Southwest min da oj (Matrix fast Power)

My name can't be so hard to rememberTime limit (normal/java): ms/3000 MS Run memory limit: 65536 KByte
Total Submissions: 16 Tested by: 9

Describe

Nirvava:hi,misaki, said ZC to leave.

Misaki: Good to go not to send, wish smooth sailing ...

Nirvana: But he left a lot of Doge for US ...

Misaki: ...

Nirvana: and they have a name, the name is as follows.

It is said that one of the reasons for ZC's departure is that nth doge always ask him if he can remember his name.

Misaki: Remember someone else's name or something, so annoying da, Zc certainly can't remember.

> Sure, it's his pet, after all. > Baffled, because he counted the doge > The spirit is polluted.

Misaki: That's a very difficult question. However, they should be able to solve.

PS: This problem is not a bit ripe ♂, too lazy to think about the surface, the previous problem surface took out.

Input

Multiple sets of inputs

One N per group (0<n<=10^9)

Output

How long is the name of Nth Doge, because it may be too long, you only need to output the result of length%1000000007 (1e9+7).

Sample input

1
2
3
4
500

Sample output

1
3
7
17
875025602

Tips

name[1]= "X"name[2]= "XXY" name[3]= "Xxyxxyx" name[4]= "Xxyxxyxxxyxxyxxxy" ... And so on                                Analysis: Set A[i] is the total length, b[i] is the number of x, C[i] is the number of Y, then there is a[i]=3*b[i-1]+c[i-1],b[i]=2*b[i-1]+c[i-1],c[i]=b[i-1]. The matrix can be constructed |0,0,0| |                                A[i-1], b[i-1],c[i-1]|*|3,2,1|=|a[i],b[i],c[i]| |1,1,0|

#include <cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>#include<cstdlib>#include<stack>#include<vector>#include<Set>#include<map>#defineLL Long Long#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineN 1000010using namespacestd;structmatrix{LL m[3][3];};    LL N;matrix mult (Matrix A,matrix b) {matrix C; memset (C.M,0,sizeof(C.M));  for(intI=0;i<3; i++)     for(intk=0;k<3; k++)    {        if(a.m[i][k]==0)Continue;  for(intj=0;j<3; j + +)        {            if(b.m[k][j]==0)Continue; C.M[I][J]+=a.m[i][k]*b.m[k][j]%MoD; C.M[I][J]%=MoD; }    }    returnC;}    Matrix Quickmod (Matrix A,ll n) {matrix temp; memset (temp.m,0,sizeof(temp.m));  for(intI=0;i<3; i++) temp.m[i][i]=1;  while(n) {if(n&1) temp=mult (temp,a); A=mult (a,a); N>>=1; }    returntemp;}intMain () { while(SCANF ("%lld", &n)! =EOF) {        if(n==1) {puts ("1");Continue;}        Matrix ans; memset (ANS.M,0,sizeof(ANS.M)); ans.m[0][0]=0; ans.m[0][1]=0; ans.m[0][2]=0; ans.m[1][0]=3; ans.m[1][1]=2; ans.m[1][2]=1; ans.m[2][0]=1; ans.m[2][1]=1; ans.m[2][2]=0; Ans=quickmod (ans,n-1); printf ("%lld\n", ans.m[1][0]); }}

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Southwest min da oj (Matrix fast Power)