SMU 2024 spring 天梯賽3
7-1 重要的話說三遍 - SMU 2024 spring 天梯賽3 (pintia.cn)
I'm gonna WIN!
I'm gonna WIN!
I'm gonna WIN!
7-2 兩小時學完C語言 - SMU 2024 spring 天梯賽3 (pintia.cn)
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,k,m;
cin >> n >> k >> m;
cout << max(n - k * m,0) << '\n';
return 0;
}
7-3 拯救外星人 - SMU 2024 spring 天梯賽3 (pintia.cn)
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a,b,ans = 1;
cin >> a >> b;
for(int i = 1;i <= a + b;i ++)
ans *= i;
cout << ans << '\n';
return 0;
}
7-4 誰能進圖書館 - SMU 2024 spring 天梯賽3 (pintia.cn)
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a, b, c, d;
cin >> a >> b >> c >> d;
if (d >= b && c < a) {
printf("%d-Y %d-Y\nqing 2 zhao gu hao 1", c, d);
} else if (c >= b && d < a) {
printf("%d-Y %d-Y\nqing 1 zhao gu hao 2", c, d);
}else if(c >= a && d >= a){
printf("%d-Y %d-Y\nhuan ying ru guan", c, d);
}else if(c < a && d < a){
printf("%d-N %d-N\nzhang da zai lai ba", c, d);
}else if(c >= a){
printf("%d-Y %d-N\n1: huan ying ru guan", c, d);
}else{
printf("%d-N %d-Y\n2: huan ying ru guan", c, d);
}
return 0;
}
7-5 試試手氣 - SMU 2024 spring 天梯賽3 (pintia.cn)
每個骰子從6往下列舉n個,遇見有過的跳過即可;
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
vector<int> a(6);
for (auto &i : a)cin >> i;
cin >> n;
for(int i = 0;i < 6;i ++){
for(int j = 6,m = n;j >= 1 && m;j --){
if(j == a[i]) continue;
m--;
if(!m) cout << j << " \n"[i == 5];
}
}
return 0;
}
7-6 查驗身份證 - SMU 2024 spring 天梯賽3 (pintia.cn)
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
const int w[] = {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
const char M[] = {'1','0','X','9','8','7','6','5','4','3','2'};
vector<string> ans;
while(n --){
string s;
cin >> s;
int num = 0,f = 0;
for(int i = 0;i < 17;i ++){
if(s[i] < '0' || s[i] > '9'){
ans.push_back(s);
f = 1;
break;
}
num += (s[i] - '0') * w[i];
}
if(f) continue;
num %= 11;
if(M[num] != s.back())
ans.push_back(s);
}
if(!ans.size()) cout << "All passed\n";
else{
for(auto i : ans)
cout << i << '\n';
}
return 0;
}
7-8 連續因子 - SMU 2024 spring 天梯賽3(補題) (pintia.cn)
按照滑動視窗的思想每次連乘,大於n時除掉前面的數;
維護連續乘的個數,大於ans時更新答案;
資料範圍弱,可以直接列舉到n,因其有一個資料點最大因子超過\(\sqrt n\),也可以列舉到\(2 \times \sqrt n\),不過要注意特判素數;
素數判斷時也超時,可以改下\(i \times i \leq n\)換成$ i \leq \sqrt n$,可能有奇效;
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
i64 n;
cin >> n;
auto ss = [](int x) {
if (x < 2) return false;
if (x == 2) return true;
for (int i = 2; i <= sqrt(x); i ++)
if (x % i == 0) return false;
return true;
};
if (ss(n)) {
cout << 1 << '\n' << n << '\n';
return 0;
}
vector<i64> ans;
i64 num = 0, res = 1, head = 2;
for (int i = 2; i <= sqrt(n) * 2; i ++) {
res *= i;
num ++;
if(!ans.size() && n % i == 0){
ans.push_back(i);
}
while (res > n) {
res /= head;
head ++;
num --;
}
if (n % res == 0) {
if (num > ans.size()) {
vector<i64>().swap(ans);
for (int j = head; j <= i; j ++)
ans.push_back(j);
}
}
}
cout << ans.size() << '\n';
for (auto i : ans)
cout << i << "*\n"[i == ans.back()];
return 0;
}
7-8 出租 - SMU 2024 spring 天梯賽3 (pintia.cn)
模擬
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s;
cin >> s;
