Description
How good an opportunity that Gardon can not give up! The "Problem GF" told by Angel is actually "Gauss Fibonacci".
As we know ,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly, and Fibonacci is the crazy man who invented some numbers.
Arithmetic progression:
g(i)=k*i+b;
We assume k and b are both non-nagetive integers.
Fibonacci Numbers:
f(0)=0
f(1)=1
f(n)=f(n-1)+f(n-2) (n>=2)
The Gauss Fibonacci problem is described as follows:
Given k,b,n ,calculate the sum of every f(g(i)) for 0<=i<n
The answer may be very large, so you should divide this answer by M and just output the remainder instead.
Input
Each of them will not exceed 1,000,000,000.
Output
Sample Input
2 1 4 100
2 0 4 100
Sample Output
21
12
题意：g(i)=k*i+b , 求全部的f(g(i))在n的范围里
思路：借鉴：构造矩阵：
|1 1| | f(2) f(1)|
A= |1 0| = | f(1) f(0)|
|1 1| ^b | f(b+1) f(b)|
A^b =|1 0| = | f(b) f(b-1)|
f(b) = matrix[0][1]=matrix[1][0];
首项是：A^b
公比是：A^k
项数是：N
能够把问题进一步简化
由于矩阵的加法对乘法也符合分配律，我们提出一个A^b来，形成这种式子：
A^b*( I + A^k + (A^k)^2 + .... + (A^k)^(N-1) )
A^b 和 A^k 显然都能够用我们之前说过的方法计算出来，这剩下一部分累加怎么解决呢
设A^k=B
要求 G(N)=I + ... + B^(N-1)，
i=N/2
若N为偶数，G(N)=G(i)+G(i)*B^i = G(i) *( I+B^(i));
若N为奇数，G(N)=I+ G(i)*B + G(i) * (B^(i+1)) = G(N-1)+B^N; (前一个等式可能要快点，可是后面更简练)
我们来设置这样一个矩阵
B I
O I
当中O是零矩阵，I是单位矩阵
将它乘方，得到
B^2 I+B
O I
乘三方，得到
B^3 I+B+B^2
O I
乘四方，得到
B^4 I+B+B^2+B^3
O I
既然已经转换成矩阵的幂了，继续用我们的二分或者二进制法，直接求出幂就能够了
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
typedef long long ll;
using namespace std;
const int maxn = 2; int m;
struct Matrix {
ll v[maxn][maxn];
Matrix() {}
Matrix(int x) {
init();
for (int i = 0; i < maxn; i++)
v[i][i] = x;
}
void init() {
memset(v, 0, sizeof(v));
}
Matrix operator *(Matrix const &b) const {
Matrix c;
c.init();
for (int i = 0; i < maxn; i++)
for (int j = 0; j < maxn; j++)
for (int k = 0; k < maxn; k++)
c.v[i][j] = (c.v[i][j] + (v[i][k]*b.v[k][j])%m) % m;
return c;
}
Matrix operator ^(int b) {
Matrix a = *this, res(1);
while (b) {
if (b & 1)
res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
} u, em; Matrix Add(Matrix a, Matrix b) {
for (int i = 0; i < maxn; i++)
for (int j = 0; j < maxn; j++)
a.v[i][j] = (a.v[i][j]+b.v[i][j]) % m;
return a;
} Matrix BinarySum(Matrix a, int n) {
if (n == 1)
return a;
if (n & 1)
return Add(BinarySum(a, n-1), a^n);
else return BinarySum(a, n>>1) * Add(u, a^(n>>1));
} int main() {
int k, b, n;
u.init(), em.init();
u.v[0][0] = 1, u.v[0][1] = 0, u.v[1][0] = 0, u.v[1][1] = 1;
em.v[0][0] = 1, em.v[0][1] = 1, em.v[1][0] = 1, em.v[1][1] = 0;
while (scanf("%d%d%d%d", &k, &b, &n, &m) != EOF) {
Matrix t1, t2, ans;
t1 = em^b;
t2 = em^k;
ans = Add(u, BinarySum(t2, n-1)) * t1;
cout << ans.v[0][1] << endl;
}
return 0;
}