Hoj 2060 Fibonacci Problem Again

题目链接：http://acm.hit.edu.cn/hoj/problem/view?id=2060

题意：求Fibs和。

f[n] = f[n-1] + f[n-2]。

使用矩阵乘法很容易求出

f[n] = [0 1

           1 1]

这个矩阵的n-1次方，然后这个二维矩阵的左下角和右下角的数值相加即可。

对于本题，我们要求sigma(f[i]) ,0<=i<=n

则，我们可以观察到：

F(3) = F(1) + F(2) 

F(4) = F(2) + F(3) = 1 * F(1) + 2 * F(2) 

F(5) = F(3) + F(4) = 2 * F(1) + 3 * F(2)

 F(6) = F(4) + F(5) = 3 * F(1) + 5 * F(2)

 F(7) = F(5) + F(6) = 5 * F(1) + 8 * F(2)

 F(8) = F(6) + F(7) = 8 * F(1) + 13 * F(2) 

S(3) = 2 * F(1) + 2 * F(2) 

S(4) = 3 * F(1) + 4 * F(2) 

S(5) = 5 * F(1) + 7 * F(2)

 S(6) = 8 * F(1) + 12 *F(2) 

S(7) = 13 *F(1) + 20 *F(2) 
不难发现，S(n) = F(n + 2) - F(2) 
因此题目就转换为了求 F(b + 2) - F(a + 2 - 1)

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <map>#include <queue>#include <algorithm>using namespace std;#define MOD 1000000000#define MATRIX_SIZE 3struct Matrix{   long long int elem[MATRIX_SIZE][MATRIX_SIZE];   int size;   Matrix(){memset(elem,0,sizeof(elem));}   void setSize(int _size)   {      size = _size;   }   Matrix operator = (const Matrix & other)   {      setSize(other.size);      for(int i=0;i<size;i++)      {         for(int j= 0;j<size;j++)         {            elem[i][j] = other.elem[i][j];         }      }      return *this;   }   Matrix operator * (const Matrix & other)   {      Matrix temp;      temp.setSize(size);      for(int i=0;i<size;i++)      {         for(int j=0;j<size;j++)         {            for(int k=0;k<size;k++)            {               temp.elem[i][j] += elem[i][k] * other.elem[k][j];               if(temp.elem[i][j]>=MOD) temp.elem[i][j] %= MOD;            }         }      }      return temp;   }   void Power(int exp)   {      Matrix E;      E.setSize(size);      for(int i=0;i<size;i++) E.elem[i][i] = 1;      while(exp)      {         if(exp & 1) E = E * (*this);         *this = (*this) * (*this);         exp >>= 1;      }      *this = E;   }};Matrix m;void init(){   m.setSize(2);   memset(m.elem,0,sizeof(m.elem));   m.elem[0][1] = 1;m.elem[1][0] = 1;m.elem[1][1] = 1;}//S(n) = f(n+2) - f(2) + f(0) = f(n+2) - 1int main(){   #ifndef ONLINE_JUDGE      freopen("in.txt","r",stdin);   #endif   int a,b;   int sa,sb;   while(scanf(" %d %d",&a,&b)!=EOF)   {      if(a == 0 && b == 0 ) break;      init();      m.Power(b+1);      sb = (m.elem[1][0] + m.elem[1][1] - 1)%MOD;      init();      m.Power(a);      sa = (m.elem[1][0] + m.elem[1][1] - 1)%MOD;      printf("%d\n", (sb-sa + MOD)%MOD);   }   return 0;}