HDU 4565 So Easy!（数学+矩阵快速幂）（2013 ACM-ICPC长沙赛区全国邀请赛）

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.

 

题目大意：已知a、b、n、m，求图中Sn，向上取整后求雨。

思路：http://www.klogk.com/posts/hdu4565/

PS：哪里简单了……至于负数比较难搞果断一开始把它弄成正数了……

 

代码（62MS）：

 #include <cstdio>
 #include <queue>
 #include <utility>
 #include <iostream>
 #include <cstring>
 using namespace std;
 typedef long long LL;

 const int N = ;

 int a, b, n, m;

 #define MOD(x) ((x)%m)

 struct Mat {
     LL v[N][N];
     Mat operator * (const Mat &rhs) {
         Mat ret;
         for(int i = ; i < N; ++i)
             for(int j = ; j < N; ++j) {
                 ret.v[i][j] = ;
                 for(int k = ; k < N; ++k) ret.v[i][j] += MOD(v[i][k] * rhs.v[k][j]);
                 ret.v[i][j] %= m;
             }
         return ret;
     }
 };

 Mat mul(Mat x, int p) {
     Mat ans;
     ans.v[][] = ans.v[][] = ;
     ans.v[][] = ans.v[][] = ;
     while(p > ) {
         if(p & ) ans = ans * x;
         x = x * x;
         p >>= ;
     }
     return ans;
 }

 int main() {
     while(scanf("%d%d%d%d", &a, &b, &n, &m) != EOF) {
         Mat tmp;
         tmp.v[][] =  * a;
         tmp.v[][] = - (a * a - b);
         while(tmp.v[][] < ) tmp.v[][] += m;
         tmp.v[][] = ;
         tmp.v[][] = ;
         tmp = mul(tmp, n);
         LL ans = MOD( * a * tmp.v[][] +  * tmp.v[][]);
         cout<<ans<<endl;
     }
 }