codeforces - 185 - A - Plant 【经典矩阵快速幂】

A. Plant

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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 in n years.

Input

The first line contains a single integer n (0 ≤ n ≤ 1018) — the number of full years when the plant grew.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

Print a single integer — the remainder of dividing the number of plants that will point “upwards” in n years by 1000000007 (109 + 7).

Examples

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.

题意 : 根据图片的规律，每年会分裂一次，让你求出第n年后的图形中有几个”朝上”的三角形。范围： 0 ≤ n ≤ 1018

分析： 一开始做这个题看了第二个图和第三个图后直接断定递推式就是F[n]=3∗F[n−1]+1，然后就wa了，后来当推了后面的时候发现不对，少算了不少，然后就是慢慢找递推式，还是找不到，后来偶然发现无论怎么在第n次后，朝上的三角形+朝下的三角形 = 4n，然后又推了下发现，每一次朝上的”分裂”一次后会产生三个朝上的和一个朝下的，而一个朝下的“分裂”后会产生三个朝下的和一个朝上的，所以就可以得到递推式了，从而根据矩阵快速幂得到详细看代码

参考代码

#include<bits/stdc++.h>
#define ll long long
#define mod(x) ((x)%MOD)

using namespace std;

const ll MOD = 1e9 + 7;

struct mat{
ll m[3][3];
}a,ans,unit;

void init() {
memset(unit.m,0,sizeof(unit.m));
memset(a.m,0,sizeof(a.m));
unit.m[0][0] = 1;
unit.m[1][1] = 1;
a.m[0][0] = 3;
a.m[0][1] = 1;
a.m[1][0] = 1;
a.m[1][1] = 3;
}

mat operator * (mat m1,mat m2) {
mat t;
ll r;
for(int i = 0;i < 3;i++) {
for(int j = 0;j < 3;j++) {
r = 0;
for(int k = 0;k < 3;k++) {
r = mod(r*1ll + mod(mod(m1.m[i][k])*1ll*mod(m2.m[k][j])));
}
t.m[i][j] = r;
}
}
return t;
}

mat quick_pow(ll x) {
mat t = unit;
while(x) {
if(x & 1) {
t = t*a;
}
a = a*a;
x >>= 1;
}
return t;
}

int main(){
ios_base::sync_with_stdio(0);
ll n;while(cin>>n){
init();
if(n == 0) {
cout<<1<<endl;
return 0;
}
if(n == 1) {
cout<<3<<endl;
return 0;
}
ans = quick_pow(n);
cout<<mod(ans.m[0][0]*1ll)<<endl;}
return 0;

}

如有错误或遗漏，请私聊下UP，thx