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
099999999991000000000-1
Sample Output
0346266875
1.使用一个结构体存下矩阵，再写一个二维矩阵乘法函数
2.然后求[1 1 1 0]的n次方？当然不是。
注意：0 ≤ n ≤ 1,000,000,000
如果这样直接乘以n次肯定会超时
可以使用二进制求快速幂
利用二进制求指数幂
举例：
3 ^ 999 = 3 * 3 * 3 * … * 3
直接乘要做998次乘法。但事实上可以这样做，先求出2^k次幂：
3 ^ 2 = 3 * 3
3 ^ 4 = (3 ^ 2) * (3 ^ 2)
3 ^ 8 = (3 ^ 4) * (3 ^ 4)
3 ^ 16 = (3 ^ 8) * (3 ^ 8)
3 ^ 32 = (3 ^ 16) * (3 ^ 16)
3 ^ 64 = (3 ^ 32) * (3 ^ 32)
3 ^ 128 = (3 ^ 64) * (3 ^ 64)
3 ^ 256 = (3 ^ 128) * (3 ^ 128)
3 ^ 512 = (3 ^ 256) * (3 ^ 256)
再相乘：
3 ^ 999
= 3 ^ (512 + 256 + 128 + 64 + 32 + 4 + 2 + 1)
= (3 ^ 512) * (3 ^ 256) * (3 ^ 128) * (3 ^ 64) * (3 ^ 32) * (3 ^ 4) * (3 ^ 2) * 3
把999转为2进制数：1111100111，其个位就是要乘的数。
1 pow ← 1
2 while (n > 0)
3 do if (n mod 2 = 1)
4 then pow ← pow * x
5 x ← x * x
6 n ← n / 2
7 return pow
#include"iostream"#include"cstdio"using namespace std;typedef struct{ int m[2][2];}node;node work(node a,node b){ node c; c.m[0][0]=(a.m[0][0]*b.m[0][0]+a.m[0][1]*b.m[1][0])%10000; c.m[0][1]=(a.m[0][0]*b.m[0][1]+a.m[0][1]*b.m[1][1])%10000; c.m[1][0]=(a.m[1][0]*b.m[0][0]+a.m[1][1]*b.m[1][0])%10000; c.m[1][1]=(a.m[1][0]*b.m[0][1]+a.m[1][1]*b.m[1][1])%10000; return c;}void caculate(int c){ node ans,base; base.m[0][0]=base.m[1][0]=base.m[0][1]=1; base.m[1][1]=0; ans.m[0][0]=ans.m[1][1]=1; ans.m[0][1]=ans.m[1][0]=0; while(c) { if(c&1) ans=work(ans,base); base=work(base,base); c>>=1; } cout< < >n&&n>=0) { caculate(n); }}