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1002: [FJOI2007]轮状病毒

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

　　轮状病毒有很多变种，所有轮状病毒的变种都是从一个轮状基产生的。一个N轮状基由圆环上N个不同的基原子
和圆心处一个核原子构成的，2个原子之间的边表示这2个原子之间的信息通道。如下图所示

　　N轮状病毒的产生规律是在一个N轮状基中删去若干条边，使得各原子之间有唯一的信息通道，例如共有16个不
同的3轮状病毒，如下图所示

　　现给定n(N<=100)，编程计算有多少个不同的n轮状病毒

Input

　　第一行有1个正整数n

Output

　　计算出的不同的n轮状病毒数输出

Sample Input

3

Sample Output

16

学习：2007周冬《生成树计数》 论文

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

代码：

/**************************************************************
    Problem: 1002
    User: Jstyle
    Language: C++
    Result: Accepted
    Time:44 ms
    Memory: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++) ;  //去前导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;  
}