n<=100的形如的图有多少生成树。不取模。
$f(i)=3*f(i-1)-f(i-2)+2$,VFK的题解
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #include<stdlib.h> 5 //#include<queue> 6 #include<math.h> 7 //#include<time.h> 8 //#include<iostream> 9 using namespace std; 10 11 int n; 12 #define maxn 311 13 const int base=100000000; 14 struct LLL 15 { 16 int num[maxn],len; 17 LLL() {len=0; memset(num,0,sizeof(num));} 18 void operator = (int x) 19 { 20 if (x==0) {num[len=1]=0; return;} 21 while (x) {num[++len]=x%base; x/=base;} 22 } 23 LLL operator + (const LLL &b) 24 { 25 LLL ans; 26 for (int i=1,top=ans.len=max(len,b.len);i<=top;i++) 27 { 28 ans.num[i]+=(i<=len?num[i]:0)+(i<=b.len?b.num[i]:0); 29 if (ans.num[i]>=base) 30 { 31 ans.num[i]-=base; 32 ans.num[i+1]++; 33 } 34 } 35 ans.len+=(ans.num[ans.len+1]>0); 36 return ans; 37 } 38 LLL operator - (const LLL &b) 39 { 40 LLL ans; 41 for (int i=1,top=ans.len=len;i<=top;i++) 42 { 43 ans.num[i]+=num[i]-(i<=b.len?b.num[i]:0); 44 if (ans.num[i]<0) 45 { 46 ans.num[i]+=base; 47 ans.num[i+1]--; 48 } 49 } 50 ans.len-=(ans.num[ans.len]==0); 51 return ans; 52 } 53 LLL operator + (int b) 54 { 55 LLL tmp; tmp=b; 56 return *this+tmp; 57 } 58 void print() 59 { 60 printf("%d",num[len]); 61 for (int i=len-1;i;i--) printf("%08d",num[i]); 62 } 63 }; 64 65 LLL a[maxn]; 66 int main() 67 { 68 scanf("%d",&n); 69 a[1]=1; a[2]=5; 70 for (int i=3;i<=n;i++) a[i]=a[i-1]+a[i-1]+a[i-1]-a[i-2]+2; 71 a[n].print(); 72 return 0; 73 }