A
按照题意输出即可。
/*********************************************************************程序名:作者: xAlec日期: 2025-10-01 15:45说明: sakana ~
*********************************************************************/
#include <bits/stdc++.h>
#define int long long
using namespace std;int x, y;
char ch;signed main() {cin >> x >> ch >> y;if (y < 8) cout << x << ch << y + 1 << '\n';else if (x < 8 && y == 8)cout << x + 1 << ch << 1 << '\n';
}
B
按照题意模拟即可。
/*********************************************************************程序名:作者: xAlec日期: 2025-10-01 15:45说明: sakana ~
*********************************************************************/
#include <bits/stdc++.h>
#define int long long
using namespace std;
int h, w;
char s[25][25];signed main() {scanf ("%lld %lld", &h, &w);for (int i = 1; i <= h; i++)scanf ("%s", s[i] + 1);bool flag = true;for (int i = 1; i <= h; i++) {for (int j = 1; j <= w; j++) {if (s[i][j] == '#') {int cnt = 0;cnt += (s[i - 1][j] == '#');cnt += (s[i][j - 1] == '#');cnt += (s[i + 1][j] == '#');cnt += (s[i][j + 1] == '#');if (cnt != 2 && cnt != 4) {flag = false;break;}}}if (!flag)break;}if (flag)puts("Yes");elseputs("No");return 0;
}
C
看完题第一反应是 \(\min(cnt_A,cnt_C)\)，然后它 WA 了。
严肃思考后发现 \(cnt_B = 0\) 的时候需要特殊考虑，一般化就是因为三个位置要均分，最大的个数只能是 \(\lfloor\frac{n_A + n_B + n_C}{3}\rfloor\)。
#include <bits/stdc++.h>
#define int long long
using namespace std;
int test;
int na, nb, nc;void sol() {scanf ("%lld %lld %lld", &na, &nb, &nc);int up = max(na, nc), down = min(na, nc);printf ("%lld\n", min(down, (na + nb + nc) / 3));
}signed main() {scanf ("%lld", &test);while (test--)sol();return 0;
}
D
读完题意可以发现，其实操作是线段树建树。由于总和均分，答案只有可能是 0/1。
#include <bits/stdc++.h>
#define int long long
using namespace std;
constexpr int MAXN = (1 << 20) + 3;
int n, K, a[MAXN], U;inline int qpow(int x, int exp) {int res = 1;for (; exp; exp /= 2) {if (exp & 1)res = res * x;x = x * x;}return res;
}void build(int l, int r, int sum) {if (l == r) {a[l] = sum;return;}int mid = (l + r) / 2;build(l, mid, sum / 2);build(mid + 1, r, sum - sum / 2);
}signed main() {scanf ("%lld %lld", &n, &K);n = qpow(2, n);build(1, n, K);for (int i = 1; i < n; i ++) {if (a[i] != a[i + 1]) {U = 1;break;}}printf ("%lld\n", U);for (int i = 1; i <= n; i++)printf ("%lld%c", a[i], " \n"[i == n]);return 0;
}
E
考虑两点 \((x_1, y_1), (x_2,y_2)\) 确定一条直线则有 \(ax_1 + by_1 + c = ax_2 + by_2 + c\)，解得 \(\begin{cases}a = y_2 - y_1 \\ b = x_1 - x_2 \\ c = x_2y_1 - x_1y_2\end{cases}\)。
考虑一波随机化，每次随机两个点确定一条直线来判，好像概率挺高的，高达 \(25\%\)。
#include <bits/stdc++.h>
#define int long long
using namespace std;
random_device seed;
mt19937 rd(seed());
constexpr int MAXN = 5e5 + 10;
int n, a = -1, b = -1, c = -1;
pair<int,int> p[MAXN];signed main() {scanf ("%lld", &n);for (int i = 1; i <= n; i++)scanf ("%lld %lld", &p[i].first, &p[i].second);for (int _ = 1; _ <= 100; _++) {int x = rd() % n + 1, y = rd() % n + 1;// cout << x << ' ' << y << '\n';if (x != y) {int ta = p[y].second - p[x].second,tb = p[x].first - p[y].first,tc = p[y].first * p[x].second - p[x].first * p[y].second,tot = 0;for (int i = 1; i <= n; i++) tot += (ta * p[i].first + tb * p[i].second + tc == 0);if (tot >= n / 2 + 1) {a = ta, b = tb, c = tc;break;}}}if (a == -1 && b == -1 && c == -1)printf ("No\n");else printf ("Yes\n %lld %lld %lld\n", a, b, c);return 0;
}
F
对于 \(u\) 节点的 \(W_u\)，如果此时从它开始还有 \(x\) 步走到目的地，则它会被计算 \(x\) 次。
考虑记 \(f_{u,x}\) 表示当前在 \(u\) 节点，还有 \(x\) 步走到目的地的燃料最小花费。有转移：\(f_{v,x - 1} = \max\{f_{u,x} + w_u \times x\}\)。
答案即为每个 \(f_{i,0}\)。
#include <bits/stdc++.h>
#define FASTIO
#define int long long
using namespace std;
#ifdef FASTIOstatic int ostk[33];char buf[1 << 23], *p1, *p2;#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 23, stdin), p1 == p2) ? EOF : *p1++)inline int read() {int res = 0, f = 1;char ch = getchar();while (!isdigit(ch))f = ch == '-' ? -1 : 1, ch = getchar();while (isdigit(ch))res = res * 10 + (ch ^ 48), ch = getchar();return res * f;}inline void write(int x) {int top = 0;if (x < 0)x = -x, putchar('-');do {ostk[top++] = x % 10;x /= 10;} while(x);while(top)putchar(ostk[--top] + '0');}
#endifconstexpr int MAXN = 5005;
int n, m, idx;
int head[MAXN], w[MAXN];
int f[MAXN][MAXN];
struct graph {int v, nxt;
} e[MAXN << 1];inline void addedge(int u, int v) {e[idx].v = v;e[idx].nxt = head[u];head[u] = idx++;
}signed main() {n = read();m = read();memset(head, -1, sizeof(head));for (int i = 1; i <= n; i++)w[i] = read();for (int i = 1; i <= m; i++) {int u = read(), v = read();addedge(u, v), addedge(v, u);}memset(f, 0x3f, sizeof(f));for (int u = 0; u <= n; u++)f[1][u] = 0;for (int j = n; j >= 1; j--) {for (int u = 1; u <= n; u++) {for (int i = head[u]; ~i; i = e[i].nxt) {int v = e[i].v;f[v][j - 1] = min(f[v][j - 1], f[u][j] + w[u] * j);}}}for (int i = 1; i <= n; i++)write(f[i][0]), putchar('\n');return 0;
}
