#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN=2,mod=1000000007 ;
struct Matrix
{
ll mat[MAXN][MAXN];
Matrix() {}
Matrix operator*(Matrix const &b)const
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0 ;i < MAXN; i++)
for (int j = 0; j < MAXN; j++)
for (int k = 0; k < MAXN; k++)
res.mat[i][j] = (res.mat[i][j]+this->mat[i][k] * b.mat[k][j])%mod;
return res;
}
};
Matrix pow_mod(Matrix base, int n)
{
Matrix res;
memset(res.mat, 0, sizeof(res.mat));
for (int i = 0; i < MAXN; i++)
res.mat[i][i] = 1;
while (n > 0)
{
if (n & 1) res = res*base;
base = base*base;
n >>= 1;
}
return res;
}
Matrix base,fi;
int main()
{
base.mat[0][0] = 1;
base.mat[0][1] = 1;
base.mat[1][0] = 1;
fi.mat[0][0]=1;
fi.mat[1][0]=1;
int n;
while (~scanf("%d", &n)&&n)
{
if(n<=2)
{
printf("%d\n",1);
continue;
}
Matrix ans = pow_mod(base, n-2);
ans=ans*fi;
printf("%lld\n", ans.mat[0][0]);
}
return 0;
}
斐波那契数大数求法(模板)
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转载自blog.csdn.net/qq_41286356/article/details/89437065
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