A Short problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2461 Accepted Submission(s): 864
Problem Description
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 <= 1018), You should solve for
g(g(g(n))) mod 109 + 7
where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0
Hence they prefer problems short, too. Here is a short one:
Given n (1 <= n <= 1018), You should solve for
where
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).
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
三层复合函数。写完交上去TLE,不解。查了题解发现有循环节。循环节的话用set::find函数来查找是否存在过,若存在则输出循环节,不存在则插入当前数值继续循环。
然后就是矩阵快速幂了。递推式:(当某一层n为0的时候加上mod否则可能出错)
代码:
#include<iostream> #include<algorithm> #include<cstdlib> #include<sstream> #include<cstring> #include<cstdio> #include<string> #include<deque> #include<stack> #include<cmath> #include<queue> #include<set> #include<map> using namespace std; typedef long long LL; #define INF 0x3f3f3f3f const long long mod=1000000007; struct mat { LL pos[2][2]; mat(){memset(pos,0,sizeof(pos));} }; inline mat mul(const mat &a,const mat &b,const LL &mod) { mat c; for (int i=0; i<2; i++) { for (int j=0; j<2; j++) { for (int k=0; k<2; k++) { c.pos[i][j]+=((a.pos[i][k])%mod*(b.pos[k][j])%mod)%(mod); } } } return c; } inline mat matpow(mat a,LL b,const LL &mod) { mat r; r.pos[1][1]=r.pos[0][0]=1; mat bas=a; while (b!=0) { if(b&1) r=mul(r,bas,mod); bas=mul(bas,bas,mod); b>>=1; } return r; } int main(void) { mat one,t; LL n,ans; one.pos[0][0]=1; one.pos[0][1]=0; t.pos[0][0]=3; t.pos[0][1]=t.pos[1][0]=1; mat poww,initt; while (cin>>n) { if(n==0) { cout<<"0"<<endl; continue; } poww=t,initt=one; poww=matpow(poww,n-1,183120); initt=mul(initt,poww,183120); ans=initt.pos[0][0]; ans+=183120; poww=t,initt=one; poww=matpow(poww,ans-1,222222224); initt=mul(initt,poww,222222224); ans=initt.pos[0][0]; ans+=222222224; poww=t,initt=one; poww=matpow(poww,ans-1,1000000007); initt=mul(initt,poww,1000000007); cout<<initt.pos[0][0]%1000000007<<endl; } return 0; }