斐波那契数列poj（矩阵快速幂）

Fibonacci Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 18040   Accepted: 12541 Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, … An alternative formula for the Fibonacci sequence is . Given an integer n, your goal is to compute the last 4 digits of Fn. Input The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1. Output For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000). Sample Input 0 9 999999999 1000000000 -1 Sample Output 0 34 626 6875 Hint As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by . Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix: . Source Stanford Local 2006          以前写斐波那契数列用递推公式写，对以前的n不大，但碰见n比较大的时候就会超时，比较尴尬，这个时候矩阵快速幂来了。 首先构造f1,f0矩阵（此题比较特殊，不用构造） 然后类似于1构造一个单位矩阵（单位矩阵就是那么一个矩阵 这个矩阵乘以任何一个矩阵都等于他乘的那个矩阵）sec 如下图，构造矩阵fir 然后判断n如果n<2直接输出,否则fir的（n-1)次方，在这个过程中注意取余操作；然后再乘以初始矩阵（此题比较特殊，你会发现直接输出乘方后的pos[0][0]就行了。 第一次写这个东西，多多指教，代码很烂。       #include #include #include using namespace std; struct mat { int pos[2][2]; mat() { memset(pos,0,sizeof(pos)); } }; mat fir,sec; mat Cheng(mat a,mat b) { mat c; for(int i=0;i<2;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++) c.pos[i][j]=(c.pos[i][j]+a.pos[i][k]*b.pos[k][j])%10000; return c; } void mul(mat b,int n) { while(n>0) { if(n%2==1) sec=Cheng(sec,b); n=n/2; b=Cheng(b,b); } } int main() { int n; while(scanf("%d",&n)&&n>=0) { if(n==0) cout<<0<