1002: [FJOI2007]輪狀病毒

Time Limit: 1 Sec??Memory Limit: 162 MB
Submit: 4746??Solved: 2597
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Description

　　輪狀病毒有很多變種，所有輪狀病毒的變種都是從一個輪狀基產生的。一個N輪狀基由圓環上N個不同的基原子
和圓心處一個核原子構成的，2個原子之間的邊表示這2個原子之間的信息通道。如下圖所示

　　N輪狀病毒的產生規律是在一個N輪狀基中刪去若干條邊，使得各原子之間有唯一的信息通道，例如共有16個不
同的3輪狀病毒，如下圖所示

現給定n(N<=100)，編程計算有多少個不同的n輪狀病毒

Input

　　第一行有1個正整數n

Output

　　計算出的不同的n輪狀病毒數輸出

Sample Input

3

Sample Output

16

學習：2007周冬《生成樹計數》 論文

題解：推公式，f[n] = 3*f[n-1] - f[n-2] + 2。高精度模擬

代碼：

/**************************************************************Problem: 1002User: JstyleLanguage: C++Result: AcceptedTime:44 msMemory:1456 kb ****************************************************************/#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <bitset> #include <vector> #include <queue> #include <stack> #include <cmath> #include <list> #include <set> #include <map> #define rep(i,a,b) for(int (i) = (a);(i) <= (b);++ (i)) #define per(i,a,b) for(int (i) = (a);(i) >= (b);-- (i)) #define mem(a,b) memset((a),(b),sizeof((a))) #define FIN freopen("in.txt","r",stdin) #define FOUT freopen("out.txt","w",stdout) #define IO ios_base::sync_with_stdio(0),cin.tie(0) #define mid ((l+r)>>1) #define ls (id<<1) #define rs ((id<<1)|1) #define N 100000+5 #define INF 0x3f3f3f3f #define INFF 0x3f3f3f3f3f3f3f typedef long long ll; const ll mod = 20071027; const ll eps = 1e-12; using namespace std;const int MAXN = 410; struct bign { int len, s[MAXN]; bign () { memset(s, 0, sizeof(s)); len = 1; } bign (int num) { *this = num; } bign (const char *num) { *this = num; } bign operator = (const int num) { char s[MAXN]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { for(int i = 0; num[i] == '0'; num++) ; //去前導0 len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } bign operator + (const bign &b) const //+ { bign c; c.len = 0; for(int i = 0, g = 0; g || i < max(len, b.len); i++) { int x = g; if(i < len) x += s[i]; if(i < b.len) x += b.s[i]; c.s[c.len++] = x % 10; g = x / 10; } return c; } bign operator += (const bign &b) { *this = *this + b; return *this; } void clean() { while(len > 1 && !s[len-1]) len--; } bign operator * (const bign &b) //* { bign c; c.len = len + b.len; for(int i = 0; i < len; i++) { for(int j = 0; j < b.len; j++) { c.s[i+j] += s[i] * b.s[j]; } } for(int i = 0; i < c.len; i++) { c.s[i+1] += c.s[i]/10; c.s[i] %= 10; } c.clean(); return c; } bign operator *= (const bign &b) { *this = *this * b; return *this; } bign operator - (const bign &b) { bign c; c.len = 0; for(int i = 0, g = 0; i < len; i++) { int x = s[i] - g; if(i < b.len) x -= b.s[i]; if(x >= 0) g = 0; else { g = 1; x += 10; } c.s[c.len++] = x; } c.clean(); return c; } bign operator -= (const bign &b) { *this = *this - b; return *this; } bign operator / (const bign &b) { bign c, f = 0; for(int i = len-1; i >= 0; i--) { f = f*10; f.s[0] = s[i]; while(f >= b) { f -= b; c.s[i]++; } } c.len = len; c.clean(); return c; } bign operator /= (const bign &b) { *this = *this / b; return *this; } bign operator % (const bign &b) { bign r = *this / b; r = *this - r*b; return r; } bign operator %= (const bign &b) { *this = *this % b; return *this; } bool operator < (const bign &b) { if(len != b.len) return len < b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] < b.s[i]; } return false; } bool operator > (const bign &b) { if(len != b.len) return len > b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] > b.s[i]; } return false; } bool operator == (const bign &b) { return !(*this > b) && !(*this < b); } bool operator != (const bign &b) { return !(*this == b); } bool operator <= (const bign &b) { return *this < b || *this == b; } bool operator >= (const bign &b) { return *this > b || *this == b; } string str() const { string res = ""; for(int i = 0; i < len; i++) res = char(s[i]+'0') + res; return res; } }; istream& operator >> (istream &in, bign &x) { string s; in >> s; x = s.c_str(); return in; } ostream& operator << (ostream &out, const bign &x) { out << x.str(); return out; } int n; bign dp[105]; void fuc(){bign a = "3",b = "2";dp[1] = "1";dp[2] = "5";rep(i, 3, 100)dp[i] = a*dp[i-1]-dp[i-2]+b; } int main() {fuc();while(cin >> n)cout << dp[n] << endl;return 0; }

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