UVA 1647 - Computer Transformation（大数 找规律）

这个题 就是 找规律。 ?由于位数太多 直接手写会乱的。 ?并且 效率太低。 我就 写了个暴利的程序。

找出规律 ： n为奇数的时候 ?a【n】 = a【n-1】 - 1， 偶数的时候 a【n】 = a【n-1】+1. ?便可以解决了

暴利的程序是 ?：

#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
#include <cmath>
#include <algorithm>
#include <cstdlib>
using namespace std;
#define ll long long
#define maxn 2000 + 10
int main (){

    for(int i = 1; i <= 10; i++){
        vector <char> q;
        q.push_back('1');
        for(int j = 1; j <= i; j++){
            vector <char> p;
            for(int k = 0; k < q.size(); k++){
                if(q[k] == '1'){
                    p.push_back('0');
                    p.push_back('1');
                }
                if(q[k] == '0'){
                    p.push_back('1');
                    p.push_back('0');
                }
            }
            q = p;
        }
        int sum = 0;
        for(int j = 0; j < q.size()-1 ; j ++){
            if(q[j] == '0' && q[j+1] == '0')
            sum ++;
        }
        printf("i = %d  sum = %dn",i,sum);
    }
    return 0;
}

AC的程序是 ：

 #include<iostream>
#include<string>
#include<iomanip>
#include<algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
	int a[500];
	int len;
public:
	BigNum(){ len = 1;memset(a,sizeof(a)); }
	BigNum(const int);
	BigNum(const char*);
	BigNum(const BigNum &);
	BigNum &operator=(const BigNum &);

	friend istream& operator>>(istream&,BigNum&);
	friend ostream& operator<<(ostream&,BigNum&);

	BigNum operator+(const BigNum &) const;
	BigNum operator-(const BigNum &) const;
	BigNum operator*(const BigNum &) const;
	BigNum operator/(const int   &) const;

	BigNum operator^(const int  &) const;
	int    operator%(const int  &) const;
	bool   operator>(const BigNum & T)const;
	bool   operator>(const int & t)const;

	void print();
};
BigNum::BigNum(const int b)
{
	int c,d = b;
	len = 0;
	memset(a,sizeof(a));
	while(d > MAXN)
	{
		c = d - (d / (MAXN + 1)) * (MAXN + 1);
		d = d / (MAXN + 1);
		a[len++] = c;
	}
	a[len++] = d;
}
BigNum::BigNum(const char*s)
{
	int t,k,index,l,i;
	memset(a,sizeof(a));
	l=strlen(s);
	len=l/DLEN;
	if(l%DLEN)
		len++;
	index=0;
	for(i=l-1;i>=0;i-=DLEN)
	{
		t=0;
		k=i-DLEN+1;
		if(k<0)
			k=0;
		for(int j=k;j<=i;j++)
			t=t*10+s[j]-'0';
		a[index++]=t;
	}
}
BigNum::BigNum(const BigNum & T) : len(T.len)
{
	int i;
	memset(a,sizeof(a));
	for(i = 0 ; i < len ; i++)
		a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)
{
	int i;
	len = n.len;
	memset(a,sizeof(a));
	for(i = 0 ; i < len ; i++)
		a[i] = n.a[i];
	return *this;
}
istream& operator>>(istream & in,BigNum & b)
{
	char ch[MAXSIZE*4];
	int i = -1;
	in>>ch;
	int l=strlen(ch);
	int count=0,sum=0;
	for(i=l-1;i>=0;)
	{
		sum = 0;
		int t=1;
		for(int j=0;j<4&&i>=0;j++,i--,t*=10)
		{
			sum+=(ch[i]-'0')*t;
		}
		b.a[count]=sum;
		count++;
	}
	b.len =count++;
	return in;

}
ostream& operator<<(ostream& out,BigNum& b)
{
	int i;
	cout << b.a[b.len - 1];
	for(i = b.len - 2 ; i >= 0 ; i--)
	{
		cout.width(DLEN);
		cout.fill('0');
		cout << b.a[i];
	}
	return out;
}

