hdu4291 A Short problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2383    Accepted Submission(s): 833

Problem Description

　　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for 
g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

 

Input

　　There are several test cases. For each test case there is an integer n in a single line.
　　Please process until EOF (End Of File).

 

Output

　　For each test case, please print a single line with a integer, the corresponding answer to this case.

 

Sample Input

0 1 2

 

Sample Output

0 1 42837

 

这题要找循环节然后再用快速幂，第一层的循环节为1e9+7,第二层为222222224，第三层为183120,然后三个快速幂就行了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
ll MOD;
struct matrix{
    ll n,m,i;
    ll data[99][99];
    void init_danwei(){
        for(i=0;i<n;i++){
            data[i][i]=1;
        }
    }
};

matrix multi(matrix &a,matrix &b){
    ll i,j,k;
    matrix temp;
    temp.n=a.n;
    temp.m=b.m;
    for(i=0;i<temp.n;i++){
        for(j=0;j<temp.m;j++){
            temp.data[i][j]=0;
        }
    }
    for(i=0;i<a.n;i++){
        for(k=0;k<a.m;k++){
            if(a.data[i][k]>0){
                for(j=0;j<b.m;j++){
                    temp.data[i][j]=(temp.data[i][j]+(a.data[i][k]*b.data[k][j])%MOD )%MOD;
                }
            }
        }
    }
    return temp;
}

matrix fast_mod(matrix &a,ll n){
    matrix ans;
    ans.n=a.n;
    ans.m=a.m;
    memset(ans.data,0,sizeof(ans.data));
    ans.init_danwei();
    while(n>0){
        if(n&1)ans=multi(ans,a);
        a=multi(a,a);
        n>>=1;
    }
    return ans;
}

int main()
{
    ll n,m,i,j;
    while(scanf("%lld",&n)!=EOF)
    {
        if(n==0){
            printf("0
");continue;
        }
        if(n==1){
            printf("1
");continue;
        }
        matrix a;
        matrix ans;
        if(n>=2){
            MOD=183120;
            a.n=a.m=2;
            a.data[0][0]=3;a.data[0][1]=1;
            a.data[1][0]=1;a.data[1][1]=0;
            ans=fast_mod(a,n-1);
            n=ans.data[0][0];
        }
        if(n>=2){
            MOD=222222224;
            a.n=a.m=2;
            a.data[0][0]=3;a.data[0][1]=1;
            a.data[1][0]=1;a.data[1][1]=0;
            ans=fast_mod(a,n-1);
            n=ans.data[0][0];
        }
        if(n>=2){
            MOD=1000000007;
            a.n=a.m=2;
            a.data[0][0]=3;a.data[0][1]=1;
            a.data[1][0]=1;a.data[1][1]=0;
            ans=fast_mod(a,n-1);
            n=ans.data[0][0];
        }
        printf("%lld
",n);

    }
    return 0;
}