M斐波那契數列
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2416 Accepted Submission(s): 701
Problem Description M斐波那契數列F[n]是一種整數數列,它的定義如下:
F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )
現在給出a, b, n,你能求出F[n]的值嗎?
Input 輸入包含多組測試資料;
每組資料占一行,包含3個整數a, b, n( 0 <= a, b, n <= 10^9 )
Output 對每組測試資料請輸出一個整數F[n],由于F[n]可能很大,你隻需輸出F[n]對1000000007取模後的值即可,每組資料輸出一行。
Sample Input
0 1 0
6 10 2
Sample Output
0
60
思路:
先找規律 f(0)=a (1,0)
f(1)=b; (0,1)
f(2)=ab (1,1)
f(3)=abb (1,2)
f(4)=abbab (2,3)
f(5)=abbababb (3,5)
f(6)=abbababbabbab (5,8)
.......
由上面的規律可以看出 f[n]=a^m0*b^m1%M.(m0,m1都為斐波那契數列)
是以先求出m0,m1,再求出 a^m0*b^m1%M就行了。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 2
#define M 1000000007
using namespace std;
ll a,b,n;
struct mat
{
ll m[N][N];
};
mat A=
{
1,1,
1,0
};
mat I=
{
1,0,
0,1
};
mat multi(mat a,mat b)//兩矩陣相乘
{
int i,j,k;
mat c;
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
{
c.m[i][j]=0;
for(k=0;k<N;k++)
{
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j]%(M-1))%(M-1);
}
}
}
return c;
}
mat power(mat A,int k)//矩陣快速幂
{
mat ans=I,p=A;
while(k)
{
if(k&1)
ans=multi(ans,p);
p=multi(p,p);
k>>=1;
}
return ans;
}
ll ppower(ll k,ll m)//整數快速幂
{
ll ans=1;
while(m)
{
if(m&1)
ans=(ans*k%M);
k=(k*k%M);
m>>=1;
}
return ans;
}
int main()
{
while(scanf("%d%d%d",&a,&b,&n)!=EOF)
{
if(n==0)
{
printf("%lld\n",a);
continue;
}
mat ans=power(A,n-1);
ll ansa=ppower(a,ans.m[0][1]);
ll ansb=ppower(b,ans.m[0][0]);
ll s=(ansa*ansb)%M;
printf("%lld\n",s);
}
return 0;
}