jzoj5765 【省選模擬8.5】相互再歸的鵝媽媽 (集合劃分,斯特林反演)
阿新 • • 發佈:2019-02-08
解法
先考慮可以有相同怎麼做:
列舉一個第一個脫離限制的位置,然後用一個脫離限制的數來安排使得異或和為0,其他數可以任意取(要分是否脫離限制確定方案數)。這樣可以計算出g(n)表示n個可以相同的數,異或和為0的答案。
斯特林反演式子:
證明非常簡單,右側是圓周排列加上一個正負1的係數
等式右邊也就是
不難得出的生成函式就是
令x=1,那麼等式右側就是,即
我們要求的就是滿足根據這個反演一波即可。
#include <cstdio>
#include <iostream>
#include <cstring>
#define lowbit(x) ((x) & -(x))
using namespace std;
typedef long long ll;
const ll mo = 1e9 + 7;
ll n, m, k, len;
char s[50100];
ll a[5000100],suf[5000100],mi[5000100];
ll f[8][2][2],g[8],ans,jc[8],mir[8];
ll q[8],size[8];
void dfs(ll x,ll c) {
if (x > n) {
memset(size,0,sizeof size);
for (ll i = 1; i <= n; i++) size[q[i] ]++;
ll eve = 0, xs = 1;
for (ll i = 1; i <= c; i++) {
if (size[i] % 2 == 0)
eve++;
xs = xs * jc[size[i] - 1] % mo * ((size[i]-1)%2==0?1:-1) % mo;
}
ll odd = c - eve;
(ans += xs * g[odd] % mo * mir[eve]) %= mo;
return;
}
q[x] = c + 1;
dfs(x + 1, c + 1);
for (ll i = 1; i <= c; i++) {
q[x] = i;
dfs(x + 1, c);
}
}
ll ksm(ll x,ll y) {
ll ret = 1;
for (; y; y>>=1) {
if (y & 1) ret = ret * x % mo;
x = x * x % mo;
}
return ret;
}
int main() {
freopen("mothergoose.in","r",stdin);
freopen("mothergoose.out","w",stdout);
cin>>n>>k;
jc[0] = 1;
for (ll i = 1; i <= n; i++) jc[i] = jc[i - 1] * i;
scanf("%s",s + 1); m = strlen(s + 1);
for (ll i = 1; i <= k; i++) for (ll j = 1; j <= m; j++) {
a[++len] = s[j] - '0';
}
ll w = 1; mi[len+1] = 1;
for (ll i = len; i; i--, w = w * 2 % mo) {
suf[i] = (suf[i+1] + w * a[i]) % mo;
mi[i] = w * 2 % mo;
}
mir[0] = 1;
for (ll i = 1; i <= n; i++) mir[i] = mir[i - 1] * suf[1] % mo;
ll hasone = 0;
for (ll i = 1; i <= len; i++) if (a[i] == 1) {
memset(f,0,sizeof f);
f[0][0][0] = 1;
for (ll j = 1; j <= n; j++) {
for (ll z = 0; z < 2; z++) //odd or even
for (ll e = 0; e < 2; e++) { //has or not
(f[j][z^1][e] += f[j-1][z][e] * suf[i+1]) %=mo;
(f[j][z][1] += f[j-1][z][e] * (e == 1 ? mi[i+1] : 1)) %= mo;
}
if (j%2==0 || !hasone) {
(g[j] += f[j][0][1]) %= mo;
}
}
hasone = 1;
}
g[0] = 1;
dfs(1,0); ans = (ans + mo) % mo;
cout<<ans * ksm(jc[n], mo - 2) % mo<<endl;
}