smu2024蓝桥杯训练1
A
查看代码
void solve() {
int n;
cin >> n;
if (n <= 100) cout << 100;
else if(n <= 150) cout << 150;
else if(n <= 300) cout << 300;
else if(n <= 400) cout << 400;
else if(n <= 1000) cout << 1000;
}B
注意：数据末尾为/r/t，需要getchar两次，或者cin.ignore（100，'\n'）
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
string a, b, ss;
getline(cin, ss);
// assert( ss.back() != '\r');
a = ss.substr(0, 8), b = ss.substr(9, 8);
auto P = [](string s) {
int ans = 0;
string x = s.substr(0,2);
int y = stoi(x);
ans += y * 3600;
x = s.substr(3, 2);
y = stoi(x);
ans += y * 60;
x = s.substr(6, 2);
y = stoi(x);
ans += y;
return ans;
};
int x1 = P(a), x2 = P(b);
if (ss.size() > 18) {
x2 += (ss[20] - '0') * 24 * 60 * 60;
}
getline(cin, ss);
a = ss.substr(0, 8), b = ss.substr(9, 8);
int y1 = P(a), y2 = P(b);
if (ss.size() > 18) {
y2 += (ss[20] - '0') * 24 * 60 * 60;
}
// cout << x1 << ' ' << x2 << '|' << y1 << ' ' << y2 << '\n';
// cout << x2 - x1 << ' ' << y2 - y1 << '\n';
int c = (x2 - x1 + (y2 - y1)) / 2;
auto Print = [](int x) {
int a, b, c;
c = x % 60; x /= 60;
b = x % 60; x /= 60;
a = x;
printf("%02lld:%02lld:%02lld\n", a, b, c);
};
Print(c);
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
cin >> T;
cin.ignore(100, '\n');
// getchar();
// getchar();
while (T--) {
solve();
}
}
C
哈希
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
int n, k;
cin >> n >> k;
map<string, int> mp;
for (int i = 0, x; i < n; ++i) {
string s;
cin >> s >> x;
mp[s] = x;
}
getchar();
for (int i = 0; i < k; ++i) {
string s;
getline(cin, s);
// cout << s << '\n';
for (int j = 0; j < s.size(); ++j) {
if (s[j] == '{') {
int p = j + 1;
string x;
while (p < s.size() && s[p] != '}') {
x.push_back(s[p++]);
}
j = p;
if (mp.count(x)) cout << mp[x];
else cout << '{' << x << '}';
}
else cout << s[j];
}
cout << '\n';
}
}
signed main() {
// ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
// getchar();
while (T--) {
solve();
}
}
D
dfs搜索每种情况
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
int n;
struct E {
int l, r, c;
};
vector<E> ve;
vector<int> st, f;
bool dfs(int u) {
if (u == n) {
return true;
}
for (int i = 0; i < n; ++i) {
if (st[i]) continue;
if (f.empty()) f.push_back(ve[i].l + ve[i].c);
else {
if (f.back() > ve[i].r) continue;
f.push_back(max(f.back(), ve[i].l) + ve[i].c);
}
st[i] = 1;
if (dfs(u + 1)) return true;
f.pop_back();
st[i] = 0;
}
return false;
}
void solve() {
cin >> n;
ve = vector <E> (n);
st = vector<int> (n);
f.clear();
for (int i = 0; i < n; ++i) cin >> ve[i].l >> ve[i].r >> ve[i].c, ve[i].r += ve[i].l;
bool ok = dfs(0);
if (ok) cout << "YES\n";
else cout << "NO\n";
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
cin >> T;
while (T--) {
solve();
}
}
E
dp，f[i][j]表示前i个数中末尾的数最后一位为j的最少删除次数
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
int n;
cin >> n;
vector f(n + 1, vector <int> (10, INF));
vector<int> a(n + 1), pre(n + 1), las(n + 1);
auto P = [] (int x) {
int y;
while (x > 0) {
y = x % 10;
x /= 10;
}
return y;
};
for (int i = 1; i <= n; ++i) cin >> a[i], pre[i] = P(a[i]), las[i] = a[i] % 10;
for (int i = 0; i <= n; ++i) {
for (int j = 0; j <= 9; ++j) {
if (i == 0) f[i][j] = 0;
else {
f[i][j] = f[i - 1][j] + 1;
}
}
