According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
Hence they prefer problems short, too. Here is a short one:
Given n (1 <= n <= 10 18), You should solve for
g(g(g(n))) mod 10 9 + 7
where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0
Input
There are several test cases. For each test case there is an integer n in a single line.
Please process until EOF (End Of File).
Output
For each test case, please print a single line with a integer, the corresponding answer to this case.
Sample Input
0 1 2
Sample Output
0 1 42837
题解：题目给出递推关系式：g[n]=3*g[n-1]+g[n-2],则可构造规律矩阵a如下
3 1
1 0
初始矩阵b
1 0
0 0
运用矩阵快速幂即可，注意考虑0,1特殊情况。
#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
typedef long long ll;
ll mod1=1000000007;
ll mod2=222222224;
ll mod3=183120;
ll a[3][3],b[3][3],c[3][3],n,ans;
ll f(ll n,ll mod)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(c));
a[1][1]=3,a[1][2]=1;
a[2][1]=1,a[2][2]=0;
b[1][1]=1;
n--;
while(n)
{
if(n&1)
{
memset(c,0,sizeof(c));
for(int k=1; k<=2; k++)
{
for(int i=1; i<=2; i++)
{
//if(a[i][k])
for(int j=1; j<=2; j++)
{
c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mod;
}
}
}
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
b[i][j]=c[i][j];
}
}
}
memset(c,0,sizeof(c));
for(int k=1; k<=2; k++)
{
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
c[i][j]=(c[i][j]+a[i][k]*a[k][j])%mod;
}
}
}
for(int i=1; i<=2; i++)
{
for(int j=1; j<=2; j++)
{
a[i][j]=c[i][j];
}
}
n>>=1;
}
return b[1][1];
}
int main()
{
while(scanf("%lld",&n)!=EOF)
{
if(n==0||n==1)
{
cout<<n<<endl;
continue;
}
ans=f(n,mod3);
if(ans>=2)
ans=f(ans,mod2);
if(ans>=2)
ans=f(ans,mod1);
cout<<ans<<endl;
}
return 0;
}