vector<int> arr,vis(10),index(12);
for(int i = 0;i < 11;i ++){
int num = s[i] - '0';
if(vis[num]) continue;
arr.push_back(num);
vis[num] = 1;
}
sort(arr.begin(),arr.end(),greater<>());
for(int i = 0;i < arr.size();i ++){
index[arr[i]] = i;
}
cout << "int[] arr = new int[]{";
for(int i = 0;i < arr.size();i ++){
cout << arr[i] ;
if(i != arr.size() - 1) cout << ',';
}
cout << "};\n";
cout << "int[] index = new int[]{";
for(int i = 0;i < 11;i ++){
cout << index[s[i] - '0'] ;
if(i != 10) cout << ',';
}
cout << "};";
return 0;
}
7-9 哈利·波特的考試 - SMU 2024 spring 天梯賽3 (pintia.cn)
Floyd處理出兩兩最短長度;
列舉每隻動物能變成其餘動物最長魔咒長度,取最小值;
若沒有一隻動物能變成其餘各種動物,輸出0;
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
typedef pair<i64, i64> PII;
vector dis(n + 1, vector<i64>(n + 1, INT_MAX));
for (int i = 0; i < m; i ++) {
int u, v, w;
cin >> u >> v >> w;
dis[u][v] = w;
dis[v][u] = w;
}
for (int k = 1; k <= n; k ++) {
for (int i = 1; i <= n; i ++) {
for (int j = 1; j <= n; j ++) {
if (i == j) dis[i][j] = 0;
if (dis[i][k] + dis[k][j] < dis[i][j])
dis[i][j] = dis[i][k] + dis[k][j];
}
}
}
i64 ans = INT_MAX, f = 0, index = 0;
for (int i = 1; i <= n; i ++) {
i64 res = INT_MIN;
for (int j = 1; j <= n; j ++) {
if (dis[i][j] == INT_MAX) {
res = INT_MIN;
break;
}
res = max(res, dis[i][j]);
}
if (res < ans && res != INT_MIN) {
ans = res, index = i;
}
}
if (index) {
cout << index << ' ' << ans << '\n';
} else
cout << 0 << '\n';
return 0;
}
7-10 列車廂排程 - SMU 2024 spring 天梯賽3 (pintia.cn)
按題意模擬;
注意點是,1號車廂沒有列舉完時若此時3號車廂滿足條件,應該從3號轉到2號;
具體例子:
ABCD
CBAD
output:
1->3
1->3
1->2
3->2
3->2
1->2
code:
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s1, s2;
cin >> s1 >> s2;
stack<char> box;
vector<string> ans;
int index = s1.size() - 1;
reverse(s1.begin(), s1.end());
for (int i = 0; i < s2.size(); i ++) {
if (!box.size() || box.top() != s2[i]) {
while (index >= 0 && s1[index] != s2[i]) {
box.push(s1[index]);
ans.push_back("1->3");
index --;
}
}
if (index >= 0 && s1[index] == s2[i]) {
ans.push_back("1->2");
index --;
continue;
}
if (box.size() && box.top() == s2[i]) {
ans.push_back("3->2");
box.pop();
} else {
cout << "Are you kidding me?\n";
return 0;
}
}
for (auto i : ans)
cout << i << '\n';
return 0;
}
7-11 檔案傳輸 - SMU 2024 spring 天梯賽3 (pintia.cn)
並查集;
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
struct UFS {
vector<int> fa;
int n;
UFS(int n): n(n) {
fa.resize(n + 1);
for (int i = 0 ; i <= n; i ++)
fa[i] = i;
}
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void unin(int x, int y) {
x = find(x), y = find(y);
if (x != y) fa[x] = y;
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
UFS ufs(n);
char c;
int c1, c2;
while (cin >> c) {
if (c == 'S') break;
cin >> c1 >> c2;
if (c != 'C') {
ufs.unin(c1, c2);
} else {
c1 = ufs.find(c1);
c2 = ufs.find(c2);
if (c1 != c2) {
cout << "no\n";
} else
cout << "yes\n";
}
}
int num = 0;
for (int i = 1; i <= n; i ++)
if (ufs.find(i) == i) num ++;
if(num == 1) cout << "The network is connected.\n";
else cout << "There are " << num << " components.\n";
return 0;
}
7-12 病毒溯源 - SMU 2024 spring 天梯賽3 (pintia.cn)
