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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1005

题目大意：

给你一个递推公式，求出第n项。由于某项可能太大，所以取余7

解题思路：

矩阵二分幂的经典运用。

代码如下：

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<climits>
using namespace std;#define CLR(arr, what) memset(arr, what, sizeof(arr))
typedef long long LL;
const LL Size = 3;
const LL MOD = 7; //取余class Matrix67
{
public:Matrix67(const LL ff1, const LL ff2, const LL aa, const LL bb, const LL cc, const LL nn); //构造函数void copymat(LL mata[Size][Size], LL matb[Size][Size]); //矩阵复制:后->前void binary(LL n); //矩阵二分幂：次数void multiply(LL mata[Size][Size], LL matb[Size][Size]); //矩阵连乘LL result(); //打印结果private:LL f1, f2, a, b, c, n;LL mat[Size][Size], temp[Size][Size], mid[Size][Size]; //二分幂、递推、单位矩阵
};Matrix67::Matrix67(const LL ff1, const LL ff2, const LL aa, const LL bb, const LL cc, const LL nn)
{CLR(mat, 0); CLR(temp, 0); CLR(mid, 0);f1 = ff1, f2 = ff2, a = aa, b = bb, c = cc, n = nn;mat[0][1] = aa, mat[1][0] = 1, mat[1][1] = bb, mat[2][1] = 1, mat[2][2] = 1; //初始化单位矩阵temp[0][0] = ff1, temp[0][1] = ff2, temp[0][2] = cc; //初始化递推矩阵copymat(mid, mat);
}void Matrix67::copymat(LL mata[Size][Size], LL matb[Size][Size])
{for(int i = 0; i < Size; ++i)for(int j = 0; j < Size; ++j)mata[i][j] = matb[i][j];
}void Matrix67::binary(LL n)
{if(n == 1)return ;binary(n >> 1);multiply(mat, mat);if(n & 1) //奇次幂multiply(mat, mid);
}void Matrix67::multiply(LL mata[Size][Size], LL matb[Size][Size])
{LL sum, tempmat[Size][Size];for(int i = 0; i < Size; ++i){for(int j = 0; j < Size; ++j){sum = 0;for(int k = 0; k < Size; ++k)sum = (sum + mata[i][k] * matb[k][j]) % MOD;tempmat[i][j] = (sum + MOD) % MOD;}}copymat(mata, tempmat);
}LL Matrix67::result()
{if(n > 1){binary(n - 1);multiply(temp, mat);}return (temp[0][0] % MOD + MOD) % MOD; //temp[0][1]是f(n+1)，注意～
}int main()
{LL a, b, n;while(scanf("%lld%lld%lld", &a, &b, &n) && a && b && n){Matrix67 m(1, 1, b, a, 0, n); //af(n-1)+bf(n-2),顺序调换～～printf("%lld\n",m.result() % MOD);}return 0;
}