CF 317A(Perfect Pair-广义Fib序列p,q=1性质2&加法增长极)

最新推荐文章于 2021-06-12 23:14:57 发布

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本文介绍了一个算法问题，即如何通过一系列操作使一对整数变为m-完美对。详细解释了问题背景及输入输出要求，并给出了一个C++实现示例，包括对特殊情况的处理方法。

Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).

What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?

Input

Single line of the input contains three integers x, y and m ( - 10¹⁸ ≤ x, y, m ≤ 10¹⁸).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64dspecifier.

Output

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

Sample test(s)

input
1 2 5

output
2

input
-1 4 15

output
4

input
0 -1 5

output
-1

Note

In the first sample the following sequence of operations is suitable: (1, 2) $\text{[math]}$ (3, 2) $\text{[math]}$ (5, 2).

In the second sample: (-1, 4) $\text{[math]}$ (3, 4) $\text{[math]}$ (7, 4) $\text{[math]}$ (11, 4) $\text{[math]}$ (15, 4).

Finally, in the third sample x, y cannot be made positive, hence there is no proper sequence of operations.

首先，这题尽管我用了Sx的公式（可二分），但是可以暴力（加法增长极很大）

只需要把负数弄好就行.

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define INF (1e18)
#define MAXN (1000000)
long long x,y,m;
long long f[MAXN]={0,1,1};
int n=0;
long long work()
{
	if (max(x,y)>=m) return 0;
	long long j=0;
	if (x==0&&y==0) return -1;
	if (x<0&&y<0) return -1;
	bool b=0;
	while (max(x,y)<m)
	{
		if (x+y<=min(x,y)) return -1;
		if (x>y) swap(x,y);
		if (x>0&&y>0)
		{
			int k=1;
			while (f[k]*x+f[k+1]*y<m) k++;
			j+=k;
			return j; 
		}	
		else if (!b&&x<0&&y>0)
		{
			long long k=-x/y;
			if (m<0) k=(m-x)/y;
			j+=k;
			x+=k*y;
			b=1;
		}
		else
		{
			j++;
			x+=y;
		}	
	}
	return j;
}
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);

	for(n=3;f[n-1]<INF;n++) f[n]=f[n-1]+f[n-2];
	n--;
//	For(i,n) cout<<f[i]<<' ';
	while (cin>>x>>y>>m)
		cout<<work()<<endl;

	return 0;
}