hzau 1202 GCD（矩阵快速幂）

1202: GCD

Time Limit: 1 Sec Memory Limit: 1280 MB
Submit: 201 Solved: 31
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Description

Input

The first line is an positive integer T . (1<=T<= 10^3) indicates the number of test cases. In the next T lines, there are three positive integer n, m, p (1<= n,m,p<=10^9) at each line.

Output

Sample Input

1 
1 2 3

Sample Output

HINT

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define eps 1e-4
 5 const int N=1e6+10,M=1e6+10;
 6 
 7 ///数组大小
 8 ll MOD;
 9 
10 struct Matrix
11 {
12     ll matri[2][2];
13     Matrix()
14     {
15         memset(matri,0,sizeof(matri));
16     }
17     void init()
18     {
19         for(int i=0;i<2;i++)
20             for(int j=0;j<2;j++)
21                 matri[i][j]=(i==j);
22     }
23     Matrix operator + (const Matrix &B)const
24     {
25         Matrix C;
26         for(int i=0;i<2;i++)
27             for(int j=0;j<2;j++)
28                 C.matri[i][j]=(matri[i][j]+B.matri[i][j])%MOD;
29         return C;
30     }
31     Matrix operator * (const Matrix &B)const
32     {
33         Matrix C;
34         for(int i=0;i<2;i++)
35             for(int k=0;k<2;k++)
36                 for(int j=0;j<2;j++)
37                     C.matri[i][j]=(C.matri[i][j]+1LL*matri[i][k]*B.matri[k][j])%MOD;
38         return C;
39     }
40     Matrix operator ^ (const ll &t)const
41     {
42         Matrix A=(*this),res;
43         res.init();
44         ll p=t;
45         while(p)
46         {
47             if(p&1)res=res*A;
48             A=A*A;
49             p>>=1;
50         }
51         return res;
52     }
53 };
54 int main()
55 {
56     Matrix base;  ///初始化矩阵
57     base.matri[0][0]=1;base.matri[0][1]=1;
58     base.matri[1][0]=1;base.matri[1][1]=0;
59 
60     int T;
61     scanf("%d",&T);
62     while(T--)
63     {
64         int n,m,p;
65         scanf("%d%d%d",&n,&m,&p);
66         int x=__gcd(n+2,m+2);
67         MOD=p;
68         if(x<=2)
69             printf("%d
",1%p);
70         else
71         {
72             Matrix ans=base^(x-2);
73             printf("%lld
",(ans.matri[0][0]+ans.matri[0][1])%MOD);
74         }
75     }
76     return 0;
77 }