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Hdoj 1588 Gauss Fibonacci 【矩阵快速幂】

来源:程序员人生   发布时间:2015-06-16 08:29:58 阅读次数:1346次

Gauss Fibonacci

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2584 Accepted Submission(s): 1078

Problem Description
Without expecting, Angel replied quickly.She says: “I’v heard that you’r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. ”
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(n⑴)+f(n⑵) (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

#include <cstdio> #include <iostream> #include <cstring> using namespace std; #define LL __int64 struct node{ LL num[3][3]; }; node e, a; LL n, m, k, b; node mul(node aa, node bb){ node c; for(int i = 1; i < 3; ++i){ for(int j = 1; j < 3; ++j){ c.num[i][j] = 0; for(int k = 1; k < 3; ++k){ c.num[i][j] = (c.num[i][j]+aa.num[i][k]*bb.num[k][j])%m; } } } return c; } node fa(node a, LL n){ node b = e; while(n){ if(n&1) b = mul(a, b); n >>= 1; a = mul(a, a); } return b; } node add(node aa, node bb){ node c; for(int i = 1; i < 3; ++i){ for(int j = 1; j < 3; ++j){ c.num[i][j] = (aa.num[i][j]+bb.num[i][j])%m; } } return c; } node dg(node p, LL k){ //这里很巧 if(k == 1) return p; else if(k&1) return add(dg(p, k-1), fa(p, k)); //这里就是A^(K⑴)+A^k else return mul(dg(p, k>>1), add(fa(p, k>>1), e));//这里就是 A^k+A^(k>>1); } int main(){ e.num[1][1] = e.num[2][2] = 1; e.num[1][2] = e.num[2][1] = 0; a.num[1][1] = a.num[1][2] = a.num[2][1] = 1; a.num[2][2] = 0; while(cin >> k >> b >> n >> m){ node ak = fa(a, k); node ab = fa(a, b); node ans = dg(ak, n-1); ans = add(e, ans); ans = mul(ab, ans); cout << ans.num[1][2]<< endl; } return 0; }
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