预计得分:$30 + 0 + 0 = 30$
实际得分:$0+0+0= 0$
T1算概率的时候没模爆long long了。。。
我敢打赌这不是noip难度。。。
考虑算一个位置的概率,若想要$k$步把它干掉,那么与他距离为$1$到$k - 1$的点都必须阻塞
且距离为$k$的点至少有一个没被阻塞
概率的处理可以用前缀和优化。
这样看似是$O(n^3 logn)$,但是却不能通过,考虑在前缀和处理的时候有很多没用的状态(超出边界)
加一些剪枝即可
#include<cstdio>
#define max(a, b) (a < b ? b : a)
#define LL long long
using namespace std;
const int MAXN = 201, mod = 1e9 + 7, INF = 1e9 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M, a[MAXN][MAXN], g[MAXN][MAXN][MAXN], vis[MAXN][MAXN];
LL fastpow(LL a, LL p) {
LL base = 1;
while(p) {
if(p & 1) base = 1ll * base * a % mod;
a = 1ll * a * a % mod; p >>= 1;
}
return base;
}
LL inv(LL a) {
return fastpow(a, mod - 2);
}
int mul(int a, int b) {
if(1ll * a * b > mod) return 1ll * a * b % mod;
else return a * b;
}
void Pre() {
//cout << a[1][1] << endl;
for(int i = 1; i <= N; i++)
for(int j = 1; j <= M; j++)
g[0][i][j] = a[i][j] % mod;
for(int k = 1; k <= max(N, M); k++)
for(int i = 1; i <= N; i++)
for(int j = 1; j <= M; j++) {
if((i - k < 0) || (j - k < 0) || (i + k > N + 1) || (j + k > M + 1)) {vis[i][j] = 1; continue;}
if(vis[i][j]) continue;
g[k][i][j] = mul(g[k - 1][i - 1][j], g[k - 1][i + 1][j]);
if(k > 2) g[k][i][j] = mul(g[k][i][j], inv(g[k - 2][i][j]));
if(k >= 2) g[k][i][j] = mul(mul(g[k][i][j], inv(a[i][j + k - 2])), inv(a[i][j - k + 2]));
g[k][i][j] = mul(mul(g[k][i][j], a[i][j + k]), a[i][j - k]);
}
}
LL calc(int x, int y) {
LL ans = 0, s = a[x][y];
for(int i = 1; i <= max(N, M); i++) {
if((x - i < 0) || (y - i < 0) || (x + i > N + 1) || (y + i > M + 1)) break;
int now = g[i][x][y];
ans = (ans + mul(mul(i, (1 - now + mod)), s)) % mod;
s = mul(s, now);
}
return ans;
}
int main() {
// freopen("a.in", "r", stdin);
N = read(); M = read();
for(LL i = 1; i <= N; i++) {
for(LL j = 1; j <= M; j++) {
LL x = read(), y = read();
a[i][j] = mul(x, inv(y));
}
}
Pre();
for(LL i = 1; i <= N; i++, puts(""))
for(LL j = 1; j <= M; j++)
printf("%lld ", calc(i, j) % mod);
return 0;
}
考场上根本就没时间做。。
题目给出的模型太难处理了,考虑转化成一个较为普通的模型
遇到这种每个点有两个状态的题不难想到拆点,分别表示赢 / 输
当$a$赢了$b$,就从$a$赢向$b$输连边。
这样会得到一个新的无环图,可以证明,两个图中的环是等价的。
直接暴力找最小环即可,时间复杂度:$O(n^2 T)$
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 10005, BIT = 13;
int N, M, dep[MAXN], fa[MAXN][21], MC, U, V;
vector<int> v[MAXN];
int LCA(int x, int y) {
if(dep[x] < dep[y]) swap(x, y);
for(int i = BIT; i >= 0; i--) if(dep[fa[x][i]] >= dep[y]) x = fa[x][i];
if(x == y) return x;
for(int i = BIT; i >= 0; i--) if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
return fa[x][0];
}
void bfs(int x) {
queue<int> q;
q.push(x); dep[x] = 1; fa[x][0] = 0;
while(!q.empty()) {
int p = q.front(); q.pop();
for(int i = 0, to; i < v[p].size(); i++) {
if(!dep[to = v[p][i]]) {
fa[to][0] = p; dep[to] = dep[p] + 1;
for(int j = 1; j <= BIT; j++) fa[to][j] = fa[fa[to][j - 1]][j - 1];
q.push(to);
}
else if(to != fa[p][0]) {
int lca = LCA(p, to), dis = dep[p] + dep[to] - 2 * dep[lca] + 1;
if(dis < MC) MC = dis, U = p, V = to;
}
}
}
}
int main() {
int meiyong; scanf("%d", &meiyong);
while(scanf("%d", &N) && N) {//tag
scanf("%d", &M);
MC = MAXN;
for(int i = 1; i <= 2 * N; i++) v[i].clear();
memset(dep, 0, sizeof(dep));
memset(fa, 0, sizeof(fa));
for(int i = 1; i <= M; i++) {
int x, y; scanf("%d %d", &x, &y);
v[x].push_back(y + N); v[y + N].push_back(x);
}
for(int i = 1; i <= 2 * N; i++) if(!dep[i]) bfs(i);
if(MC == MAXN) {puts("-1");continue;}
int lca = LCA(U, V);
vector<int> ans; ans.clear();
printf("%d\n", MC);
while(U != lca) printf("%d ", (U - 1) % N + 1), U = fa[U][0];
printf("%d", (lca - 1) % N + 1);
while(V != lca) ans.push_back((V - 1) % N + 1), V = fa[V][0];
for(int i = ans.size() - 1; i >= 0; i--) printf(" %d", ans[i]);
puts("");
}
return 0;
}
/*
*/
$k=2$的时候是斐波那契博弈
$k \not = 2$的时候是神仙结论
考虑$k \not = 2$怎么做。
结论:
若$l = n$时先手必输,那么我们找到一个$m = \frac{n}{k} >= n$,且最小的$m$,当$l = n+m$先手也一定必输
证明:
我们把多着的$m$个单独考虑,若$n < l < n+m$时,先手拿走多余的$m$个,后手必败。
但当$l = n +m$时,先手不能拿走$m$个,因为此时后手可以一步拿走剩余的。
不断往下推就行了
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 1e7 + 10;
int T, k, top;
LL l, sta[MAXN];
int main() {
cin >> T;
while(T--) {
cin >> k >> l;
LL now = 1;
sta[top = 1] = 1;
while(now < l) {
int nxt = lower_bound(sta + 1, sta + top + 1, (now % k == 0) ? now / k : (now / k + 1)) - sta;
sta[++top] = (now = (now + sta[nxt]));
}
puts(now == l ? "DOG" : "GOD");
}
return 0;
}
/*
1
2 21
*/