【Codeforces】450 B(div2)
题目链接：http://codeforces.com/problemset/problem/450/B
题意：
求这个的第n项。
题解：$f_{i+1} = f_i - f_{i-1} $
\begin{pmatrix} 1 & 1\\ -1 & 0 \end{pmatrix} *
\begin{pmatrix} f(n)& f(n-1) \end{pmatrix} =
\begin{pmatrix} f(n) - f(n-1) & f(n) \end{pmatrix}=
\begin{pmatrix} f(n+1) & f(n) \end{pmatrix}
代入前两项即可。
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define ll long long 6 const int maxn = 3; 7 const ll mod = 1e9+7; 8 9 ll n,p; 10 11 struct Matrix{ 12 ll a[maxn][maxn]; 13 void init(){ 14 memset(a, 0, sizeof(a)); 15 for(int i = 1; i <= maxn;i++){ 16 a[i][i] = 1; 17 } 18 } 19 }; 20 21 //矩阵乘法 22 Matrix mul(Matrix a, Matrix b){ 23 Matrix ans; 24 for(int i = 1;i <= 2;++i){ 25 for(int j = 1;j <= 2;++j){ 26 ans.a[i][j] = 0; 27 for(int k = 1;k <= 2;++k){ 28 ans.a[i][j] = ans.a[i][j] % mod + a.a[i][k] * b.a[k][j] % mod; 29 ans.a[i][j] %= mod; 30 } 31 } 32 } 33 return ans; 34 } 35 36 //矩阵快速幂 37 Matrix qpow(Matrix a,ll b){ 38 Matrix ans; 39 ans.init(); 40 while(b){ 41 if(b & 1) 42 ans = mul(ans,a); 43 a = mul(a,a); 44 b >>= 1; 45 } 46 return ans; 47 } 48 49 void print(Matrix a){ 50 for(int i = 1; i <= n;++i){ 51 for(int j = 1;j <= n;++j){ 52 cout << a.a[i][j]%mod<< " "; 53 } 54 cout << endl; 55 } 56 } 57 58 int main(){ 59 Matrix base; 60 Matrix ans; 61 int x,y,n; 62 cin>>x>>y>>n; 63 if(n == 1){ 64 cout<<(mod + x) % mod<<endl; 65 return 0; 66 } 67 if( n == 2){ 68 cout<<(mod + y) % mod<<endl; 69 return 0; 70 } 71 72 ans.a[1][1] = y;ans.a[1][2] = x; 73 base.a[1][1] = 1;base.a[1][2] = 1; 74 base.a[2][1] = -1;base.a[2][2] = 0; 75 76 ans = mul(ans,qpow(base,n-2)); 77 ll res = (mod + ans.a[1][1]) % mod; 78 cout<<res<<endl; 79 return 0; 80 }
