Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1951    Accepted Submission(s): 750

Problem Description

Let us define a sequence as below

⎧⎩⎨⎪⎪⎪⎪⎪⎪F1F2Fn===ABC⋅Fn−2+D⋅Fn−1+⌊Pn⌋

Your job is simple, for each task, you should output Fn module 109+7.

Input

The first line has only one integer T, indicates the number of tasks.

Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.

1≤T≤200≤A,B,C,D≤1091≤P,n≤109

Sample Input

2 3 3 2 1 3 5 3 2 2 2 1 4

Sample Output

36 24

Source

2018 Multi-University Training Contest 7

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学习了一个新的递推求数的方法

通过构造矩阵 用矩阵递推 用矩阵快速幂优化 可以用来求一个随机的很大的数的值

矩阵快速幂的写法 O(logn)

这道题还有一点特别的是 后面的常数项是会变化的 但是他是分段的

所以就分块的来求 因为要知道乘的次数所以每次要求一下这个区间有多少个数

c++TLE g++AC

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#define inf 1e18
using namespace std;long long t, a, b, c, d, p, n;
const int mod = 1e9 + 7;
struct mat{long long m[3][3];mat(){memset(m, 0, sizeof(m));}void init(){memset(m, 0, sizeof(m));for(int i = 0; i < 3; i++){m[i][i] = 1;}}friend mat operator * (mat a, mat b){mat c;for(int i = 0; i < 3; i++){for(int j = 0; j < 3; j++){for(int k = 0; k < 3; k++){c.m[i][j] += a.m[i][k] * b.m[k][j];c.m[i][j] %= mod;}}}return c;}
};mat pow_mat(mat a, int b)
{mat c;c.init();while(b){if(b & 1){c = c * a;}a = a * a;b >>= 1;}return c;
}int main()
{scanf("%lld", &t);while(t--){scanf("%lld%lld%lld%lld%lld%lld", &a, &b, &c, &d, &p, &n);if(n == 1){printf("%lld\n", a);continue;}mat f;f.m[0][0] = d;f.m[1][0] = c;f.m[2][0] = f.m[0][1] = f.m[2][2] = 1;mat g;g.m[0][0] = b;g.m[0][1] = a;if(p >= n){for(long long i = 3, j; i <= n; i = j + 1){j = p / (p / i);//个数g.m[0][2] = p / i;mat po = pow_mat(f, min(j - i + 1, n - i + 1));g = g * po;}}else{for(long long i = 3, j; i <= p; i = j + 1){j = p / (p / i);g.m[0][2] = p / i;mat po = pow_mat(f, j - i + 1);g = g * po;}mat po;g.m[0][2] = 0;if(p < 3){po = pow_mat(f, n - 2);}else{po = pow_mat(f, n - p);}g = g * po;}printf("%lld\n", g.m[0][0]);}return 0;
}