Logu P1306 Fibonacci common count, p1306 Fibonacci

Logu P1306 Fibonacci common count, p1306 Fibonacci
Description

You should be familiar with the series of Fibonacci:, 13 ~~~ But now there is a very simple question: what is the maximum number of public appointments between the n and m items?

Input/Output Format

Input Format:

 

Two positive integers n and m. (N, m <= 10 ^ 9)

Note: The data is large.

 

Output Format:

 

The maximum number of common Fn and Fm.

Since reading big numbers makes you dizzy, you only need to output the last 8 digits.

 

Input and Output sample input sample #1: Copy

4 7

Output example #1: Copy

1

Description

Recursion and recurrence will time out.

The formula also times out.

 

 

 

Extended Euclidean has a very important nature

$ Gcd (F [I], F [j]) = F [gcd (I, j)] $

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #define LL long long  5 using namespace std; 6 const LL mod=100000000; 7 const LL MAXN=5000001; 8 inline LL read() 9 {10     char c=getchar();LL x=0,flag=1;11     while(c<'0'||c>'9')    {if(c=='-')    flag=-1;c=getchar();}12     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*flag;13 }14 LL dp[MAXN];15 LL gcd(LL a,LL b)16 {17     return b==0?a:gcd(b,a%b);18 }19 int main()20 {21     LL n=read(),m=read();22     dp[1]=1;dp[2]=1;23     for(LL i=3;i<=5000000;i++)24         dp[i]=(dp[i-1]+dp[i-2])%100000000;25         26     LL p=gcd(n,m);27     printf("%lld",dp[p]%100000000);28     return 0;29 }