蓝桥杯-02-2023蓝桥杯c/c++省赛B组题目
蓝桥杯-02-2023蓝桥杯c/c++省赛B组题目
用户2225445
发布于 2024-03-21 11:05:40
发布于 2024-03-21 11:05:40
参考
2023 年第十四届蓝桥杯 C/C++ B组省赛题解 2023蓝桥杯c/c++省赛B组题目(最全版)：
A：日期统计
这题方法应该很多，没有和别人讨论想法。我的解法思路是：先 load 函数生成所有这一年的合法日期，然后枚举所有可以从数据中拿到的八位日期，检查它是否在合法日期的集合中。
这里枚举到四位，即年份的时候要判断一下，不然可能数据太多要跑很久。
注意要去重！不知道有多少冤大头痛失这 5 分。（原题里特意加粗了不重复）
// 日期统计
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 1e5+5;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9+7;
// 235
set<int> S; // 记录一年中合法的日期
set<int> M; // 记录出现的日期，自带去重
// 生成数据
int mon[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
void load() {
for (int i = 1; i <= 12; i++) {
for (int j = 1; j <= mon[i-1]; j++) {
S.insert(2023 * 10000 + i * 100 + j);
}
}
}
void solve() {
const int n = 100;
int a[256];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < n; i++) {
cout << a[i] << ' ';
}
cout << endl;
for (int _1 = 0; _1 < n; _1++) {
for (int _2 = _1 + 1; _2 < n; _2++) {
for (int _3 = _2 + 1; _3 < n; _3++) {
for (int _4 = _3 + 1; _4 < n; _4++) {
int x = 0;
x = x * 10 + a[_1];
x = x * 10 + a[_2];
x = x * 10 + a[_3];
x = x * 10 + a[_4];
if (x != 2023) continue; // 这里先判断一下合法，否则程序可能跑不动
for (int _5 = _4 + 1; _5 < n; _5++) {
for (int _6 = _5 + 1; _6 < n; _6++) {
for (int _7 = _6 + 1; _7 < n; _7++) {
for (int _8 = _7 + 1; _8 < n; _8++) {
int x = 0;
x = x * 10 + a[_1];
x = x * 10 + a[_2];
x = x * 10 + a[_3];
x = x * 10 + a[_4];
x = x * 10 + a[_5];
x = x * 10 + a[_6];
x = x * 10 + a[_7];
x = x * 10 + a[_8];
// 如果这个日期是这年的合法日期
if (S.count(x)) {
M.insert(x);
}
}
}
}
}
}
}
}
}
cout << M.size() << endl;
}
int32_t main() {
// 该文件是复制下来的题目给的数据
freopen("A.txt", "r", stdin);
load();
cout << S.size() << endl;
solve();
return 0;
}B: 01 串的熵
用Excel做比较方便，让我看看有谁？哈哈哈哈哈
答案当然就是 11027421了！！！！！
C: 冶炼金属
// 冶炼金属
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 1e5 + 5;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9 + 7;
void solve() {
int n;
cin >> n;
int min_ = -INF - 1, max_ = INF;
for (int i = 0; i < n; i++) {
int x, y;
cin >> x >> y;
min_ = max(min_, x / (y + 1) + 1); // 求左边界的并
max_ = min(max_, x / y); // 求右边界的并
}
cout << min_ << " " << max_ << endl;
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
solve();
return 0;
}#include <iostream>
using namespace std;
#include<cstdio>
int main(){
long long n,a,b,t,t1;
long long max=0,min=9999999999;
cin >> n;
for(int i = 0;i <n;i++){
cin >> a >> b;
t=a/b;
if(t <= min){
min = t;
}
t1 = a/(b+1);
if(t1 >= max){
max = t1;
}
}
printf("%lld %lld",max+1,min);
return 0;
}
D: 飞机降落
// 飞机降落
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 15;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9+7;
struct Air {
// 最早降落时间, 最迟降落时间, 降落过程耗时
int st, ed, len;
} air[MAX];
int n;
bool vis[MAX];
bool dfs(int x, int pos) {
if (x >= n) return true;
for (int i = 0; i < n; i++) {
// 如果 i 没有用过，且 i 最迟降落时间在前者的完成这个降落流程之后
if (not vis[i] and pos <= air[i].ed) {
vis[i] = true;
// 在条件允许的情况下后者尽早降落为后者提供，要两类讨论，这里算有一点点贪心
