题解 你猜是不是找规律
这题做的历尽坎坷
首先有个 \(k=3\) 的部分分
既然 \(k\) 这么小,那就枚举恰好交换 \(k\) 次然后组合数求好了
一次操作有两种可能,两两交换或多个轮换(瞎起名字.jpg)
枚举给二者分配多少个数以及多少次操作即可
考完尝试拓展这种方法,但花了一下午喜提10pts
然后我康到了一位大佬的博客
方法巨强,膜拜@Cyber_Tree大佬!
首先错排的部分可以与整个序列独立出来
然后考虑处理出 \(f_{i, j}\) 表示有 \(i\) 个数,恰好 \(j\) 次操作归位的方案数
递推的话就和错排一样,考虑第一个位置上的数在哪即可
然后就组合数枚举这些错排的数在序列的什么位置
最终答案为
\[ans = \sum\limits_{i=0}^{2k}\binom{n}{i}\sum\limits_{j=0}^{k}f_{i, j}
\]
- 当组合数 \(n\) 很大,\(k\) 较小的时候可以转为下降幂的形式爆算
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 3010
#define ll long long
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n, k;
const ll mod=1e9+7;
namespace force{
int p[N], ans;
bool vis[N];
bool dfs(int u) {
for (int i=1; i<=n; ++i) if (p[i]!=i) goto jump;
return 1;
jump:
if (u>k) return 0;
for (int i=1; i<=n; ++i) if (p[i]!=i) {
for (int j=i+1; j<=n; ++j) if (p[i]==j||p[j]==i) {
swap(p[i], p[j]);
if (dfs(u+1)) {swap(p[i], p[j]); return 1;}
swap(p[i], p[j]);
}
}
return 0;
}
void solve() {
for (int i=1; i<=n; ++i) p[i]=i;
do {
// for (int i=1; i<=n; ++i) cout<<p[i]<<' '; cout<<endl;
if (dfs(1)) {
++ans;
// for (int i=1; i<=n; ++i) cout<<p[i]<<' '; cout<<endl;
}
} while (next_permutation(p+1, p+n+1));
printf("%d\n", ans);
// exit(0);
}
}
namespace task1{
ll d[N], fac[N], inv[N], ans[N], g[N];
inline ll C(ll n, ll k) {
if (n<k) return 0ll;
else if (k==1) return n;
else if (k==2) return n*(n-1)%mod*inv[2]%mod;
else if (k==3) return n*(n-1)%mod*(n-2)%mod*inv[6]%mod;
else if (k==4) return n*(n-1)%mod*(n-2)%mod*(n-3)%mod*inv[24]%mod;
else if (k==5) return n*(n-1)%mod*(n-2)%mod*(n-3)%mod*(n-4)%mod*inv[120]%mod;
else if (k==6) return n*(n-1)%mod*(n-2)%mod*(n-3)%mod*(n-4)%mod*(n-5)%mod*inv[720]%mod;
}
void solve() {
int lim=1000;
d[1]=0; d[2]=1; inv[0]=inv[1]=1; g[0]=1;
for (int i=3; i<=lim; ++i) d[i]=(i-1)*(d[i-1]+d[i-2])%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) g[i]=g[i-2]*(i-1)%mod;
ans[0]=1;
ans[1]=C(n, 2);
ans[2]=(C(n, 3)*d[3]%mod + C(n, 4)*g[4]%mod)%mod;
ans[3]=(C(n, 4)*(d[4]-g[4])%mod + C(n, 6)*g[6]%mod + C(n, 5)*C(5, 2)%mod*d[3]%mod)%mod;
ll sum=0;
for (int i=0; i<=k; ++i) sum=(sum+ans[i])%mod;
printf("%lld\n", sum);
exit(0);
}
}
namespace task2{
ll d[N], fac[N], inv[N], ans[N], g[N], t[3010][3010], f[3010][3010], h[3010][3010];
inline ll C(int n, int k) {return n<k?0ll:fac[n]*inv[k]%mod*inv[n-k]%mod;}
void solve() {
int lim=1000;
d[1]=0; d[2]=1; fac[0]=fac[1]=1; inv[0]=inv[1]=1; g[0]=1;
