LOJ1070(SummerTrainingDay05-B 矩阵快速幂)

Algebraic Problem

Given the value of a+b and ab you will have to find the value of aⁿ+bⁿ. a and bnot necessarily have to be real numbers.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains three non-negative integers, p, q and n. Here p denotes the value of a+b and q denotes the value of ab. Each number in the input file fits in a signed 32-bit integer. There will be no such input so that you have to find the value of 0⁰.

Output

For each test case, print the case number and (aⁿ+bⁿ) modulo 2⁶⁴.

Sample Input

2

10 16 2

7 12 3

Sample Output

Case 1: 68

Case 2: 91

令Sn =  aⁿ+bⁿ  ，则有递推公式Sn = p×S_n-1 - q×S_n-2 ，构造矩阵[[p, 1], [-q, 0]]，初始矩阵为[S₁, S₀]。

 //2017-08-05
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define ll unsigned long long 

 using namespace std;

 ll p, q;
 const int N = ;

 struct Matrix{
     ll a[N][N];
     int r, c;
 }ori, res;  

 void init(){
     memset(res.a, , sizeof(res.a));
     res.r = ; res.c = ;
     res.a[][] = p;
     res.a[][] = ;
     ori.r = ; ori.c = ;//构造矩阵
     ori.a[][] = p;
     ori.a[][] = ;
     ori.a[][] = -q;
     ori.a[][] = ;
 }

 Matrix multi(Matrix x, Matrix y)//矩阵乘法
 {
     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++)//加速优化
         {
             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]));
         }
     }
     return z;
 }

 void Matrix_pow(int n)//矩阵快速幂
 {
     while(n){
         if(n & )
             res = multi(res, ori);
         ori = multi(ori, ori);
         n >>= ;
     }
     printf("%llu\n", res.a[][]);
 }  

 int main(){
      int T, n;
      scanf("%d", &T);
      int kase = ;
      while(T--){
         scanf("%lld%lld%d", &p, &q, &n);
         init();
         printf("Case %d: ", ++kase);
         if(n == )printf("2\n");
         else if(n == )printf("%llu\n", p);
         else Matrix_pow(n-);
      }

     return ;
 }