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!
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; }