试题 G: 最长不下降子序列
预计得分10% ~ 30%
暴力思路:
枚举修改位置,二分法 O ( n l o g n ) O(nlogn) O(nlogn)求最长不下降子序列,(跳过中间那段长度为 k k k的数组)
总时间复杂度 O ( n 2 l o g n ) O(n^2logn) O(n2logn)
参考代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int a[N], dp[N];
int n, k;
int LIS(int L, int R) //二分求LIS
int len = 0;
dp[len] = -0x3f3f3f3f;
for (int i = 1; i <= n; i++)
if (i == L) //跳过k个
i = R;
continue;
if (a[i] >= dp[len])
++len;
dp[len] = a[i];
int p = upper_bound(dp + 1, dp + 1 + len, a[i]) - dp;
dp[p] = a[i];
return len;
void solve()
scanf("%d %d", &n, &k);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
if (n - 1 <= k)
//修改k个必然能全部满足
printf("%d\\n", n);
return;
int ans = k + 1;
for (int i = 1; i + k - 1 <= n; i++)
ans = max(ans, k + LIS(i, i + k - 1));
printf("%d\\n", ans);
int main()
solve();
return 0;
试题 H: 扫描游戏
预计得分10% ~ 30%
暴力思路:
极角排序。之后每次暴力扫描,循环直到扫描不到物品。
总时间复杂度 O ( n 2 ) O(n^2) O(n2)
参考代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int INF = 0x3f3f3f3f;
struct node
long long x, y, z, id;
double theta;
bool operator<(const node &tmp) const return theta < tmp.theta;
;
vector<node> vec;
int out[N];
node tmp;
void solve()
int n;
long long L;
scanf("%d %lld", &n, &L);
for (int i = 1; i <= n; i++)
out[i] = -1;
tmp.id = i;
scanf("%lld %lld %lld", &tmp.x, &tmp.y, 2022 第十三届蓝桥杯大赛软件赛省赛(第二场),C/C++ 大学B组题解
第十三届蓝桥杯大赛软件赛省赛 C/C++ 大学 B 组思考+总结
2022年4月第十三届蓝桥杯C/C++程序设计A组(省赛)试题及题解