第14届蓝桥杯B组国赛
子2023
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
void solve() {
vector<int> Q;
for (int i = 1 ; i <= 2023 ; ++i) {
int x = i;
vector<int> tmp;
while (x) {
int y = x % 10;
if (y == 2 || y == 0 || y == 3) {
tmp.push_back(y);
}
x /= 10;
}
reverse(tmp.begin() , tmp.end());
for (auto y : tmp) {
Q.push_back(y);
}
}
int SZ = Q.size();
vector<LL> dp3(SZ+1),dp23(SZ+1),dp023(SZ+1),dp2023(SZ+1);
for (int i = SZ - 1 ; i >= 0 ; --i) {
dp3[i] = dp3[i + 1] + (Q[i] == 3);
}
for (int i = SZ - 1 ; i >= 0 ; --i) {
dp23[i] = dp23[i + 1] + (Q[i] == 2) * dp3[i + 1];
}
for (int i = SZ - 1 ; i >= 0 ; --i) {
dp023[i] = dp023[i + 1] + (Q[i] == 0) * dp23[i + 1];
}
for (int i = SZ - 1 ; i >= 0 ; --i) {
dp2023[i] = dp2023[i + 1] + (Q[i] == 2) * dp023[i + 1];
}
cout << dp2023[0] << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
双子数
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 5E6 + 5;
int vaild[N];
vector<int> pri;
void init(int n) {
vaild[1] = 1;
for (int i = 2 ; i <= n ; ++i) {
if (!vaild[i]) {
pri.push_back(i);
}
for (int j = 0 ; j < pri.size() && i <= n / pri[j] ; ++j) {
vaild[i * pri[j]] = 1;
if (i % pri[j] == 0) {
break;
}
}
}
}
LL func(LL tar) {
LL ans = 0;
int SZ = pri.size();
for (int i = 0 ; i + 1 < SZ ; ++i) {
if (1LL * pri[i] * pri[i] > tar) break;
auto check = [&] (int md) {
LL x = 1LL * pri[i] * pri[md];
if (x <= tar / x) return true;
else return false;
};
int l = i + 1 , r = SZ - 1;
while (l < r) {
int md = (l + r + 1) / 2;
if (check(md)) l = md;
else r = md - 1;
}
if (check(l)) {
ans += l - i;
}
}
return ans;
}
void solve() {
LL l = 2333 , r = 23333333333333;
int n = sqrt(r) + 5;
init(n);
LL L,R;
L = func(l - 1);
R = func(r);
// cout << L << " " << R << "\n";
cout << R - L << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
班级活动
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
void solve() {
int n;
cin >> n;
vector<int> a(n + 1) , num(n + 1);
for (int i = 1 ; i <= n ; ++i) {
cin >> a[i];
num[a[i]] += 1;
}
int cnt1 = 0 , cnt2 = 0;
for (int i = 1 ; i <= n ; ++i) {
if (num[i]) {
if (num[i] > 2) cnt2 += num[i] - 2;
else if (num[i] == 1) cnt1 += 1;
}
}
int ans = 0;
if (cnt1 <= cnt2) {
ans = cnt2;
} else {
ans += cnt2;
cnt1 -= cnt2;
ans += cnt1 / 2;
}
cout << ans << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
合并数列
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
void solve() {
int n,m;
cin >> n >> m;
map<LL,int> mp;
vector<LL> a(n + 1) , b(m + 1) , sa(n + 1) , sb(m + 1);
for (int i = 1 ; i <= n ; ++i) {
cin >> a[i];
sa[i] = sa[i - 1] + a[i];
mp[sa[i]] += 1;
}
for (int i = 1 ; i <= m ; ++i) {
cin >> b[i];
sb[i] = sb[i - 1] + b[i];
mp[sb[i]] += 1;
}
int ans = 0;
for (auto f : mp) {
if (f.second == 1) {
ans += 1;
}
}
cout << ans << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
AB路线
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 1005 , inf = 1E9;
const int dx[] = {0 , 0 , -1 , 1} , dy[] = {1 , -1 , 0 , 0};
int n,m,K;
int s[N][N];
int dis[N][N][11];
void solve() {
cin >> n >> m >> K;
for (int i = 1 ; i <= n ; ++i) {
for (int j = 1 ; j <= m ; ++j) {
char c;
cin >> c;
if (c == 'B') s[i][j] = 1;
}
}
memset(dis , -1 , sizeof dis);
queue<array<int,3>> Q;
Q.push({1 , 1 , K - 1});
dis[1][1][K - 1] = 0;
while (!Q.empty()) {
auto f = Q.front();
Q.pop();
int x = f[0] , y = f[1] , rem = f[2];
for (int k = 0 ; k < 4 ; ++k) {
