Codeforces 450B Jzzhu and Sequences

题目链接：http://codeforces.com/problemset/problem/450/B

题意：给出f1,f2，fi = fi-1 + fi+1 ,求fn mod 1e9+7.

思路：fi+1 = fi - fi-1 , 由于n很大，我们可以构造一个转移矩阵（fi , fi+1）×  

「  0   -1          =>  ( fi+1 , fi+2 )

      1    1 」

一开始我们知道f1,f2 那么只需要再转移n-1次，就可以得到fn了。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod 1000000007int x , y;int n;struct matrix{    LL a[2][2];    void init(LL A = 0,LL B = 0,LL C = 0,LL D = 0)    {        a[0][0] = A , a[0][1] = B , a[1][0] = C , a[1][1] = D;    }    void E()    {        a[0][0] = a[1][1] = 1;        a[1][0] = a[0][1] = 0;    }};matrix multi( matrix & tx , matrix & ty ){    matrix ans;    ans.init();    rep( i , 0 , 1 )        rep( j , 0 , 1 )            rep( k , 0 , 1 )            ans.a[i][j] = ( ( ans.a[i][j] + tx.a[i][k] * ty.a[k][j] ) % mod + mod ) % mod ;    return ans;}int solve(){    matrix ans,temp;    temp.init(0,-1,1,1);    ans.E();    if ( n == 1 ) return ( x % mod + mod ) % mod;    if ( n == 2 ) return ( y % mod + mod ) % mod;    n--;    while( n )    {        if ( n & 1 ) ans = multi( ans , temp );        temp = multi( temp , temp );        n >>= 1;    }    LL t = ( ( x * ans.a[0][0] + y * ans.a[1][0] ) % mod + mod ) % mod;    return t;}int main(){    cin>>x>>y>>n;    cout<<solve()<<endl;    return 0;}