Problem Description
A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input
1 1 3 1 2 10 0 0 0
Sample Output
2 5
题解：矩阵乘优化斐波那契的变形题，原本的公式是
它实际上是
每次进行一次
这样的变换。
在这道题中我们将变换稍微改变一下，变成每次
CODE:
#include#include #include #define REP(i, s, n) for(int i = s; i <= n; i ++)#define REP_(i, s, n) for(int i = n; i >= s; i --)#define MAX_N 3#define mod 7using namespace std;int a, b, n;struct node{ int Mtx[MAX_N][MAX_N];}tmp, coe;node operator+(node a,node b){ node c; REP(i, 1, 2) REP(j, 1, 2){ c.Mtx[i][j] = (a.Mtx[i][j] + b.Mtx[i][j]) % mod; } return c;}node operator*(node a,node b){ node c; REP(i, 1, 2) REP(j, 1, 2){ c.Mtx[i][j] = 0; REP(k, 1, 2) c.Mtx[i][j] = (c.Mtx[i][j] + a.Mtx[i][k] * b.Mtx[k][j]) % mod; } return c;} node operator^(node a,int k){ if(k == 0){ memset(a.Mtx, 0, sizeof(a.Mtx)); REP(i, 1, 2) a.Mtx[i][i] = 1; return a; } if(k == 1) return a; node c = a ^ (k >> 1); if(k & 1) return c * c * a; return c * c;} int main(){ while(scanf("%d%d%d", &a, &b, &n) != EOF){ if(a == 0 && b == 0 && n == 0) break; if(n == 1) { cout << 1 << endl; continue; } if(n == 2) { cout << 1 << endl; continue; } else { tmp.Mtx[1][1] = a * 1 + b * 1; tmp.Mtx[1][2] = 1; tmp.Mtx[2][1] = 1; tmp.Mtx[2][2] = 1; coe.Mtx[1][1] = a; coe.Mtx[1][2] = 1; coe.Mtx[2][1] = b; coe.Mtx[2][2] = 0; n -= 3; coe = coe ^ n; tmp = tmp * coe; cout << tmp.Mtx[1][1] << endl; } } return 0;}