hdu 6198（矩阵快速幂）

number number number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 175    Accepted Submission(s): 119

暴力发现当4 12 33 88 232

和斐波那契数列对比  答案为 第2*k+3个数减1

直接用矩阵快速幂求的F[2*k+3]  然后减1

A=1,B=0;

然后矩阵快速幂2*k+3-1次得到F[2*k+3]

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string.h>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<cstdlib>
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int INF=0x3f3f3f3f;
const int num=100;
const int mod=998244353;
int N;
struct Mat{
    ll a[num][num];
    void init(){
        memset(a,0,sizeof(a));
        for(int i=0;i<num;i++)
            a[i][i]=1;
    }
};
Mat mul(Mat a,Mat b){
    Mat ans;
    for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
        ans.a[i][j]=0;
        for(int k=0;k<N;k++){
            ans.a[i][j]+=a.a[i][k]*b.a[k][j];
        }
        ans.a[i][j]=ans.a[i][j]%mod;
    }
    }
    return ans;
}
Mat power(Mat a,int n){
    Mat ans;
    ans.init();
    while(n){
        if(n&1){
            ans=mul(ans,a);
        }
        n=n>>1;
        a=mul(a,a);
    }
    return ans;
}
int main(){
    int k;
    N=2;
    while(scanf("%d",&k)!=EOF){
        k=k*2+3;
        Mat aa;
        aa.a[0][0]=1;
        aa.a[0][1]=1;
        aa.a[1][0]=1;
        aa.a[1][1]=0;
        Mat ans=power(aa,k-1);
        ll t=((ans.a[0][0]-1)%mod+mod)%mod;
        cout<<t<<endl;
    }
    return 0;
}