题目链接：
Dwarfs have planted a very interesting plant, which is a triangle directed "upwards". This plant has an amusing feature. After one year a triangle plant directed "upwards" divides into four triangle plants: three of them will point "upwards" and one will point "downwards". After another year, each triangle plant divides into four triangle plants: three of them will be directed in the same direction as the parent plant, and one of them will be directed in the opposite direction. Then each year the process repeats. The figure below illustrates this process.
Help the dwarfs find out how many triangle plants that point "upwards" will be innyears.
Input
The first line contains a single integern(0≤n≤1018)— the number of full years when the plant grew.
Please do not use the%lldspecifier to read or write 64-bit integers in С++. It is preferred to usecin,coutstreams or the%I64dspecifier.
Output
Print a single integer — the remainder of dividing the number of plants that will point "upwards" innyears by1000000007(109+7).
Sample test(s)
input
1
output
3
input
2
output
10
Note
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
PS:
上三角个数：△-> 3*△ +1*▽
下三角个数：▽-> 1*△ +3*▽
矩阵快速幂；
等比矩阵为：
3 1
1 3
代码如下：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define LL __int64struct Matrix{LL m[5][5];} I,A,B,T;LL a, b, n;int ssize = 2;#define mod 1000000007Matrix Mul(Matrix a,Matrix b){int i, j, k;Matrix c;for(i = 1; i <= ssize; i++){for(j = 1; j <= ssize; j++){c.m[i][j]=0;for(k = 1; k <= ssize; k++){c.m[i][j]+=(a.m[i][k]*b.m[k][j]);c.m[i][j]%=mod;}}}return c;}Matrix quickpagow(LL n){Matrix m = A, b = I;while(n){if(n & 1)b = Mul(b,m);n = n >> 1;m = Mul(m,m);}return b;}int main(){//a1 向下，，b1向上LL c;while(~scanf("%I64d",&n)){memset(I.m,0,sizeof(I.m));memset(A.m,0,sizeof(A.m));memset(B.m,0,sizeof(B.m));for(int i = 1; i <= ssize; i++){//单位矩阵I.m[i][i]=1;}B.m[1][1] = 0, B.m[2][1] = 1;A.m[1][1] = 3;//初始化等比矩阵A.m[1][2] = 1;A.m[2][1] = 1;A.m[2][2] = 3;T = quickpagow(n);T = Mul(B,T);printf("%I64d\n",T.m[2][1]%mod);}return 0;}/*1231000000100000000000000000010*/
一个人最大的破产是绝望，最大的资产是希望。
