CodeForces

http://codeforces.com/problemset/problem/226/A

题意：有①—②—③三个点如此连接。有n个数在位置③，要全部移动到位置①。数A可以从一个点移动到另一个点，当且仅当A大于这两个点的所有数。

题解：多写几个可以发现递推式：a[n]=3*a[n-1]+2。

法一：构造矩阵进行矩阵快速幂。

| a[n] 1 |   | 3 0 |

|  0    0 |   | 2 1 |

法二：等比数列求和：a[n]+1=3*(a[n-1]+1)，所以a[n]=3^n-1。

矩阵快速幂：

#include<set>#include<map>#include<stack>#include<queue>#include<vector>#include<string>#include<bitset>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<iostream>#define debug cout<<"aaa"<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 2;LL mod;struct node{LL m[N][N];};node Mul(node a,node b){node c;mem(c.m,0);for(int k=0;k<N;k++){for(int i=0;i<N;i++){if(a.m[i][k]){for(int j=0;j<N;j++){c.m[i][j]=(c.m[i][j]+(a.m[i][k]*b.m[k][j])%mod+mod)%mod;}}}}return c;} node quickmod(node a,LL b){node res;mem(res.m,0);for(int i=0;i<N;i++){res.m[i][i]=1;}while(b){if(b&1){res=Mul(res,a);}a=Mul(a,a);b>>=1;}return res;}int main(){LL n;node a,b,ans;cin>>n>>mod;mem(a.m,0),mem(b.m,0); b.m[0][0]=0,b.m[0][1]=1;a.m[0][0]=3,a.m[1][0]=2,a.m[1][1]=1;a=quickmod(a,n);ans=Mul(b,a);cout<<ans.m[0][0]<<endl;return 0;}

等比数列：

#include<bits/stdc++.h>#define debug cout<<"aaa"<<endl#define d(a) cout<<a<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 100000 + 5;const int mod = 1000000000 + 7;const double eps = 1e-8;LL quickMod(LL a,LL b,LL m){LL ans=1;a%=m;while(b){if(b&1)ans=(ans*a)%m;a=(a*a)%m;b>>=1;}return ans;} int main(){LL n,m,ans;cin>>n>>m;ans=quickMod(3,n,m)-1;ans=(ans+m)%m;cout<<ans<<endl;return 0;}