HDU 3936 FIB Query

题目链接:http://henuly.top/?p=680

题目:

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

$p[1]+p[2]+...+p[n]$

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

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

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。

$\left[\begin{matrix}f(n)\\f(n-1)\end{matrix}\right]=\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\times\left[\begin{matrix}f(n-1)\\f(n-2)\end{matrix}\right] \tag{矩阵快速幂求斐波那契数列递推式}$

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;
}