HDU 6198(2017 ACM/ICPC Asia Regional Shenyang Online)

Posted 2020-10-06 欲儿很轻狂

思路：找规律发现这个数是斐波那契第2*k+3项-1，数据较大矩阵快速幂搞定.

 

快速幂入门第一题QAQ

#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt  (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define exp 1e-8
//***************************************************
#define eps             1e-8
#define inf             0x3f3f3f3f
#define INF             2e18
#define LL              long long
#define ULL             unsigned long long
#define PI              acos(-1.0)
#define pb              push_back
#define mk              make_pair

#define all(a)          a.begin(),a.end()
#define rall(a)         a.rbegin(),a.rend()
#define SQR(a)          ((a)*(a))
#define Unique(a)       sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a)     __typeof__(a.begin())
#define rIterator(a)    __typeof__(a.rbegin())
//#define FastRead      ios_base::sync_with_stdio(0);cin.tie(0)
#define FastRead        ios::sync_with_stdio(false);
#define CasePrint       pc(‘C‘); pc(‘a‘); pc(‘s‘); pc(‘e‘); pc(‘ ‘); write(qq++,false); pc(‘:‘); pc(‘ ‘)
#define vi              vector <int>
#define vL              vector <LL>
#define For(I,A,B)      for(int I = (A); I < (B); ++I)
#define FOR(I,A,B)      for(int I = (A); I <= (B); ++I)
#define rFor(I,A,B)     for(int I = (A); I >= (B); --I)
#define Rep(I,N)        For(I,0,N)
#define REP(I,N)        FOR(I,1,N)
using namespace std;
const int maxn=1e5+10;
#define mod 998244353


struct Matirx {
    int h,w;
    LL a[5][5];
}ori,res,it;
LL f[5]={0,1,1};
void iint(){
    it.w=2;it.h=1;it.a[1][1]=1;it.a[1][2]=0;
    res.w=res.h=2;
    memset(res.a,0,sizeof(res.a));
    res.a[1][1]=res.a[2][2]=1;
    ori.w=ori.h=2;
    memset(ori.a,0,sizeof(ori.a));
    ori.a[1][1]= 1;ori.a[1][2]= 1;
    ori.a[2][1]= 1;ori.a[2][2]= 0;
}
Matirx multy(Matirx x,Matirx y){
    Matirx z;z.w=y.w;z.h=x.h;
    memset(z.a,0,sizeof(z.a));
    for(int i=1;i<=x.h;i++){
        for(int k=1;k<=x.w;k++){
            if(x.a[i][k]==0) continue;
            for(int j=1;j<=y.w;j++)
                z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j]%mod)%mod;
        }
    }
    return z;
}

LL Matirx_mod(LL n){
    if(n<2) return f[n];
    else n-=1;
    while(n){
        if(n&1) res=multy(ori,res);
        ori=multy(ori,ori);
        n>>=1;
    }
    res=multy(it,res);
    return res.a[1][1]%mod;
}

int main()
{
    LL n;
    while(scanf("%I64d",&n)!=EOF)
    {
        iint();
        printf("%I64d\n",Matirx_mod(2*n+3)-1);
    }
    return 0;
}