Problem description.
The Fibonacci numbers defined as f(n) = f(n-1) + f(n-2) where f0 = 0 and f1 = 1.
We define a function as follows D(n,x) = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 +...+f(n)x^n
Given two integers n and x, you need to compute D(n,x) since the output can be very large output the result modulo 1000000007 (1e9+7) .
Input
Input description.
- The first line of the input contains an integer T denoting the number of test cases.
The description of T test cases follows. - The first line of each test case contains two integers n and x as described above.
Output
Output description.
- For each test case, output D(n,x)%1000000007 in a seperate line.
Constraints
Should contain all the constraints on the input data that you may have. Format it like:
- 1 ≤ T ≤ 1000
- 0 ≤ n ≤ 10^15
- 0 ≤ x ≤ 10^15
Example
Input: 1 7 11 Output: 268357683
题意：f(n)是斐波拉契数列，g(n)=f(n)*x^n；求前N项的g(n)的累加和。
思路：易得g(n)=g(n-1)*x+g(n-2)*x^2，
可以得到g(n)的矩阵求解方程： g(n)=base^N*g(0); 其中，base[1][1]=X; base[1][2]=X^2; base[2][1]=1（由递推式得到）;
前缀和可以由大矩阵得到：A[1][1]=base; A[1][2]=1; A[2][2]=1（有前缀和求和公式得到） ；
大概地解释了一下，不是很清楚，可以看代码。前缀和可以参考这里：http://www.cnblogs.com/hua-dong/p/8479103.html
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
const int Mod=1e9+7;struct mat
{
ll mp[5][5];
mat(){memset(mp,0,sizeof(mp)); }
mat friend operator *(mat a,mat b)
{
mat res;
for(int k=1;k<=4;k++)
for(int i=1;i<=4;i++)
for(int j=1;j<=4;j++)
res.mp[i][j]=(res.mp[i][j]+(a.mp[i][k]*b.mp[k][j])%Mod)%Mod;
return res;
}
mat friend operator ^(mat a,ll x)
{
mat res;
for(int i=1;i<=4;i++) res.mp[i][i]=1;
while(x){
if(x&1) res=res*a; a=a*a; x>>=1;
} return res;
}
};
int main()
{
ll T,N,X;
scanf("%lld",&T);
while(T--){
scanf("%lld%lld",&N,&X); X%=Mod;
mat base;
base.mp[1][1]=X; base.mp[1][2]=X*X%Mod; base.mp[2][1]=1;
for(int i=1;i<=2;i++) base.mp[i][i+2]=base.mp[i+2][i+2]=1;
base=base^N;
printf("%lld\n",base.mp[1][3]*X%Mod);
}
return 0;
}
原文地址：https://www.cnblogs.com/hua-dong/p/8893887.html