求：简单又困难的 “斐波那契数列”代码

 1 #include <cstdio>
 2 #include <cstring>
 3 const int M = 1E9 + 7;
 4 struct Matrix {
 5   long long a[2][2];
 6   Matrix() { memset(a, 0, sizeof(a)); } //构造矩阵
 7   //定义矩阵乘法
 8   Matrix operator * (const Matrix y) {  
 9     Matrix ans;
10     for (int i = 0; i <= 1; ++ i)
11       for (int j = 0; j <= 1; ++ j)
12         for (int k = 0; k <= 1; ++ k)
13           ans.a[i][j] += a[i][k] * y.a[k][j];
14     for (int i = 0; i <= 1; ++i)
15       for (int j = 0; j <= 1; ++ j)
16         ans.a[i][j] %= M;
17     return ans;
18   }
19   //重载等于号
20   void operator = (const Matrix b) {
21     for (int i = 0; i <= 1; ++i)
22       for (int j = 0; j <= 1; ++j)
23         a[i][j] = b.a[i][j];
24   }
25 };
26 //矩阵快速幂
27 int fast_pow(long long x) {
28   Matrix ans, trs;
29   ans.a[0][0] = ans.a[1][1] = 1;  //单位矩阵
30   trs.a[0][0] = trs.a[1][0] = trs.a[0][1] = 1; //神奇的"斐波那契矩阵"
31   //同数字的快速幂一样
32   while (x) {
33   //若指数为奇数，则用单位矩阵来纪录
34     if (x & 1)  
35       ans = ans * trs;
36     trs = trs * trs;
37     x >>= 1;
38   }
39   //因为最后指数一定会变为1，且矩阵遵守结合律
40   //所以开始的单位矩阵在最后就是结果
41   return ans.a[0][0];
42 }
43 int main() {
44   int n;
45   scanf("%d", &n);
46   //结果矩阵左上角是斐波那契数列的第n+1位，所以要算第n-1位
47   printf("%d\n", fast_pow(n - 1));
48   return 0;
49 }