for (int i=3; i<=lim; ++i) d[i]=(i-1)*(d[i-1]+d[i-2])%mod;
for (int i=2; i<=lim; ++i) fac[i]=fac[i-1]*i%mod;
for (int i=2; i<=lim; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=lim; ++i) inv[i]=inv[i-1]*inv[i]%mod;
for (int i=2; i<=lim; ++i) g[i]=g[i-2]*(i-1)%mod;
// for (int i=0; i<=n; ++i) h[i][0]=1;
for (int i=2; i<=k; ++i) h[i*2-1][i]=(h[i*2-1][i]+d[i*2-1]-g[i*2-1])%mod;
for (int i=1; i<=n; ++i) {
for (int j=1; j<=k; ++j) {
for (int s=2; s<j; ++s) if (i>=2*s-1) {
// cout<<i<<' '<<s<<' '<<2*s-1<<endl;
h[i][j]=(h[i][j]+C(i, 2*s-1)*h[i-(2*s-1)][j-s])%mod;
}
// printf("h[%d][%d]=%lld\n", i, j, h[i][j]);
}
}
for (int i=0; i<=n; ++i) t[i][0]=1;
for (int i=1; i<=n; ++i) {
for (int j=1; j<=k; ++j) {
// for (int s=2; s<j; ++s)
// t[i][j]=(t[i][j]+t[i-(2*s-1)][j-s])%mod;
// t[i][j]=(t[i][j]+i*t[i-1][j]%mod)%mod;
for (int l=1; l<=i; ++l)
t[i][j]=(t[i][j]+C(i, l)*h[l][j]%mod)%mod;
// printf("t[%d][%d]=%lld\n", i, j, t[i][j]);
}
}
for (int i=1; i<=n; ++i) f[i][0]=1;
for (int i=1; i<=n; ++i) {
for (int j=1; j<=k; ++j) {
for (int s=0; s<=j; ++s) if (2*s<=i) {
f[i][j]=(f[i][j]+C(i, 2*s)*g[2*s]%mod*t[i-2*s][j-s]%mod)%mod;
// printf("f[%d][%d]+=C(%d, %d)(%lld)*g[%d](%lld)*t[%d][%d](%lld) (+=%lld)\n", i, j, i, 2*s, C(i, 2*s), 2*s, g[2*s], i-2*s, j-s, t[i-2*s][j-s], C(i, 2*s)*g[2*s]%mod*t[i-2*s][j-s]);
}
f[i][j]=(f[i][j]+f[i][j-1])%mod;
// printf("f[%d][%d]=%lld\n", i, j, f[i][j]);
}
}
// cout<<"dlt: "<<d[4]-g[4]<<endl;
printf("%lld\n", f[n][k]);
exit(0);
}
}
namespace task{
int f[N<<1][N];
ll inv[N<<1], ans;
ll C(ll n, ll k) {
if (n<k) return 0ll;
ll ans=inv[k];
for (ll i=n; i>n-k; --i) ans=ans*i%mod;
return ans;
}
void solve() {
// cout<<double(sizeof(f))/1024/1024<<endl;
inv[0]=inv[1]=1; f[0][0]=1;
for (int i=2; i<=k*2; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
for (int i=2; i<=k*2; ++i) inv[i]=inv[i-1]*inv[i]%mod;
for (int j=1; j<=k; ++j) for (int i=2; i<=k*2; ++i) f[i][j]=1ll*(i-1)*(f[i-1][j-1]+f[i-2][j-1])%mod; //, printf("f[%d][%d]=%lld\n", i, j, f[i][j]);
for (int i=0; i<=k*2; ++i) {
ll tem=0;
// cout<<"tem: "<<tem<<endl;
for (int j=0; j<=k; ++j) tem=(tem+f[i][j])%mod;
ans=(ans+C(n, i)*tem%mod)%mod;
}
printf("%lld\n", ans);
exit(0);
}
}
signed main()
{
freopen("guess.in", "r", stdin);
freopen("guess.out", "w", stdout);
n=read(); k=read();
// if (k<=3) task1::solve();
// else force::solve();
task::solve();
return 0;
}