华中农业大学第五届程序设计大赛网络同步赛-D

Problem D: GCD

Time Limit: 1 Sec  Memory Limit: 1280 MB
Submit: 179  Solved: 25
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Description

XiaoMingfoundthecomputetimeof

Input

The first line is an positive integer T. (

Output

In each test case, output the compute result of

Sample Input

1
1 2 3

Sample Output

1

HINT

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define LL long long
 #define MAXN 100

 using namespace std;

 int MOD;

 struct Matrix
 {
     LL a[MAXN][MAXN];
     int r, c;
 };  

 Matrix ori, res; 

 void init()
 {
     memset(res.a, , sizeof(res.a));
     res.r = ; res.c = ;
     for(int i = ; i <= ; i++)
         res.a[i][i] = ;
     ori.r = ; ori.c = ;
     ori.a[][] = ori.a[][] = ori.a[][] = ;
     ori.a[][] = ;
 }  

 Matrix multi(Matrix x, Matrix y)
 {
     Matrix z;
     memset(z.a, , sizeof(z.a));
     z.r = x.r, z.c = y.c;
     for(int i = ; i <= x.r; i++)
     {
         for(int k = ; k <= x.c; k++)
         {
             if(x.a[i][k] == ) continue;
             for(int j = ; j<= y.c; j++)
                 z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
         }
     }
     return z;
 }
 void Matrix_mod(int n)
 {
     while(n)
     {
         if(n & )
             res = multi(ori, res);
         ori = multi(ori, ori);
         n >>= ;
     }
     printf("%lld\n", res.a[][] % MOD);
 }  

 int main()
 {
     int T, n, m;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d%d", &n, &m, &MOD);
         LL pos = __gcd(n+, m+);
         init();
         Matrix_mod(pos);
     }
     return ;
 }