根節點不一定從0開始,但一定唯一,所以可以先找出根節點;
跑一遍dfs計算出最大深度;
然後對每個點的相連的點做一個排序;
再跑一遍dfs統計答案,當第一次跑到最大深度時即是最小序列;
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector g(n + 1, vector<int>());
vector<int> fa(n);
iota(fa.begin(),fa.end(),0);
for (int i = 0; i < n; i ++) {
int k;
cin >> k;
if (k) {
for (int j = 0; j < k; j ++) {
int x;
cin >> x;
g[i].push_back(x);
fa[x] = i;
}
}
}
int root = 0;
for(int i = 0;i< n;i ++){
if(fa[i] == i) {
root = i;
break;
}
}
for (int i = 0; i < n; i ++) {
sort(g[i].begin(), g[i].end());
}
int mx = 0;
auto dfs = [&](auto dfs, int u, int now) -> void{
mx = max(now, mx);
for (auto i : g[u]) {
dfs(dfs, i, now + 1);
}
};
dfs(dfs, root, 1);
vector<int> ans;
auto dfs1 = [&](auto dfs, int u, int now) -> void{
if (now == mx) {
cout << mx << '\n';
for (auto i : ans)
cout << i << " \n"[i == ans.back()];
exit(0);
}
for (auto i : g[u]) {
ans.push_back(i);
dfs(dfs, i, now + 1);
ans.pop_back();
}
};
ans.push_back(root);
dfs1(dfs1, root, 1);
return 0;
}
7-14 天梯地圖 - SMU 2024 spring 天梯賽3(補題) (pintia.cn)
dijkstra進階做法,包含路徑記錄,以及按權重統計路徑條件等;
不過最開始我一直將優先佇列開的最大堆,但是一直過不了自己的例子,後來改成最小堆並且路徑值改成負數存進去就對了,再後來我發現改成最大堆也可以,不過要把點值改成負數,最後才煥然大悟,這題路徑取最短的同時,節點值也應該儘量的小(
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
struct DIJ {
using i64 = long long;
using PII = pair<i64, i64>;
vector<i64> dis, path, node;
vector<vector<array<int, 3>>> G;
int n;
DIJ() {}
DIJ(int n): n(n) {
node.resize(n + 1, 1);
dis.assign(n + 1, 1e18);
G.resize(n + 1);
path.resize(n + 1, -1);
}
void add(int u, int v, int w, int val) {
G[u].push_back({v, w, val});
}
void dijkstra(int s) {
priority_queue<PII,vector<PII>,greater<PII>> que;
dis[s] = 0;
que.push({0, -s});
while (!que.empty()) {
auto p = que.top();
que.pop();
int u = -p.second;
if (dis[u] < p.first) continue;
for (auto [v, w, val] : G[u]) {
if (dis[v] > dis[u] + w) {
node[v] = node[u] + val;
dis[v] = dis[u] + w;
que.push({ dis[v], -v});
path[v] = u;
} else if (dis[v] == dis[u] + w) {
if (node[v] > node[u] + val) {
node[v] = node[u] + val;
path[v] = u;
}
}
}
}
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m;
cin >> n >> m;
DIJ time(n), way(n);
while (m --) {
int u, v, c, t, w;
cin >> u >> v >> c >> w >> t;
time.add(u, v, t, w);
way.add(u, v, w, 1);
if (!c) {
time.add(v, u, t, w);
way.add(v, u, w, 1);
}
}
int st, ed, ok;
cin >> st >> ed;
time.dijkstra(st);
way.dijkstra(st);
ok = ed;
vector<int> ans, ans1;
while (ok != -1) {
ans.push_back(ok);
ok = time.path[ok];
}
ok = ed;
while (ok != -1) {
ans1.push_back(ok);
ok = way.path[ok];
}
if (ans1 == ans) {
cout << "Time = " << time.dis[ed] << "; ";
cout << "Distance = " << way.dis[ed] << ": ";
for (int i = ans1.size() - 1; i >= 0; i --) {
cout << ans1[i] ;
if (i) cout << " => ";
else cout << '\n';
}
} else {
cout << "Time = " << time.dis[ed] << ": ";
for (int i = ans.size() - 1; i >= 0; i --) {
cout << ans[i] ;
if (i) cout << " => ";
else cout << '\n';
}
cout << "Distance = " << way.dis[ed] << ": ";
for (int i = ans1.size() - 1; i >= 0; i --) {
cout << ans1[i] ;
if (i) cout << " => ";
else cout << '\n';
}
}
return 0;
}