int nx = x + dx[k] , ny = y + dy[k];
if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
if (rem > 0) {
if (s[nx][ny] == s[x][y]) {
if (dis[nx][ny][rem - 1] == -1) {
dis[nx][ny][rem - 1] = dis[x][y][rem] + 1;
Q.push({nx , ny , rem - 1});
}
}
} else {
if (s[nx][ny] != s[x][y]) {
if (dis[nx][ny][K - 1] == -1) {
dis[nx][ny][K - 1] = dis[x][y][rem] + 1;
Q.push({nx , ny , K - 1});
}
}
}
}
}
const int inf = 2E9;
int ans = inf;
for (int i = 0 ; i <= K ; ++i) {
if (dis[n][m][i] != -1) {
ans = min(ans , dis[n][m][i]);
}
}
if (ans == inf) {
ans = -1;
}
cout << ans << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
删边问题
注意要考虑原图本身就存在两个及以上连通块的情况
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 2E5 + 5;
LL w[N],s[N];
vector<array<int,2>> g[N],que;
int tot,dfn[N],low[N];
void dfs(int u,int id) {
s[u] = w[u];
dfn[u] = low[u] = ++tot;
for (auto f : g[u]) {
int v = f[0] , id2 = f[1];
if (!dfn[v]) {
dfs(v , id2);
s[u] += s[v];
low[u] = min(low[u] , low[v]);
if (low[v] > dfn[u]) {
que.push_back({u , v});
}
} else if (id != id2) {
low[u] = min(low[u] , dfn[v]);
}
}
}
void solve() {
int n,m;
cin >> n >> m;
LL sum = 0;
for (int i = 1 ; i <= n ; ++i) {
cin >> w[i];
sum += w[i];
}
for (int i = 1 ; i <= m ; ++i) {
int u,v;
cin >> u >> v;
g[u].push_back({v , i});
g[v].push_back({u , i});
}
int scc = 0 , p1 = 0 , p2 = 0;
for (int i = 1 ; i <= n ; ++i) {
if (!dfn[i]) {
++scc;
dfs(i , 0);
if (p1 == 0) p1 = i;
else if (p2 == 0) p2 = i;
}
}
if (scc >= 2) {
if (scc > 2) {
cout << "-1\n";
} else if (scc == 2) {
if ((int)que.size() == m) {
cout << "-1\n";
} else {
cout << abs(s[p1] - s[p2]) << "\n";
}
}
return;
}
const LL inf = 1E18;
LL ans = inf;
for (auto f : que) {
int u = f[0] , v = f[1];
LL mn = min(s[u] , s[v]);
ans = min(ans , abs(sum - mn - mn));
}
if (ans == inf) {
ans = -1;
}
cout << ans << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
数三角
基本思路就是枚举等腰三角形的顶点，把它想象成圆心，转圈，注意去除三点共线的情况
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
void solve() {
int n;
cin >> n;
set<array<int,2>> S;
vector<array<int,2>> Node(n + 1);
for (int i = 1 ; i <= n ; ++i) {
int x,y;
cin >> x >> y;
Node[i] = {x , y};
S.insert({x , y});
}
auto get = [&] (int x1,int y1,int x2,int y2) -> LL {
LL ans = 1LL * (x1 - x2) * (x1 - x2) + 1LL * (y1 - y2) * (y1 - y2);
return ans;
};
LL ans = 0;
for (int i = 1 ; i <= n ; ++i) {
auto tmp = S;
int x1 = Node[i][0] , y1 = Node[i][1];
tmp.erase({x1 , y1});
unordered_map<LL,int> mp;
int cnt = 0;
for (auto f : tmp) {
int x2 = f[0] , y2 = f[1];
LL dis = get(x1 , y1 , x2 , y2);
ans += mp[dis];
mp[dis] += 1;
int x3 = 2 * x1 - x2 , y3 = 2 * y1 - y2;
if (tmp.count({x3 , y3})) ++cnt;
}
ans -= cnt / 2;
}
cout << ans << "\n";
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
抓娃娃
注意到性质 \(max(r_i - l_i) \leq min(R_i - L_i)\).
那么只要区间包含线段的中心即可.
注意到中心除 \(2\) 可能会有精度问题，那么我们坐标乘 \(2\) 即可.
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
void solve() {
int n,m;
cin >> n >> m;
vector<int> x;
for (int i = 1 ; i <= n ; ++i) {
int l,r;
cin >> l >> r;
x.push_back(l + r);
}
sort(x.begin() , x.end());
for (int i = 1 ; i <= m ; ++i) {
int L,R;
cin >> L >> R;
L = L * 2 , R = R * 2;
int l = lower_bound(x.begin() , x.end() , L) - x.begin();
int r = upper_bound(x.begin() , x.end() , R) - x.begin();
cout << r - l << "\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
while (T--) {
solve();
}
return 0;
}
\(n^22^n\)
