D1T1 煎蛋的疑惑excatalan

Input 1
2 1

Output 1
3

Input 2
9 3

Output 2
9996

Input 3
996 223

Output 3
361421692

Input 4
514223 0

Output 4
287888483

#include<bits/stdc++.h>
#define ll long long
using namespace std;const ll mod = 998244353;
const int N = 2e6+10;
ll M(ll a) {return a%mod;}
int n, m;
ll fac[N], inv[N];ll Pow(ll a, ll b) {ll c=1;while(b){if(b&1) c=M(c*a);b>>=1;a=M(a*a);}return c;}ll C(ll x, ll y) {if(x<y || x<0 || y<0) return 0;return M(M(fac[x]*inv[y])*inv[x-y]);
}ll Ca(ll a, ll b) {if(b<0) return 0;return M(C(a<<1, a) - C((a<<1), a-b-1) + mod);
}void work() {fac[0] = 1;for(int i = 1; i <= (n<<1); i++)fac[i] = M(fac[i-1]*i);inv[n<<1] = Pow(fac[n<<1], mod-2);for(int i = (n<<1); i; i--)inv[i-1] = M(inv[i]*i);
}int main() {scanf("%d%d", &n, &m);work();printf("%lld\n", M(Ca(n,m)-Ca(n,m-1)+mod));return 0;
}

D1T2 蘑菇shimeji

Output 1
748683268

Output 2
132579339

Input 3
514
每一条边都是i与i+1相连（0<i<514）

Output 2
831885280

#include<bits/stdc++.h>
#define ll long long
using namespace std;const ll mod = 998244353;
const int N = 1e6+5;
struct psx{int y, next;} e[N<<1];
int lin[N], len=0, n;
ll f[N];//该点所在连通块被选择
ll g[N];//该点所在连通块没有被选择ll M(ll a){return a%mod;}//取模
ll add(ll a, ll b) {return M(a+b);}//加法
ll Pow(ll a, ll b) {ll c=1;while(b){if(b&1) c=M(c*a); b>>=1; a=M(a*a);}return c;}//快速幂void insert(int xx, int yy) {e[++len].next = lin[xx];lin[xx] = len;e[len].y = yy;
}void dfs(int k, int fa) {f[k] = g[k] = 1;//本身for(int i = lin[k]; i; i = e[i].next) {int y = e[i].y;if(y == fa) continue;dfs(y, k);f[k] = add(f[k]*g[y],g[k]*f[y]);//k被选择，可以由选择y和选择k的其他儿子（或本身)两种情况传递g[k] = M(g[k]*g[y]);//k不被选择，那么儿子y必然不被选择}g[k] = add(g[k], f[k]); //如果k与fa[k]的边断了，相当于不选择k//如果没断，那么选择k就可以传递到选择fa[k]
}int main() {scanf("%d", &n);for(int i = 1; i < n; i++) {int u, v;scanf("%d%d", &u, &v);insert(u, v);insert(v, u);}dfs(1, 0);printf("%lld\n", M(f[1]*Pow((mod+1)/2,n-1)));//因为所求为期望，所以还需要乘2^(n-1)//(mod+1)/2为模mod意义下2的逆元return 0;
}

D1T3 墙wall

Input 1
3 514223
2 1 3

Output 1
4

Input 2
8 514223
1 5 6 3 4 8 7 2

Output 2
32

Input 3
223 514223
墙的排列为1~223

Output 3
223

#include<bits/stdc++.h>
#define ll long long
using namespace std;ll mod, ans=0, a, n;
ll f1[6005], f2[6005], p[6005];ll M(ll a) {return a%mod;}
void add(ll &a, ll b){a = M(a+b);}void solve() {for(int i = 1; i <= n; i++)for(int j = i-1; j >= 1; j--) if(p[i] < p[j]) add(f2[j], f1[i]);else add(f1[i], f2[j]);
}int main() {scanf("%lld%lld", &n, &mod);for(int i = 1; i <= n; i++) {scanf("%lld", &a);p[a] = i;} for(int i = 1; i <= n; i++)f1[i] = 1, f2[i] = 0;solve();for(int i = 1; i <= n; i++)add(ans, f1[i]);for(int i = 1; i <= n; i++)f1[i] = 0, f2[i] = 1;solve();for(int i = 1; i <= n; i++)add(ans, f2[i]);//由样例可知，单点有贡献，以上两种情况单点记录两次，要减去add(ans, mod-n);printf("%lld\n", M(ans+mod));return 0;
}

D2T1 a

Input 1
5
1
3
5
7
11

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int N = 1e7+10;
const ll mod = 998244353;
ll M(ll a) { return a%mod; }
int q;
ll f[N];快读dread()快输print()略ll Pow(ll a,ll b){ll c=1;while(b){if(b&1)c=M(c*a);b>>=1;a=M(a*a);}return c;}int main() {q = dread();ll ans=1;while( q-- ) {int n = dread();ll k = Pow(n, mod-2);f[n] = 1;for(int i = n-1; i>=0; i--)f[i] = M( M(f[i+1]*(n-i))*k+1 ); print(f[0]);}return 0;
}

D2T2 b

Output 1
1
-1
2
3

D2T3 c

Input 1
3
2 3
4 5
6 7

Output 1
1
6
960

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国庆集训1027-1028（未完成）

D1T1 煎蛋的疑惑excatalan

D1T2 蘑菇shimeji

D1T3 墙wall

D2T1 a

D2T2 b

D2T3 c

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