HDU4565(SummerTrainingDay05-C 矩阵快速幂)

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4673    Accepted Submission(s): 1539

Problem Description

　　A sequence S_n is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S_n.
　　You, a top coder, say: So easy! 

 

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹.The input will finish with the end of file.

 

Output

　　For each the case, output an integer S_n.

 

Sample Input

2 3 1 2013
2 3 2 2013
2 2 1 2013

 

Sample Output

4
14
4

 

根据条件可证(a-√b)ⁿ小于1，(a+√b)ⁿ向上取整即为求(a+√b)ⁿ + (a-√b)ⁿ，问题又转换为a^n+b^n的形式。

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

 using namespace std;

 ll a, b, n;
 ll p, q;
 const int N = ;
 ll MOD;

 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 = ;
     ori.a[][] = p;
     ori.a[][] = ;
     ori.a[][] = -q;
     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) % MOD;
         }
     }
     return z;
 }  

 void Matrix_pow(int n)
 {
     while(n)
     {
         if(n & )
             res = multi(res, ori);
         ori = multi(ori, ori);
         n >>= ;
     }
     printf("%lld\n", res.a[][] % MOD);
 }  

 int main()
 {
     while(scanf("%lld%lld%lld%lld", &a, &b, &n, &MOD)!=EOF){
         p = *a;
         q = a*a-b;
         init();
         if(n == )printf("2\n");
         else if(n == )printf("%lld\n", p);
         else Matrix_pow(n-);
     }

     return ;
 }

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现