bool tmp = dfs(x+1, max(air[i].st, pos) + air[i].len);
vis[i] = false;
if (tmp) return true;
}
}
return false;
}
void solve() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> air[i].st >> air[i].ed >> air[i].len;
air[i].ed += air[i].st;
}
bool res = dfs(0, 0);
if (res) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int t;
cin >> t;
while (t--)
solve();
return 0;
}E: 接龙数列
// 接龙序列
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
int res[10] = {0};
void solve() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
int tail = x % 10; // 计算末尾
int head;
while (x) {
head = x % 10;
x /= 10;
} // 计算首位
res[tail] = max(res[tail], res[head] + 1); // 状态转移方程
}
int max_ = 0;
for (int i = 0; i < 10; i++) {
max_ = max(max_, res[i]); // 获取最长接龙序列
}
cout << n - max_ << endl;
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
solve();
return 0;
}F: 岛屿个数
// 岛屿个数
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 55;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9 + 7;
char map_[MAX][MAX]; // 地图
bool track[MAX][MAX] = {false}; // 访问标记
bool outsea[MAX][MAX] = {false}; // 外海标记
struct XY {
int x, y;
} nxt[] = {
{1, 0},
{-1, 0},
{0, 1},
{0, -1},
{1, 1},
{-1, 1},
{1, -1},
{-1, -1},
};
void solve() {
int n, m;
cin >> n >> m;
for (int i = 0; i < n; i++) {
cin >> map_[i];
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
track[i][j] = false;
outsea[i][j] = false;
}
}
// 预处理外海
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (map_[i][j] == '1' or track[i][j]) continue;
bool yep = false;
queue<XY> que, arr;
track[i][j] = true;
que.push({i, j});
arr.push({i, j}); // 记录搜索到的所有海域
while (not que.empty()) {
XY pos = que.front();
que.pop();
// 注意：海域搜索可以往八个方向走，陆地搜索只能朝四个方向
for (int i = 0; i < 8; i++) {
XY nw = {pos.x + nxt[i].x, pos.y + nxt[i].y};
if (0 <= nw.x and nw.x < n and 0 <= nw.y and nw.y < m) {
if (map_[nw.x][nw.y] == '0' and track[nw.x][nw.y] == false) {
track[nw.x][nw.y] = true;
que.push(nw);
arr.push(nw);
}
} else {
yep = true; // 如果有一次搜索到地图外，标记此次搜索的所有区域为外海
}
}
}
if (yep) {
while (not arr.empty()) {
XY pos = arr.front();
arr.pop();
outsea[pos.x][pos.y] = true; // 标记搜索到的海域为外海
}
}
}
}
// 别忘了清空访问标记
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
track[i][j] = false;
}
}
// 处理岛屿
int cnt = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (map_[i][j] == '0' or track[i][j]) continue;
bool yep = false;
queue<XY> que;
track[i][j] = true;
que.push({i, j});
while (not que.empty()) {
XY pos = que.front();
que.pop();
for (int i = 0; i < 4; i++) {
XY nw = {pos.x + nxt[i].x, pos.y + nxt[i].y};
if (0 <= nw.x and nw.x < n and 0 <= nw.y and nw.y < m) {
if (map_[nw.x][nw.y] == '1') {
if (track[nw.x][nw.y] == false) {
track[nw.x][nw.y] = true;
que.push(nw);
}
} else {
if (outsea[nw.x][nw.y]) {
yep = true; // 搜索到外海为外岛
}
}
} else {
yep = true; // 搜索到边界外为外岛
}
}
}
if (yep) {
cnt++; // 外岛计数
}
}
}
cout << cnt << endl;
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
int t;
