2018年湘潭大学程序设计竞赛G又见斐波那契
链接：https://www.nowcoder.com/acm/contest/105/G
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld
题目描述
这是一个加强版的斐波那契数列。
给定递推式%20%3D%20%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%20%5C%5CF(i-1)%20%2B%20F(i-2)%20%2B%20i%5E3%20%2B%20i%5E2%20%2B%20i%20%2B%201%20%26%20i%3E1%20%5C%5C0%20%26%20i%20%3D%200%20%5C%5C1%20%26%20i%20%3D%201%5Cend%7Barray%7D%5Cright.)
求F(n)的值，由于这个值可能太大，请对109+7取模。
输入描述:
第一行是一个整数T(1 ≤ T ≤ 1000)，表示样例的个数。18
以后每个样例一行，是一个整数n(1 ≤ n ≤ 10
)。
输出描述:
每个样例输出一行，一个整数，表示F(n) mod 1000000007。
示例1
输入
4 1 2 3 100
输出
1 16 57 558616258
这题我必须要说句MMP才能开始讲，题目你给出T范围是1 <= T <= 1000，结果实际数据T，TMD超过了int，excuse me????真TM的坑,教会我做人了，找了2小时bug，一遍一遍排查哦。。。
知识储备：快速幂: https://www.cnblogs.com/linyujun/p/5194184.html
矩阵相乘运算：https://blog.csdn.net/wust_zzwh/article/details/52058209
不会不要紧，几句话就解释了，快速幂就是求 a ^ b次方(假装你不知道逆元)，矩阵相乘：线性代数内容,c[i][j]为A的第i行与B的第j列对应乘积的和，即:
-------------------------------------------------------------正式开始讲题目-------------------------------------------------------------------------------------------------
注：本题一开始我想的是 2018年东北农业大学春季校赛 的wyh的数列那题，找循环节：https://www.nowcoder.com/acm/contest/93/K
但是发现找不到，事后才知道有矩阵快速幂这一说（别想着暴力，10^18次方你装不下的，还有T卡你呢!!!）所以，那就开始矩阵快速幂了：
步骤1> 先了解斐波拉契数列的基本矩阵相乘和快速幂和混合运用
Fi[i] = 1*Fi[i-1]+F*fi[i-2] Fi[i-1] = 1*F[i-1] + 0*f[i-2];即
我们这里不妨把[1 1 当成一个数，也就是求res ^ （n- 1）,那么就是直接快速幂了，怎么转化成数？结构体里面写好了
1, 0]
观察上列公式，这里求出来的res ^ (n - 1)就是直接是答案了,因为F[n]是res[0][0] * F(1),而F(1)=1.F[0] = 0，则不考虑res[0][1]，代码如下:
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JgGgBQDOSJss77r 67 :r;::: , PH r, ,;5: ,::::;7Ls7Lc7;: ,:7JP17rJUs 68 .:J: :; .,:K6BQS7:.,.,. 69 :r7Ji;r7;::;;. 70 . **/ 71 72 #include <bits/stdc++.h> 73 using namespace std; 74 75 #define FF(i, a, b) for(int i = a; i < b; i++) 76 #define RR(i, a, b) for(int i = a; i > b; i++) 77 #define ME(a, b) memset(a, b, sizeof(a)) 78 #define SC(x) scanf("%d", &x) 79 #define PR(x) printf("%d\n", x) 80 #define INF 0x3f3f3f3f 81 #define MAX 1001 82 #define MOD 1000000007 83 #define E 2.71828182845 84 #define PI 3.14159265358979 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 typedef pair<int, int> Author; 89 vector<pair<string, int> > VP; 90 91 struct Matrix{ 92 LL a[2][2]; 93 Matrix(){ME(a, 0);} 94 Matrix operator * (const Matrix y){ 95 Matrix ans; 96 //¾ØÕóÏà³Ë 97 for(int i = 0; i <= 1; i++)for(int j = 0; j <= 1; j++)for(int k = 0; k <= 1; k++)ans.a[i][j] += a[i][k] * y.a[k][j]; 98 return ans; 99 } 100 void operator = (const Matrix b){ 101 for(int i = 0; i <= 1; i++)for(int j = 0; j <= 1; j++)a[i][j] = b.a[i][j]; 102 } 103 }; 104 int Solve(LL x){ 105 Matrix ans, trs; 106 ans.a[0][0] = ans.a[1][1] = 1; //µ¥Î»¾ØÕó 107 trs.a[0][0] = trs.a[1][0] = trs.a[0][1] = 1; //¹¹Ôì¾ØÕó 108 while(x){ 109 if(x & 1) ans = ans * trs; 110 trs = trs * trs; 111 x >>= 1; 112 } 113 return ans.a[0][0]; 114 } 115 116 int main(void){ 117 #ifdef LOCAL 118 freopen("in.txt", "r", stdin); 119 freopen("out.txt", "w", stdout); 120 #endif 121 ios::sync_with_stdio(false); cin.tie(0); 122 int n; 123 cin>>n; 124 125 cout<<Solve(n - 1)<<endl; //ÇóµÚF[n] 126 127 return EXIT_SUCCESS; 128 }
步骤2> 你了解了基本F[n] = F[n - 1] + F[n - 1]后，但是这题不一样呀，那怎么做？其实道理一样，只是重新构建res矩阵即可
上述代码中，ans = ans * trs，这里其实乘以单元矩阵就是其本身，就好像快速幂里面要把trs初始化为1一样，这里我们可以不需要乘单元矩阵了，因为trs需要改变成新的矩阵，而,ans也要改变，其余代码均不变！！！
我们继续来分析上上个图的公式n - 1可通过乘矩阵来求F[n]，延伸一下
我要求本题的
F[n], F[n - 1], (i + 1) ^ 3, (i + 1) ^ 2, (i + 1), 1(即把i都增加1，和第一个公式一样) 所以就根据矩阵
f[i] = 1 1 1 1 1 1 f[i-1] 1
f[i-1] = 1 0 0 0 0 0 f[i-2] 0
(i+1)^3 = 0 0 1 3 3 1 i^3 8
(i+1)^2 = 0 0 0 1 2 1 * i^2 4
i + 1 = 0 0 0 0 1 1 i 2
1 = 0 0 0 0 0 1 1 1
想比这时你发现做矩阵快速幂的规律了，就是把所给的公式i + 1在右边列出，原本公式右边列出，再求矩阵即可，而这即为ans,那么trs呢？就是开头的[1, 0]，再把每行都加起来（或者说是2 ^ n次）！
直接上代码
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总结：从入门到崩溃，回想起来才发现没什么，之前一直没发现矩阵相乘可以运用到编程中，这次终于运用到了，加油吧，Acmer们.
我曾七次鄙视自己的灵魂：
第一次，当它本可进取时，却故作谦卑；
第二次，当它在空虚时，用爱欲来填充；
第三次，在困难和容易之间，它选择了容易；
第四次，它犯了错，却借由别人也会犯错来宽慰自己；
第五次，它自由软弱，却把它认为是生命的坚韧；
第六次，当它鄙夷一张丑恶的嘴脸时，却不知那正是自己面具中的一副；
第七次，它侧身于生活的污泥中，虽不甘心，却又畏首畏尾。
