HDU3306 矩阵快速幂
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/*****************
Sn-1 1 X^2 Y^2 2XY Sn
An-1^2 * 0 X^2 Y^2 2XY = An^2
An-2^2 0 1 0 0 An-1^2
An-1An-2 0 X 0 Y AnAn-1
用自己写的矩阵的类模板居然会超时,改成简单的面向过程就过了
*******************/
Another kind of Fibonacci
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1355 Accepted Submission(s): 534
Problem Description
As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.
Input
There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231– 1
Y : 2<= Y <= 231 – 1
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231– 1
Y : 2<= Y <= 231 – 1
Output
For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.
Sample Input
2 1 1 3 2 3
Sample Output
6196
# include<iostream># include<stdio.h># include<string.h>using namespace std;const int mod=10007;struct Matrix{ int element[4][4];};Matrix Multiply(Matrix a, Matrix b){ Matrix ans; for(int i=0; i<4; i++) for(int j=0; j<4; j++) { ans.element[i][j]=0; for(int k=0; k<4; k++) { ans.element[i][j]+=(a.element[i][k]*b.element[k][j])%mod; ans.element[i][j]%=mod; } } return ans;}Matrix quickpow(Matrix a, int n){ Matrix result; for(int i=0; i<4; i++) for(int j=0; j<4; j++) if(i==j) result.element[i][j]=1; else result.element[i][j]=0; while(n) { if(n&1) result = Multiply(result, a); n=n>>1; a=Multiply(a, a); } return result;}int main(){ int n, x, y; while(~scanf("%d%d%d", &n, &x, &y)) { x=x%mod; y=y%mod; Matrix a; for(int i=0; i<4; i++) for(int j=0; j<4; j++) a.element[i][j]=0; a.element[0][0]=a.element[2][1]=1; a.element[0][1]=a.element[1][1]=(x*x)%mod; a.element[0][2]=a.element[1][2]=(y*y)%mod; a.element[0][3]=a.element[1][3]=(2*x*y)%mod; a.element[3][1]=x; a.element[3][3]=y; a=quickpow(a, n-1); int ans=(a.element[0][0]*2+a.element[0][1]+a.element[0][2]+a.element[0][3])%mod; printf("%d\n", ans); } return 0;}
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