hdu 4565 (快速矩阵幂)

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So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4047 Accepted Submission(s): 1338

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 20132 3 2 20132 2 1 2013

Sample Output

4144
题目大意：
　　给出a，b，n，m，求出的值，
思路：取模前有更号，所以无法直接计算，我们发现
0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹
^{所以 0 <| a+sqrt( b ) | < 1}
可得表达式：
构造矩阵：
s[n]          2a   b-a*a        s[2]
           =                      *
s[n-1]       1      0              s[1]
#include <bits/stdc++.h>using namespace std;const int N=2;long long int mod;struct M{long long int m[N][N];};M I={                    //单位矩阵 1,0,0,1};M fun(M a,M b)            //a*b矩阵 {M c;for(int i=0;i<N;i++){for(int j=0;j<N;j++){c.m[i][j]=0;for(int k=0;k<N;k++){c.m[i][j]+=a.m[i][k]*b.m[k][j]%mod;}c.m[i][j]%=mod;}}return c;}M power(M A,long long int k)                   //快速矩阵幂 {M ans=I,p=A;while(k){if(k&1){ans=fun(ans,p);k--;}k>>=1;p=fun(p,p);}return ans;}int main(){long long int a,b,n,ret;M A,ans;                                  //A矩阵为所求矩阵 while(~scanf("%lld %lld %lld %lld",&a,&b,&n,&mod)){A.m[0][0]=2*a%mod;A.m[0][1]=((b%mod-a*a%mod)%mod+mod)%mod;A.m[1][0]=1;A.m[1][1]=0;if(n==1)                         //特殊情况 {cout<<2*a%mod<<"\n";continue;}ans=power(A,n-2);             //注意  是n-2 ret=(ans.m[0][0]%mod*2*(a*a%mod+b%mod)%mod+2*a%mod*ans.m[0][1]%mod)%mod;cout<<ret<<"\n";}return 0; }

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