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1. 传统的方法 递推公式 f[n]=f[n-1]+f[n-2] n>2
2. 运用矩阵乘法求解
#include
using namespace std;
const int N=2;
const int mod=10000007;
struct M{
long long int m[N][N];
};
M A= //需要根据题目求的矩阵
{
1,1,
1,0
};
M B= //单位矩阵
{
1,0,
0,1
};
M fun(M a,M b) //矩阵乘法
{
M c;
for(int i=0;i>=1;
p=fun(p,p);
}
return ans;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
M ans=power(n-1);
printf("%lld
",ans.m[0][0]);
}
return 0;
}