cin >> t;
while (t--)
solve();
return 0;
}G: 子串简写
// 子串简写
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
#ifdef _DEBUG
#pragma GCC optimize(2)
#endif
using namespace std;
const int MAX = 5e5 + 5;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9 + 7;
int aa[MAX], cnt_aa = 0;
int bb[MAX], cnt_bb = 0;
// binary_search 二分查找
int bs(int x) {
int l = 0, r = cnt_bb - 1;
while (l <= r) {
int mid = l + r >> 1;
if (bb[mid] < x) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return l;
}
void solve() {
int k;
cin >> k;
string s;
cin >> s;
int n = s.size();
char a, b;
cin >> a >> b;
for (int i = 0; i < n; i++) {
if (s[i] == a) {
aa[cnt_aa++] = i; // 转储 c1 位置序列
}
if (s[i] == b) {
bb[cnt_bb++] = i; // 转储 c2 位置序列
}
}
int res = 0;
for (int i = 0; i < cnt_aa; i++) {
res += cnt_bb - bs(aa[i] + k - 1); // 记录有效贡献
}
cout << res << endl;
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
solve();
return 0;
}#include<iostream>
using namespace std;
int main(){
string s;
string a,b;
int n;
int ans = 0;
cin >> n;
cin >> s >> a >> b;
for(int i = 0;i < s.length();i++){
for(int j = i+3;j<s.length();j++){
if(s[i]==a[0]){
if(s[j] == b[0]){
ans++;
}
}
}
}
cout << ans;
return 0;
}输出为：
仅供参考，个人看法。反正结果正确，理解万岁，大家有什么更好的想法，欢迎评论区交流。。评论区见。
H: 整数删除
// 整数删除
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
#ifdef _DEBUG
// #pragma GCC optimize(2)
#endif
using namespace std;
const int MAX = 5e5 + 5;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9 + 7;
int a[MAX];
int left_[MAX], right_[MAX]; // 仿链表结构
using pii = pair<int, int>;
// 小顶堆, 手写堆方法与 prority_queue 相仿，可跳至 第121行 void solve();
namespace heap {
pii heap[MAX << 1]; // 堆 (树状数组)
int a2h[MAX << 1]; // 数组索引 -> 堆索引
int h2a[MAX << 1]; // 堆索引 -> 数组索引
int total = 0; // 总节点 (含已 pop 节点)
int siz = 0; // 堆大小
// C 风格比较函数
bool cmp(pii x, pii y) { return x > y; }
// 维护映射的交换
void swap_(int u, int v) {
swap(a2h[h2a[u]], a2h[h2a[v]]);
swap(h2a[u], a2h[v]);
swap(heap[u], heap[v]);
if (u == siz) {
a2h[h2a[u]] = -1;
h2a[u] = -1;
}
if (v == siz) {
a2h[h2a[v]] = -1;
h2a[v] = -1;
}
}
// 向上维护
void up(int node) {
while (node and cmp(heap[(node - 1) / 2], heap[node])) {
swap_(node, (node - 1) / 2);
node = (node - 1) / 2;
}
}
// 向下维护
void down(int node) {
int parent = node; // 父节点下标
int child = 2 * node + 1; // 子节点下标
while (child < siz) {
// 判断子节点哪个大, 大的与父节点比较
if (child + 1 < siz and cmp(heap[child], heap[child + 1]))
child++;
if (cmp(heap[parent], heap[child])) // 判断父节点是否小于子节点
{
swap_(parent, child); // 交换父节点和子节点
parent = child; // 子节点下标 赋给 父节点下标
}
child = child * 2 + 1; // 换行, 比较下面的父节点和子节点
}
}
// 线性初始化堆
void build(int n) {
total = siz = n;
for (int i = 0; i < siz; i++) {
a2h[i] = i;
h2a[i] = i;
}
for (int i = siz / 2 - 1; i >= 0; i--)
// 必须自下而上
down(i);
}
// 插入节点
void push(pii x) {
heap[siz] = x;
a2h[total] = siz;
h2a[siz] = total;
up(siz);
total++;
siz++;
}
// 返回顶值
pii top() { return heap[0]; }
// 删除顶点
pii pop() {
siz--;
swap_(0, siz);
down(0);
