Fibonacci Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 10072Accepted: 7191 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
做的第一道矩阵，先从别人那里盗个模板。。 矩阵快速幂用来计算矩阵的n次方的。将时间复杂度降到log（n），，原理和快速幂类似，二分的思想（想不到当年学的线性代数用到了，orz）；主要在于构造矩阵；
;const int MOD = 10000;struct node{int m[2][2];}ans,base;int n;node multi(node a,node b){node tmp;for(int i=0;i<2;i++)for(int j=0;j<2;j++){tmp.m[i][j] = 0;for(int k=0;k<2;k++){tmp.m[i][j] +=(a.m[i][k] * b.m[k][j]);tmp.m[i][j] %= MOD;}}return tmp;}int fast_mod(int n)// 求矩阵 base 的 n 次幂 {base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;base.m[1][1] = 0;ans.m[0][0] = ans.m[1][1] = 1;// ans 初始化为单位矩阵ans.m[0][1] = ans.m[1][0] = 0;while (n){if (n&1) //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t ans = multi(ans,base);base = multi(base,base);n>>=1;}return ans.m[0][1];}int main(){while (scanf(“%d”,&n)!=EOF){if (n == -1)break;int ans = fast_mod(n);printf(“%d\n”,ans);}return 0;}
坚硬的城市里没有柔软的爱情，生活不是林黛玉，
