Codeforces Round #188 (Div. 2) C. Perfect Pair 数学
B. Strings of Power
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/318/problem/C
Description
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?
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 %I64d specifier.
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 %I64d specifier.
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 Input
1 2 5
Sample Output
2
HINT
题意
给你x,y,每次你可以选择一个数,让他变成x+y,然后问最少多少步,可以使得x,y中的最大值大于等于k
题解:
这道题类似于fib数,所以我们只要把(x,y)变成(y,x+y)就好
注意,全程爆ll
代码:
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 4000001 #define mod 10007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //************************************************************************************** int main() { ll x=read(),y=read(),k=read(); if(x>=k||y>=k) { cout<<"0"<<endl; return 0; } if(x<k&&y<k&&x<=0&&y<=0) { cout<<"-1"<<endl; return 0; } ll ans=0; if(x>y) swap(x,y); if(x<0) { ans=(y-x)/y; x+=ans*y; } while(y<k) { ll tmp=x; x=y; y=tmp+y; ans++; } cout<<ans<<endl; }