11.09模拟赛T1&T2

n为字符串长度，len为s长度

想到正难则反，先求出合法的字符串数量，再用总数$26^n$去减就好了

类似于隔板法的思想，s字符串总共有$len+1$个空位可以放字符串，长度为剩下的$n-len$

考虑在s前面加入$n-len$个字符，总共有$26^{n-len}$种

如果插入在s第一个字符后面（往后移一格）还是$26^{n-len}$种吗？

显然不是，因为前面已经有一个字符了，有一个26应该为25

往后$len$个空位都是这样，所以答案是：

$$26^n-(26^{n-len}+len\ast 25\ast 26^{n-len-1})$$

注意特判$n=len$的情况

Code：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cmath>
#include <algorithm>
#define qy 998244353
#define MAXN 10000010
#define int long long
using namespace std;

char s[MAXN];
int n, len;

inline int POW (int a, int b) {
	if (! b) return 1; if (b == 1) return a;
	int base = 1;
	while (b) {
		if (b & 1) base = base * a % qy;
		a = a * a % qy, b >>= 1;
	}
	return base;
}

signed main () {
	freopen ("magic.in", "r", stdin); freopen ("magic.out", "w", stdout);
	scanf ("%lld", &n), scanf ("%s", s + 1), len = strlen (s + 1);
	int k1 = POW (26, n), k2 = POW (26, n - len), k3 = len * 25 * POW (26, n - len - 1) % qy;
	return printf ("%lld\n", ((((k1 - k2) % qy + qy) % qy - k3) % qy + qy) % qy), 0;
}

这道题就是星际导航重题哈哈哈

刚刚做完，模拟赛给我来这道

一眼kruskal重构树切掉了

Code：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#define MAXN 400000
using namespace std;

struct fls {int to, next;} edge[MAXN << 2];
struct Node {int x, y, z;} data[MAXN];

int depth[MAXN], top[MAXN], head[MAXN << 2], father[MAXN], son[MAXN];
int size[MAXN], f[MAXN], val[MAXN], n, m, q, cnt;

inline int read () {
	register int s = 0, w = 1;
	register char ch = getchar ();
	while (! isdigit (ch)) {if (ch == '-') w = -1; ch = getchar ();}
	while (isdigit (ch)) {s = (s << 3) + (s << 1) + (ch ^ 48); ch = getchar ();}
	return s * w;
}

inline bool cmp (Node a, Node b) {return a.z < b.z;}
inline void connect (int u, int v) {edge[++cnt].to = v, edge[cnt].next = head[u], head[u] = cnt;}
int getfather (int x) {return x == f[x] ? x : f[x] = getfather (f[x]);}

inline int LCA (int x, int y) {
	while (top[x] != top[y]) {
		if (depth[top[x]] < depth[top[y]]) swap (x, y);
		x = father[top[x]];
	}
	return depth[x] > depth[y] ? y : x;
}

void DFS1 (int now, int fa, int d) {
	father[now] = fa, depth[now] = d, size[now] = 1; int maxson = -1;
	for (register int i = head[now]; i; i = edge[i].next) {
		int v = edge[i].to; if (v == fa) continue;
		DFS1 (v, now, d + 1), size[now] += size[v];
		if (size[v] > maxson) maxson = size[v], son[now] = v;
	}
}

void DFS2 (int now, int top_heavy) {
	top[now] = top_heavy; if (! son[now]) return;
	DFS2 (son[now], top_heavy);
	for (register int i = head[now]; i; i = edge[i].next) {
		int v = edge[i].to;
		if (v != son[now] && v != father[now]) DFS2 (v, v);
	}
}

int main () {
	freopen ("graph.in", "r", stdin); freopen ("graph.out", "w", stdout);
	n = read (), m = read (), q = read ();
	for (register int i = 1; i <= m; i++) 
		data[i].x = read (), data[i].y = read (), data[i].z = read ();
	sort (data + 1, data + m + 1, cmp); int now = n;
	for (register int i = 1; i <= n * 2; i++) f[i] = i;
	for (register int i = 1; i <= m; i++) {
		int x = data[i].x, y = data[i].y;
		int xx = getfather (x), yy = getfather (y);
		if (xx != yy) {
			f[xx] = f[yy] = ++now;
			connect (xx, now), connect (now, xx);
			connect (yy, now), connect (now, yy);
			val[now] = data[i].z;
		}
	}
	for (register int i = now; i; i--)
		if (! depth[i]) DFS1 (i, 0, 1), DFS2 (i, i);
	for (register int i = 1; i <= q; i++) {
		int x = read (), y = read ();
		if (getfather (x) != getfather (y)) puts ("-1");
		else printf ("%d\n", val[LCA (x, y)]);
	}
	return 0;
}