BigNum BigNum::operator+(const BigNum & T) const
{
	BigNum t(*this);
	int i,big;      //位数
	big = T.len > len ? T.len : len;
	for(i = 0 ; i < big ; i++)
	{
		t.a[i] +=T.a[i];
		if(t.a[i] > MAXN)
		{
			t.a[i + 1]++;
			t.a[i] -=MAXN+1;
		}
	}
	if(t.a[big] != 0)
		t.len = big + 1;
	else
		t.len = big;
	return t;
}
BigNum BigNum::operator-(const BigNum & T) const
{
	int i,j,big;
	bool flag;
	BigNum t1,t2;
	if(*this>T)
	{
		t1=*this;
		t2=T;
		flag=0;
	}
	else
	{
		t1=T;
		t2=*this;
		flag=1;
	}
	big=t1.len;
	for(i = 0 ; i < big ; i++)
	{
		if(t1.a[i] < t2.a[i])
		{
			j = i + 1;
			while(t1.a[j] == 0)
				j++;
			t1.a[j--]--;
			while(j > i)
				t1.a[j--] += MAXN;
			t1.a[i] += MAXN + 1 - t2.a[i];
		}
		else
			t1.a[i] -= t2.a[i];
	}
	t1.len = big;
	while(t1.a[len - 1] == 0 && t1.len > 1)
	{
		t1.len--;
		big--;
	}
	if(flag)
		t1.a[big-1]=0-t1.a[big-1];
	return t1;
}

BigNum BigNum::operator*(const BigNum & T) const
{
	BigNum ret;
	int i,up;
	int temp,temp1;
	for(i = 0 ; i < len ; i++)
	{
		up = 0;
		for(j = 0 ; j < T.len ; j++)
		{
			temp = a[i] * T.a[j] + ret.a[i + j] + up;
			if(temp > MAXN)
			{
				temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
				up = temp / (MAXN + 1);
				ret.a[i + j] = temp1;
			}
			else
			{
				up = 0;
				ret.a[i + j] = temp;
			}
		}
		if(up != 0)
			ret.a[i + j] = up;
	}
	ret.len = i + j;
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--;
	return ret;
}
BigNum BigNum::operator/(const int & b) const
{
	BigNum ret;
	int i,down = 0;
	for(i = len - 1 ; i >= 0 ; i--)
	{
		ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
		down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
	}
	ret.len = len;
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--;
	return ret;
}
int BigNum::operator %(const int & b) const
{
	int i,d=0;
	for (i = len-1; i>=0; i--)
	{
		d = ((d * (MAXN+1))% b + a[i])% b;
	}
	return d;
}
BigNum BigNum::operator^(const int & n) const
{
	BigNum t,ret(1);
	int i;
	if(n<0)
		exit(-1);
	if(n==0)
		return 1;
	if(n==1)
		return *this;
	int m=n;
	while(m>1)
	{
		t=*this;
		for( i=1;i<<1<=m;i<<=1)
		{
			t=t*t;
		}
		m-=i;
		ret=ret*t;
		if(m==1)
			ret=ret*(*this);
	}
	return ret;
}
bool BigNum::operator>(const BigNum & T) const
{
	int ln;
	if(len > T.len)
		return true;
	else if(len == T.len)
	{
		ln = len - 1;
		while(a[ln] == T.a[ln] && ln >= 0)
			ln--;
		if(ln >= 0 && a[ln] > T.a[ln])
			return true;
		else
			return false;
	}
	else
		return false;
}
bool BigNum::operator >(const int & t) const
{
	BigNum b(t);
	return *this>b;
}

void BigNum::print()
{
	int i;
	cout << a[len - 1];
	for(i = len - 2 ; i >= 0 ; i--)
	{
		cout.width(DLEN);
		cout.fill('0');
		cout << a[i];
	}
	cout << endl;
}
BigNum p[1010];
int init()
{
    p[1] = 0;
    for(int i = 2; i <= 1000; i++){
        if(i % 2 == 1)
            p[i] = p[i-1]*2 - 1;
        else
            p[i] = p[i-1]*2 + 1;
    }
}
int main ()
{
    int n;
    init();
    while(scanf("%d",&n)!=EOF)
    {
        p[n].print();
    }
    return 0;
}