HDU 3936 FIB Query

Description:

We all know the definition of Fibonacci series: fib[i]=fib[i-1]+fib[i-2],fib[1]=1,fib[2]=1.And we define another series P associated with the Fibonacci series: P[i]=fib[4*i-1].Now we will give several queries about P:give two integers L,R, and calculate ∑P[i](L <= i <= R).

Input:

There is only one test case.
The first line contains single integer Q – the number of queries. (Q<=10^4)
Each line next will contain two integer L, R. (1<=L<=R<=10^12)

Output:

For each query output one line.
Due to the final answer would be so large, please output the answer mod 1000000007.

Sample Input:

2
1 300
2 400

Sample Output:

838985007
352105429

题目链接

$p\left[1\right]+p\left[2\right]+...+p\left[n\right]$p[1]+p[2]+...+p[n]

$=f[1{]}^2+f[2{]}^2+...+f[2\times n-1{]}^2+f[2\times n{]}^2$=f[1]2+f[2]2+...+f[2×n−1]2+f[2×n]2

$=f[2\times n]\times f[2\times n+1]$=f[2×n]×f[2×n+1]

利用此公式求得P，由于数据范围很大，斐波那契数列需要用矩阵快速幂递推求得。

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AC代码:

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
bool Finish_read;
template<class T>inline void read(T &x) {
	Finish_read = 0;
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') {
			f = -1;
		}
		if (ch == EOF) {
			return;
		}
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	x *= f;
	Finish_read = 1;
};

struct Matrix {
	ll mat[2][2];
	Matrix() {}
	Matrix operator * (Matrix const &a) const {
		Matrix res;
		mem(res.mat, 0);
		for (int i = 0; i < 2; ++i) {
			for (int j = 0; j < 2; ++j) {
				for (int k = 0; k < 2 ; ++k) {
					res.mat[i][j] = (res.mat[i][j] + mat[i][k] * a.mat[k][j] % mod) % mod;
				}
			}
		}
		return res;
	}
};

Matrix operator ^ (Matrix base, ll k) {
	Matrix res;
	mem(res.mat, 0);
	res.mat[0][0] = res.mat[1][1] = 1;
	while (k) {
		if (k & 1) {
			res = res * base;
		}
		base = base * base;
		k >>= 1;
	}
	return res;
}

ll Fib(ll x) {
	Matrix base;
	base.mat[0][0] = base.mat[1][0] = base.mat[0][1] = 1;
	base.mat[1][1] = 0;
	return (base ^ x).mat[0][1];
}

ll P(ll x) {
	return Fib(2 * x + 1) * Fib(2 * x) % mod;
}

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	ll Q;
	read(Q);
	for (ll Query = 1, l, r; Query <= Q; ++Query) {
		read(l); read(r);
		printf("%lld\n", ((P(r) - P(l - 1)) % mod + mod) % mod);
	}
#ifndef ONLINE_JUDGE
	fclose(stdin);
	fclose(stdout);
	system("gedit out.txt");
#endif
    return 0;
}