A、小紅的葫蘆
水一篇
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
int n = 5;
std::vector<int> a(n);
for (int i = 0; i < n; ++i) {
std::cin >> a[i];
}
std::sort(a.begin(), a.end());
debug(a);
bool ok = false;
if (a[0] == a[1] && a[1] == a[2] && a[2] != a[3] && a[3] == a[4]) {
ok = true;
}
std::reverse(a.begin(), a.end());
if (a[0] == a[1] && a[1] == a[2] && a[2] != a[3] && a[3] == a[4]) {
ok = true;
}
std::cout << (ok ? "YES" : "NO") << "\n";
}
B、茉茉的密碼
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
int n;
std::cin >> n;
std::vector<std::string> a(n);
for (int i = 0; i < n; ++i) {
std::cin >> a[i];
}
for (char c = 'a'; c <= 'z'; ++c) {
bool ok = true;
for (int i = 0; i < n; ++i) {
ok &= a[i].find(c) != std::string::npos;
}
if (ok) {
std::cout << c << '\n';
return 0;
}
}
}
C、苗苗的氣球
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
int n;
std::cin >> n;
std::vector<int> a(n);
for (int i = 0; i < n; ++i) {
std::cin >> a[i];
}
auto b = a;
std::sort(a.begin(), a.end());
int sum = std::accumulate(a.begin(), a.end(), 0LL);
int ans = 0;
if (n == 2) {
ans += (b[0] > b[1]) + (b[0] < b[1]);
} else {
if (sum > 2LL * a[n - 1]) {
for (int i = 0; i < n; ++i) {
ans += b[i] > (sum % 2 == 0);
}
} else {
for (int i = 0; i < n; ++i) {
ans += b[i] == a[n - 1];
}
}
}
std::cout << ans << '\n';
}
D、萌萌的好數
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
#define int long long
int tt;
for (std::cin >> tt; tt--;) {
int n;
std::cin >> n;
auto check = [&](int x) {
x -= 3;
int ans = (x + 3) / 3 + (x / 10 - x / 10 / 3);
return x + 3 - ans >= n;
};
int l = 1, r = 1e18;
while (l < r) {
int mid = l + r >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
std::cout << r << '\n';
}
}
E、 茜茜的計算器
分析
水平和豎直對稱分開考慮。
對於水平對稱可以發現只能放\(0, 1, 3, 8\)四個數，考慮全部組合數量為\(4 ^ n\)。
對於豎直對稱可以發現只能放\(0, 2, 5, 8\)四個數，因為豎直對稱，因此只要考慮前\(\frac{n}{2}\)個位置如何擺放，方案數為\(4^{\frac{n}{2}}\)。對於n為奇數，可以發現中間只能擺放\(0, 8\)，因此方案數需要再乘上2。
最後減去兩種方案的重合數\(2^{\frac{n + 1}{2}}\)即可。
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
template <class T>
constexpr T power(T a, long long b) {
T res = 1;
for (; b; b /= 2, a *= a)
if (b % 2) res *= a;
return res;
}
template <int P>
struct MInt {
int x;
constexpr MInt() : x{} {}
constexpr MInt(long long x) : x{norm(x % P)} {}
constexpr int norm(int x) const {
if (x < 0) x += P;
if (x >= P) x -= P;
return x;
}
constexpr int val() const { return x; }
explicit constexpr operator int() const { return x; }
constexpr MInt operator-() const {
MInt res;
res.x = norm(P - x);
return res;
}
constexpr MInt inv() const {
assert(x != 0);
return power(*this, P - 2);
}
constexpr MInt &operator*=(MInt rhs) {
x = 1ll * x * rhs.x % P;
return *this;
}
constexpr MInt &operator+=(MInt rhs) {
x = norm(x + rhs.x);
return *this;
}
constexpr MInt &operator-=(MInt rhs) {
x = norm(x - rhs.x);
return *this;
}
constexpr MInt &operator/=(MInt rhs) { return *this *= rhs.inv(); }
friend constexpr MInt operator*(MInt lhs, MInt rhs) {
MInt res = lhs;
res *= rhs;
return res;
}
friend constexpr MInt operator+(MInt lhs, MInt rhs) {
MInt res = lhs;
res += rhs;
return res;
}
friend constexpr MInt operator-(MInt lhs, MInt rhs) {
MInt res = lhs;
res -= rhs;
return res;
}
friend constexpr MInt operator/(MInt lhs, MInt rhs) {
MInt res = lhs;
res /= rhs;
return res;
}
friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
long long v;
is >> v;
a = MInt(v);
return is;
}
friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) { return os << a.val(); }
friend constexpr bool operator==(MInt lhs, MInt rhs) { return lhs.val() == rhs.val(); }
friend constexpr bool operator!=(MInt lhs, MInt rhs) { return lhs.val() != rhs.val(); }
};
template <int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();
// constexpr int P = 998244353;
constexpr int P = 1e9 + 7;
using Z = MInt<P>;
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
int n ;
std::cin >> n;
Z lr = power(Z(4), n / 2), ud = power(Z(4), n);
Z cocide = power(Z(2), (n + 1) / 2);
if (n & 1) {
lr *= 2;
}
std::cout << lr + ud - cocide << '\n';
}
F、花花的地圖
程式碼實現
#include <bits/stdc++.h>
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
int main() {
std::cin.tie(nullptr)->sync_with_stdio(false);
int n, m;
std::cin >> n >> m;
std::vector<std::string> g(n);
for (int i = 0; i < n; ++i) {
std::cin >> g[i];
}
std::vector<int> C(m);
for (int i = 0; i < m; ++i) {
std::cin >> C[i];
}
std::vector dist(n, std::vector<int>(m, 1e9));
std::vector<std::vector<bool>> vis(n, std::vector<bool>(m));
std::priority_queue<std::array<int, 3>, std::vector<std::array<int, 3>>, std::greater<>> q;
q.push({0, 0, 0});
dist[0][0] = 0;
while (size(q)) {
auto [dis, x, y] = q.top();
q.pop();
if (x == n - 1 && y == m - 1) {
break;
}
vis[x][y] = true;
for (auto [fx, fy] : {std::pair{x + 1, y}, {x - 1, y}, {x, y + 1}, {x, y - 1}}) {
if (fx >= 0 && fx < n && fy >= 0 && fy < m && !vis[fx][fy]) {
if (g[fx][fy] == '.') {
if (dist[fx][fy] <= dis) continue;
dist[fx][fy] = dis;
q.push({dis, fx, fy});
} else {
for (int fx = 0; fx < n; ++fx) {
if (dist[fx][fy] <= dis + C[fy]) continue;
dist[fx][fy] = dis + C[fy];
q.push({dist[fx][fy], fx, fy});
}
}
}
}
}
std::cout << dist[n - 1][m - 1] << '\n';
}