return heap[siz];
}
// 修改第 u 个插入的节点
void modify_by_address(int u, pii x) {
int v = a2h[u];
heap[v] = x;
down(v);
up(v);
}
// 删除第 u 个插入的节点. 返回: 布尔型: 节点在堆中
void pop_by_address(int u) {
int v = a2h[u];
siz--;
swap_(v, siz);
down(v);
up(v);
}
} // namespace heap
void solve() {
int n, k;
cin >> n >> k;
for (int i = 0; i < n; i++)
cin >> a[i];
for (int i = 0; i < n; i++) {
if (i > 0)
left_[i] = i - 1; // 初始 i 的前驱 i - 1
else
left_[i] = -1; // 没有前驱的特殊标记
if (i < n - 1)
right_[i] = i + 1; // 初始 i 的后继 i + 1
else
right_[i] = -1;
heap::push({a[i], i}); // 放入堆中
}
int cnt = 0;
while (cnt < k) {
pii p = heap::pop(); // 取出最小的数
int idx = p.second, x = p.first; // idx 为数的索引，x 为这个数
if (a[idx] != x) continue; // 过滤脏数据
// 如果有前驱
if (left_[idx] >= 0) {
a[left_[idx]] += x; // 贡献给前驱
heap::push({a[left_[idx]], left_[idx]}); // 加入修改后数据，原数据变为脏数据，延迟过滤
right_[left_[idx]] = right_[idx]; // 更新前驱的后继
}
// 如果有后继
if (right_[idx] >= 0) {
a[right_[idx]] += x; // 贡献给后继
heap::push({a[right_[idx]], right_[idx]}); // 加入修改后数据，原数据变为脏数据，延迟过滤
left_[right_[idx]] = left_[idx]; // 更新后继的前驱
}
// 标记为无效
a[idx] = -1;
cnt++;
}
// 输出
for (int i = 0; i < n; i++)
if (a[i] >= 0)
cout << a[i] << ' ';
cout << endl;
}
int32_t main() {
#ifdef _DEBUG
// freopen("H.in", "r", stdin);
// freopen("H.out", "w", stdout);
// clock_t st = clock();
#endif
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
solve();
#ifdef _DEBUG
// cout << "COST: " << clock() - st << endl;
#endif
return 0;
}#include <iostream>
#include<cstdio>
using namespace std;
#define MAX 1000000
struct shuzu{
long a[MAX] = {0};
long len = 0;
}s;
long sort(shuzu &s){
long min = s.a[0];
long i= 0;
long biao = 0;
for(i = 1;i<s.len;i++){
if(s.a[i]<min){
min = s.a[i];
biao = i;
}
}
return biao;
}
void del(shuzu &s,int x){
long i;
for(i = x;i<s.len;i++){
s.a[i] = s.a[i+1];
}
s.len = s.len - 1;
}
int main(){
long n,k;
cin >> n >> k;
s.len = n;
for(int i = 0;i<n;i++){
cin >> s.a[i];
}
for(int i = 0;i < k;i++){
long x = sort(s);
long shu = s.a[x];
if(x == 0){
s.a[x+1] += shu;
}
else if(x == s.len){
s.a[x-1] += shu;
}
else{
s.a[x+1] += shu;
s.a[x-1] += shu;
}
del(s,x);
}
for(int i = 0;i<s.len;i++){
printf("%d ",s.a[i]);
}
return 0;
}示例结果也正确。
I: 景区导游
// 景区导游
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 1e5+5;
const int LOG = 20;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9+7;
// 建议直接去写板子，别看了
int lg2[MAX];
int depth[MAX];
int st[MAX][LOG];
int dis[MAX];
vector<pair<int, int>> edge[MAX];
void load_lg2() {
lg2[0] = -1;
for (int i = 1; i < MAX; i++) {
lg2[i] = lg2[i / 2] + 1;
}
}
void build(int x, int pre, int level) {
depth[x] = level;
st[x][0] = pre;
for (int i = 1; i <= lg2[depth[x]]; i++) {
st[x][i] = st[st[x][i-1]][i-1];
}
for (pair<int, int> p : edge[x]) {
int y = p.first, d = p.second;
if (y == pre) continue;
dis[y] = dis[x] + d;
build(y, x, level + 1);
}
}
int lca(int x, int y) {
if (depth[x] > depth[y]) swap(x, y);
while (depth[x] < depth[y]) {