if (i) f[i][las[i]] = min(f[i][las[i]], f[i - 1][pre[i]]);
}
int ans = INF;
for (int i = 0; i <= 9; ++i) ans = min (ans, f[n][i]);
cout << ans;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
}F
一次一步或者k步，跑最短路即可
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
int n, k;
cin >> n >> k;
vector<int> d(n, INF), st(n);
queue<int> q;
d[0] = 0;
q.push(0);
while (q.size()) {
auto t = q.front();
q.pop();
if (st[t]) continue;
st[t] = 1;
if (!st[(t + 1) % n] && d[(t + 1) % n] > d[t] + 1) {
d[(t + 1) % n] = d[t] + 1;
q.push((t + 1) % n);
}
if (!st[(t + k) % n] && d[(t + k) % n] > d[t] + 1) {
d[(t + k) % n] = d[t] + 1;
q.push((t + k) % n);
}
}
int ans = 0;
for (int i = 1; i < n; ++i) ans = max(ans, d[i]);
cout << ans;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
}
G
n的范围不大，n2枚举区间，根据区间大小从小到大枚举，若s[l] > s[r]答案加一，若s[l] = s[r]，答案取决于 [l + 1，r - 1]
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
string s;
cin >> s;
int n = s.size(), ans = 0;
s.insert(s.begin(), ' ');
vector f(n + 1, vector<int> (n + 1));
for (int i = 2; i <= n; ++i) {
for (int j = 1; j + i - 1 <= n; ++j) {
int l = j, r = j + i - 1;
if (s[l] > s[r]) f[l][r] = 1;
else if (s[l] == s[r] && i > 2) f[l][r] = f[l + 1][r - 1];
ans += f[l][r];
}
}
cout << ans;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
}
H
首先可以二分y，考虑如何check。可以将来回看作一共跳2x次，对于长度为y的区间内每只青蛙必须在此区间跳至少一次（跳跃极限是y，不可能一步跳过长度为y的区间），那么长度为y的区间内至少要有2x个石子（加上高度），用区间和判断即可。即check所有长度为y的区间和是否不小于2x
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f;
void solve() {
int n, x;
cin >> n >> x;
vector<int> a(n + 1);
for (int i = 1; i < n; ++i) {
cin >> a[i];
a[i] += a[i - 1];
}
a[n] = a[n - 1];
int l = 1, r = n, res;
auto check = [n, a, x](int y) {
for (int i = y; i < n; ++i) {
if (a[i] - a[i - y] < 2 * x) return false;
}
return true;
};
while (l <= r) {
int mid = l + r >> 1;
if (check(mid)) res = mid, r = mid - 1;
else l = mid + 1;
}
cout << res;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
}I
状压dp，一列的状态只与前一列和前前一列的状态和个数有关，n也不大，直接枚举个数、当前列、前列、前前列的状态即可
查看代码
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define PII pair<int, int>
#define PDI pair<double, int>
const double eps = 1e-12;
const int N = 1e4 + 5;
const int dx[4] = {0,-1}, dy[4] = {-1, 0};
const int INF = 0x3f3f3f3f, mod = 1e9 + 7;
int f[105][65][65][25];
void solve() {
int n, m, k;
cin >> n >> m >> k;
vector<int> sum(100);
auto C = [](int x) {
int ans = 0;
while (x > 0) {
if (x % 2 == 1)ans ++;
x /= 2;
}
return ans;
};
for (int i = 0; i < 1 << n; ++i) sum[i] = C(i);
for (int i = 0; i < 1 << n; ++i)
for (int j = 0; j < 1 << n; ++j) f[2][i][j][sum[i] + sum[j]] = 1;
for (int i = 2; i <= m; ++i) {
for (int c = 0; c < 1 << n; ++c) {
for (int d = sum[c]; d <= k; ++d) {
for (int b = 0; b < 1 << n; ++b) {
for (int a = 0; a < 1 << n; ++a) {
if (sum[c] + sum[b] + sum[a] > k) continue;
if (a & (c >> 1) || a & (c << 1) || a & (b >> 2) || a & (b << 2) || b & (c >> 2) || b & ( c << 2))continue;
f[i][b][c][d] = (f[i][b][c][d] + f[i - 1][a][b][d - sum[c]]) % mod;
}
}
}
}
}
int ans = 0;
for (int i = 0; i < 1 << n; ++i) {
for (int j = 0; j < 1 << n; ++j) ans = (ans + f[m][i][j][k]) % mod;
}
cout << ans;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
}