首页 > 其他 > 详细

T54037 最开始(矩阵快速幂)

时间:2019-09-04 16:22:25      阅读:58      评论:0      收藏:0      [点我收藏+]

给定a+1/a=n,求解a^m+1/a^m

一开始在原式上化简了半天,以为是直接求m次方之类的,无功而返,看了题解才发现神秘之处:

f(x)=a^x+1/a^x,可以发现f(x)*f(y)=f(x+y)-f(x-y)

令y=1,可得:

f(x)*f(1)=f(x+1)+f(x-1)

即:f(x)=f(1)*f(x-1)-f(x-2)

接下来就是矩阵快速幂了,转移矩阵如下:

n   -1

1   0

最坑的点是有-1,所以取余的时候一定得先加个mod

代码:

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=1000000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}

int n,m;
struct mat{
    int a[2][2];
    mat(){mem0(a);a[0][0]=1;a[1][1]=1;}
    mat(int x,int b,int c,int d){
        a[0][0]=x;
        a[0][1]=b;
        a[1][0]=c;
        a[1][1]=d;
    }
};

mat mul(mat a,mat b){
    mat tmp;
    mem0(tmp.a);
    fo(i,0,2){
        fo(j,0,2){
            fo(k,0,2){
                tmp.a[i][j]+=((a.a[i][k]%MOD)*(b.a[k][j]%MOD))%MOD;
                tmp.a[i][j]+=MOD;
                tmp.a[i][j]%=MOD;
            }
        }
    }
    return tmp;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n>>m;
    if(m==1) {cout<<n;return 0;}
    mat now(n%MOD,-1,1,0),ans;
    m-=1;
    while(m>0){
        if(m&1) ans=mul(now,ans);
        now=mul(now,now);
        m>>=1;
    }
    int a=ans.a[0][0]%MOD,b=ans.a[0][1]%MOD;
    cout<<((n%MOD*a%MOD)%MOD+(2LL*b%MOD)%MOD)%MOD;
    return 0;
}
//15632475 21747465

 

T54037 最开始(矩阵快速幂)

原文:https://www.cnblogs.com/oneman233/p/11459333.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!