codeforces 185A Plant(矩阵快速幂)
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.
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 %I64dspecifier.
Print a single integer — the remainder of dividing the number of plants that will point "upwards" in n years by 1000000007 (109 + 7).
1
3
2
10
The first test sample corresponds to the second triangle on the figure in the statement. The second test sample corresponds to the third one.
求 上三角有多少个;推了半天 想用打表水过试试,谁知道,会爆出runtime error; 还是用矩阵快速幂把;
Xn= 3*Xn-1 + Yn-1; Yn-1= 3*Yn-2+Xn-2;
构造矩阵 3 1
1 3
矩阵快速幂:
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
#define S1(n) scanf("%d",&n)
#define S2(n,m) scanf("%d%d",&n,&m)
#define Pr(n) printf("%d\n",n)
using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};
ll inv[maxn*2],fac[maxn];
ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}
void INV(){inv[1] = 1;for(int i = 2; i < maxn; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);}}
void Fac(){fac[0]=1;for(int i=1;i<=maxn;i++)fac[i]=(fac[i-1]*i)%MOD;}
ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
ll inv2(ll b){return qpow(b,MOD-2);}
ll cal(ll m,ll n){if(m<n)return 0;return (fac[m]*inv[fac[n]]%MOD)%MOD*inv[fac[m-n]]%MOD;}
ll cals(ll m,ll n){if(m<n)return 0;return (fac[m]*inv1(fac[n])%MOD)%MOD*inv1(fac[m-n])%MOD;}
ll n;
int MAXN=2;
const int N=10;
struct Matrix{
ll arr[N][N];
void init()
{
memset(arr,0,sizeof(arr));
for(int i=0;i<MAXN;i++)
arr[i][i]=1;//³õʼ»¯
}
void iinit()
{
memset(arr,0,sizeof(arr));
arr[0][0]=arr[1][1]=3;
arr[0][1]=arr[1][0]=1;
}
}A;
Matrix mul(Matrix X,Matrix Y)// ¾ØÕó³Ë·¨
{
Matrix ans;
for(int i=0;i<MAXN;i++)
for(int j=0;j<MAXN;j++){
ans.arr[i][j]=0;
for(int k=0;k<MAXN;k++){
ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];
ans.arr[i][j]%=MOD;
}
}
return ans;
}
Matrix Q_pow(Matrix B,ll n)// ¾ØÕó¿ìËÙÃÝ
{
Matrix ans;
ans.init();
while(n)
{
if(n&1)
ans=mul(ans,B);
n>>=1;
B=mul(B,B);
}
return ans;
}
int main()
{
while(~scanf("%I64d",&n))
{
Matrix ans;
ans.iinit();
if(n==0)
{
printf("1\n");
continue;
}
if(n==1)
{
printf("3\n");
continue;
}
ans=Q_pow(ans,n-1);
ll res=0;
res+=(3*ans.arr[0][0])%mod+ans.arr[0][1]%mod;
printf("%I64d\n",res%MOD);
}
return 0;
}