bzoj1002 [FJOI2007]轮状病毒 矩阵树定理

Description

---

求这样一个东西的生成树方案数量

Solution

---

正解可能是dp啥的，直接上矩阵树定理然后套高精度就完事儿了
如果把矩阵写出来可以发现对角线上全是3，然后两边都是-1。找一波规律可以发现f[n]=3f[n-1]-f[n-2]+2

Code

---

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (int i=st;i<=ed;++i)
#define drp(i,st,ed) for (int i=st;i>=ed;--i)
#define fill(x,t) memset(x,t,sizeof(x))

const int MOD=1000;
const int N=105;
const int L=105;

struct num {
	int s[L],len;

	inline bool operator ==(num b) {
		num a=*this;
		if (a.len!=b.len) return 0;
		drp(i,a.len,1) if (a.s[i]!=b.s[i]) return 0;
		return 1;
	}
	inline bool operator !=(num b) {
		return !(*this==b);
	}

	inline bool operator <(num b) {
		num a=*this;
		if (a.len<b.len) return 1;
		if (a.len>b.len) return 0;
		drp(i,a.len,1) {
			if (a.s[i]<b.s[i]) return 1;
			else if (a.s[i]>b.s[i]) return 0;
		}
		return 0;
	}
	inline bool operator <=(num b) {
		num a=*this;
		if (a<b||a==b) return 1;
		return 0;
	}
	inline bool operator >(num b) {
		num a=*this;
		if (a.len>b.len) return 1;
		if (a.len<b.len) return 0;
		drp(i,a.len,1) {
			if (a.s[i]>b.s[i]) return 1;
			else if (a.s[i]<b.s[i]) return 0;
		}
		return 0;
	}
	inline bool operator >=(num b) {
		num a=*this;
		if (a>b||a==b) return 1;
		return 0;
	}
	inline num operator +(num b) {
		num a=*this,c=(num) { {0},std:: max(a.len,b.len)};
		int v=0;
		rep(i,1,c.len) {
			c.s[i]=(a.s[i]+b.s[i]+v)%MOD;
			v=(a.s[i]+b.s[i]+v)/MOD;
		}
		if (v) {
			c.len+=1;
			c.s[c.len]=v;
		}
		return c;
	}
	inline num operator -(num b) {
		num a=*this,c=(num) { {0},std:: max(a.len,b.len)};
		rep(i,1,c.len) {
			c.s[i]=a.s[i]-b.s[i];
			if (c.s[i]<0) {
				c.s[i]+=MOD;
				a.s[i+1]-=1;
			}
		}
		while (!c.s[c.len]&&c.len>1) c.len-=1;
		return c;
	}
	inline num operator *(num b) {
		num a=*this,c=(num) { {0},a.len+b.len};
		rep(i,1,a.len) {
			rep(j,1,b.len) {
				c.s[i+j-1]+=a.s[i]*b.s[j];
			}
		}
		rep(i,1,a.len+b.len) {
			c.s[i+1]+=c.s[i]/MOD;
			c.s[i]%=MOD;
		}
		while (!c.s[c.len]&&c.len>1) c.len-=1;
		return c;
	}
	inline num operator /(int b) {
		num a=*this,c=(num) { {0},a.len};
		int v=0;
		drp(i,len,1) {
			int t=v*MOD+a.s[i];
			c.s[i]=t/b;
			v=t%b;
		}
		while (!c.s[c.len]&&c.len>1) c.len-=1;
		return c;
	}
	inline void read() {
		fill(s,0); len=0;
		char st[L];
		scanf("%s",st);
		int v=0,i;
		for (i=strlen(st)-1; i >= 3; i-=3) {
			rep(j,i-2,i) v=v*10+st[j]-'0';
			s[++len]=v;
			v=0;
		}
		rep(j,0,i) v=v*10+st[j]-'0';
		s[++len]=v;
	}
	inline void output() {
		num tmp=*this;
		int i=tmp.len;
		while (!tmp.s[i]&&i>1) {
			i-=1;
		}
		printf("%d",tmp.s[i]);
		drp(j,i-1,1) {
			int v=tmp.s[j];
			int f[5]= {0};
			rep(k,1,3) {
				f[k]=v%10;
				v/=10;
			}
			drp(k,3,1) printf("%d",f[k]);
		}
		printf("\n");
	}
} f[N],two,thr;

int main(void) {
	int n; scanf("%d",&n);
	two.len=1; two.s[1]=2;
	thr.len=1; thr.s[1]=3;
	f[1].len=f[1].s[1]=f[2].len=1;
	f[2].s[1]=5;
	rep(i,3,n) f[i]=f[i-1]*thr-f[i-2]+two;
	f[n].output();
	return 0;
}