Spoj FIBOSUM - Fibonacci Sum 矩阵快速幂

题目链接：https://www.spoj.com/problems/FIBOSUM/
The fibonacci sequence is defined by the following relation:

• F(0) = 0
• F(1) = 1
• F(N) = F(N - 1) + F(N - 2), N >= 2

Your task is very simple. Given two non-negative integers N and M, you have to calculate the sum (F(N) + F(N + 1) + … + F(M)) mod 1000000007.

Input

The first line contains an integer T (the number of test cases). Then, T lines follow. Each test case consists of a single line with two non-negative integers N and M.

Output

For each test case you have to output a single line containing the answer for the task.

Example

Input:

3
0 3
3 5
10 19

Output:

4
10
10857

Constraints

T <= 1000
0 <= N <= M <= $10^9$

矩阵快速幂，递推公式如下：

两个需要注意的地方：

1. 注意边界值处理。
2. 计算u到v之间的和时不是简单地计算Sum(v)-Sum(u-1)，中间计算时有取模，这样计算出来的结果可能为负值，如果Sum(v) < Sum(u-1)结果应该为Sum(v)+MOD-Sum(u-1)。

#include 
using namespace std;

typedef long long LL;
typedef vector<LL> vec;
typedef vector<vec> mat;

const LL MOD = 1000000007;

mat mul(mat &A,mat &B)
{
    mat C(A.size(),vec(B[0].size()));
    for(int i = 0; i < A.size(); i++){
    	for(int k = 0; k < B.size(); k++){
			if(A[i][k] != 0){
				for(int j = 0; j < B[0].size(); j++){
					C[i][j]= ((C[i][j] % MOD)  + (A[i][k] * B[k][j]) % MOD) % MOD;
				}
			}
		}
	}
    return C;
}


mat pow(mat A,LL n)
{
    mat B(A.size(),vec(A[0].size()));
    for(int i = 0;i < A.size(); i++)
    	B[i][i] = 1;
    while(n > 0){
		if(n & 1)
			B = mul(B,A);
            A = mul(A,A);
            n >>= 1;
        }
        return B;
}


int main(int argc, char const *argv[])
{
	LL arr[4][3] = {{0,1,0},{1,1,0},{1,1,1},{0,1,1}};
	vector<LL> temp(3);
	mat A,B;
	
	for(int i = 0;i < 3; i++){
		for(int j = 0;j < 3; j++) 
			temp[j] = arr[i][j];
		A.push_back(temp);
	}
	
	for(int i = 0;i < 3; i++){
		vector<LL> temp2;
		temp2.push_back(arr[3][i]);
		B.push_back(temp2);
	}
	
	map<int,LL> mp;
	int m,u,v;
	scanf("%d",&m);
	mp[-1] = 0;mp[0] = 0;mp[1] = 1;
	for(int i = 0;i < m; i++){
		scanf("%d%d",&u,&v);
		if(mp.count(u-1) == 0){
			mat powAns1 = pow(A,u-2);
			mp[u-1] = mul(powAns1,B)[2][0];
		}
		if(mp.count(v) == 0){
			mat powAns2 = pow(A,v-1);
			mp[v] = mul(powAns2,B)[2][0];
		}
		
		if(mp[v] > mp[u-1])
			cout << (mp[v] - mp[u-1]) % MOD << endl;
		else
			cout << (mp[v]+MOD - mp[u-1]) << endl;
	}
	
	return 0;
}