Title effect: to give you a n, m (1 <= n, m <= 1e9), seeking __gcd (F [n], F [m])% 1e8.
First of all I am here to propose a theorem of number theory: __ gcd (F [n], F [m]) = F [__ gcd (n, m)].
Knowing this fact, this problem in number theory Theorem becomes very easy, simply requires the greatest common divisor of n, m, and acceleration demand using matrix F [__ gcd (n, m)] of the modulo to 1e8.
But why the theorem is correct?
Certify as follows:
Provided n <m, and F. [N-] = A, F. [N-+. 1] = B
F[n+2]=a+b
F[n+3]=a+2*b=F[n]+2*F[n+1]
F[n+4]=2*a+3*b=2*F[n]+3*F[n+1]
F[n+5]=3*a+5*b=3*F[n]+5*F[n+1]
...
Difficult to find F [n + x] = F [x-1] * F [n] + F [x] * F [n + 1]
F[m]=F[m-n-1]*F[n]+F[m-n]*F[n+1]
所以 __gcd（F[n]，F[m]）=__gcd（F[n]，F[m-n-1]*F[n]+F[m-n]*F[n+1]）;
Since F [mn-1] * F [n]% F [n] == 0
Therefore __gcd ( F. [N-], F. [M]) = __ GCD ( F. [N-], F. [Mn] F. * [N-+. 1]);
Proof road seems stuck here, because this formula is difficult to go simplification, while the single from this equation is difficult to see anything, then we need another lemma: __ gcd (F [n] , F [n + 1]) = 1;
Proof of Lemma is relatively simple:
__gcd（F[n]，F[n+1]）=__gcd（F[n]，F[n]+F[n-1]）
=__gcd（F[n]，F[n-1]）
=__gcd（F[n-1]+F[n-2]，F[n-1]）
=__gcd（F[n-2]，F[n-1]）
......
=__gcd（F[1]，F[2]）=1
We draw __gcd (F [n], F [n + 1]) = 1 after the can continue __gcd ( F. [N-], F. [M]) = __gcd ( F. [N-], F. [Mn] * F [n + 1]) for simplification
Finally __gcd ( F. [N-], F. [M]) = __gcd ( F. [N-], F. [Mn])
=__gcd（F[n]，F[m%n]）
......
=F[__gcd（n，m）]
Thus, proof of completion ~
The following code is placed on this topic AC
#include<bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<int,int> PII; const int MAXN = 1e6+10; const double EPS = 1e-12; const ll mod = 1e8; ll n,mm; struct Mat{ ll m[5][5]; Mat(){ memset(m,0,sizeof(m)); } inline void build(){ for(int i=1;i<=2;i++)m[i][i]=1; } }a; Food Mul (Food x, Food y) { Mat c; for(int i=1;i<=2;i++) for(int j=1;j<=2;j++) for(int k=1;k<=2;k++) c.m[j][i]=(c.m[j][i]+x.m[j][k]*y.m[k][i]%mod)%mod; return c; } Mat poww(Mat x,ll y){ Mat aa;aa.build(); while(y){ if(y&1)aa=Mul(aa,x); x=Mul(x,x); y>>=1; } return aa; } int main () { scanf("%lld %lld",&n,&mm); ll p=__gcd(n,mm); a.m[1][1]=a.m[2][1]=a.m[1][2]=1; With ans = poww (a, p 1 ); printf("%lld",ans.m[1][1]); }