zufe 问题 : Matrix Fast Power

这道题就是找循环节

#include<bits/stdc++.h>
 
using namespace std;
 
int a[10000][2],book[10000];
 
int ff(int t)
{
    int num = 0 ;
    while(t)
    {
        num += t % 10;
        t /= 10;
    }
    return num;
}
 
int main()
{
    int T;
    cin >> T;
    for(int u = 1; u <= T ; u ++)
    {
        int l, r;
        long long n ;
        cin >> l >> r >> n;
        a[1][0] = l;a[1][1] = l;
        a[2][0] = r;a[2][1] = l + r;
        for(int i = 3; i < 100 ; i ++)
        {
            a[i][0] = ff(a[i-1][0]) + ff(a[i-2][0]);
            a[i][1] = a[i-1][0] + a[i][0];
        }
        //for(int i = 1; i <= 10 ; i ++)
        //  cout << a[i][0] << " ";
        //cout << endl;
        if(n < 100)
        {
            cout << "Case #" << u << ": " << a[n][0] << endl;
            continue;
        }
        int s1,s2;
        for(int i = 1; i < 100 ; i ++)
        {
            for(int j = i + 1; j < 100 ; j ++)
                if(a[i][0] == a[j][0] && a[i][1] == a[j][1])
                {
                    s1 = i - 1; s2 = i;
                }
        }
        int num = 0,t;
        for(int i = 1; i < 100 ; i ++)
        {
            if(a[i][0] == a[s1][0] && a[i+1][0] == a[s2][0])
            {
                t = i;
                break;
            }
            num ++;
        }
        n -= num;
        //cout << n << endl;
        num = 2;
        for(int i = t + 2 ; i < 100;i ++)
        {
            if(a[i][0] == a[s1][0] && a[i+1][0] == a[s2][0])
                break;
            num ++;
        }
        //cout << num << endl;
        n %= num;
        if(n == 0 )
            cout << "Case #" << u << ": " << a[s1 + num-1][0] << endl;
        else
            cout << "Case #" << u << ": " << a[s1 + n - 1][0] << endl;
    }
    return 0;
}