y = st[y][lg2[depth[y]-depth[x]]];
}
if (x == y) return x;
for (int i = lg2[depth[x]]; i >= 0; i--) {
if (st[x][i] != st[y][i]) {
x = st[x][i];
y = st[y][i];
}
}
return st[x][0];
}
int a[MAX];
int rount[MAX];
void solve() {
int n, k;
cin >> n >> k;
for (int i = 1; i < n; i++) {
int u, v, w;
cin >> u >> v >> w;
edge[u].push_back({v, w});
edge[v].push_back({u, w});
}
dis[1] = 0;
build(1, 1, 0);
for (int i = 0; i < k; i++) {
cin >> a[i];
}
int tot = 0;
for (int i = 1; i < k; i++) {
int f = lca(a[i-1], a[i]);
rount[i] = dis[a[i-1]] - dis[f] + dis[a[i]] - dis[f];
tot += rount[i];
}
cout << tot - rount[1] << ' '; // 第一个
for (int i = 1; i < k - 1; i++) {
int f = lca(a[i-1], a[i+1]);
// 中间置换
cout << tot - rount[i] - rount[i+1] + dis[a[i-1]] - dis[f] + dis[a[i+1]] - dis[f] << ' ';
}
cout << tot - rount[k-1] << endl; // 最后一个
while (true) {
int x, y;
cin >> x >> y;
cout << lca(x, y) << endl;
}
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
load_lg2();
solve();
return 0;
}J: 砍树
// 砍树
#include <bits/stdc++.h>
#define int int64_t
#define endl '\n'
using namespace std;
const int MAX = 1e5+5;
const int LOG = 20;
const int INF = 0x3f3f3f3f3f3f3f3fll;
const int MOD = 1e9+7;
// 代码较长，建议从 solve 开始看
int lg2[MAX];
int depth[MAX];
int st[MAX][LOG]; // pre[i] 即为 st[i][0]
int weight[MAX];
vector<pair<int, int>> edge[MAX]; // pre : suf, idx
void load_lg2() {
lg2[0] = -1;
for (int i = 1; i < MAX; i++) {
lg2[i] = lg2[i / 2] + 1;
}
}
void build(int x, int pre, int level) {
depth[x] = level;
st[x][0] = pre;
for (int i = 1; i <= lg2[depth[x]]; i++) {
st[x][i] = st[st[x][i-1]][i-1];
}
for (pair<int, int> p : edge[x]) {
int y = p.first;
if (y == pre) continue;
build(y, x, level + 1);
}
}
int lca(int x, int y) {
if (depth[x] > depth[y]) swap(x, y);
while (depth[x] < depth[y]) {
y = st[y][lg2[depth[y]-depth[x]]];
}
if (x == y) return x;
for (int i = lg2[depth[x]]; i >= 0; i--) {
if (st[x][i] != st[y][i]) {
x = st[x][i];
y = st[y][i];
}
}
return st[x][0];
}
void dfs(int x, int pre) {
for (pair<int, int> p : edge[x]) {
int y = p.first;
if (y == pre) continue;
dfs(y, x);
weight[x] += weight[y];
}
}
void solve() {
int n, m;
cin >> n >> m;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
edge[u].push_back({v, i});
edge[v].push_back({u, i});
}
build(1, 0, 0); // lca 初始化, 不能写 build(1, 1, 0)
for (int i = 0; i < m; i++) {
int u, v;
cin >> u >> v;
int f = lca(u, v);
// 记录差分
weight[u]++;
weight[v]++;
weight[f]--;
weight[st[f][0]]--;
}
dfs(1, 0);
// 输出
int res = -1;
for (int x = 1; x <= n; x++) {
if (weight[x] == m) {
for (pair<int, int> p : edge[x]) {
int y = p.first, idx = p.second;
if (weight[y] == m) {
res = max(res, idx);
}
}
}
}
cout << res << endl;
}
int32_t main() {
cin.tie(0);
cout.tie(0);
ios::sync_with_stdio(0);
load_lg2();
solve();
return 0;
}
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原始发表：2024-03-20，如有侵权请联系 cloudcommunity@tencent.com 删除
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