hdu4565之矩阵快速幂

 

So Easy!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 813    Accepted Submission(s): 226

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 < 2 ¹⁵, (a-1) ²< b < a ², 0 < b, n < 2 ³¹.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

详解在这: http://blog.csdn.net/ljd4305/article/details/8987823

 

另外Cn=(a+sqrt(b))^n+(a-sqrt(b))^n =>Cn*(a+sqrt(b) + a-sqrt(b))=(a+sqrt(b))^(n+1)+(a-sqrt(b))^(n+1)+(a*a-b)*(a+sqrt(n))^(n-1)+(a*a-b)*(a-sqrt(b))^(n-1)=C(n+1) + (a*a-b)C(n-1) =>C(n+1)=2a*Cn+(b-a^2)*C(n-1)

 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;const int MAX=10;
__int64 array[2][2],sum[2][2];void MatrixMult(__int64 a[2][2],__int64 b[2][2],__int64 mod){__int64 c[2][2]={0};for(int i=0;i<2;++i){for(int j=0;j<2;++j){for(int k=0;k<2;++k){c[i][j]+=a[i][k]*b[k][j];}}}for(int i=0;i<2;++i){for(int j=0;j<2;++j)a[i][j]=c[i][j]%mod;}
}__int64 Matrix(__int64 a,__int64 b,__int64 k,__int64 mod){array[0][0]=a%mod,array[0][1]=(b%mod+mod)%mod;array[1][0]=1,array[1][1]=0;sum[0][0]=sum[1][1]=1;sum[0][1]=sum[1][0]=0;while(k){if(k&1)MatrixMult(sum,array,mod);MatrixMult(array,array,mod);k>>=1;}return (a*sum[0][0]+2*sum[0][1])%mod;
}int main(){__int64 a,b,n,m;while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)!=EOF){if(n == 0)printf("I64d\n",1%m);else if(n == 1)printf("%I64d\n",2*a%m);else printf("%I64d\n",Matrix(2*a,b-a*a,n-1,m));}return 0;
}