A. Perfect Pair
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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) (3, 2) (5, 2).
In the second sample: (-1, 4) (3, 4) (7, 4) (11, 4) (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.
[cpp]
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;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)
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
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]
// For(i,n) cout<
cout<
return 0;
}
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;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)
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
return j;
}
else if (!b&&x