牛客 斐波那契数列问题的递归和动态规划3
题目大意
略。
分析
通项公式变了，要自己构造矩阵。
代码如下
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i) 6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i) 7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 24 25 #define ms0(a) memset(a,0,sizeof(a)) 26 #define msI(a) memset(a,0x3f,sizeof(a)) 27 #define msM(a) memset(a,-1,sizeof(a)) 28 29 #define MP make_pair 30 #define PB push_back 31 #define ft first 32 #define sd second 33 34 template<typename T1, typename T2> 35 istream &operator>>(istream &in, pair<T1, T2> &p) { 36 in >> p.first >> p.second; 37 return in; 38 } 39 40 template<typename T> 41 istream &operator>>(istream &in, vector<T> &v) { 42 for (auto &x: v) 43 in >> x; 44 return in; 45 } 46 47 template<typename T> 48 ostream &operator<<(ostream &out, vector<T> &v) { 49 Rep(i, v.size()) out << v[i] << " \n"[i == v.size() - 1]; 50 return out; 51 } 52 53 template<typename T1, typename T2> 54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 55 out << "[" << p.first << ", " << p.second << "]" << "\n"; 56 return out; 57 } 58 59 inline int gc(){ 60 static const int BUF = 1e7; 61 static char buf[BUF], *bg = buf + BUF, *ed = bg; 62 63 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 64 return *bg++; 65 } 66 67 inline int ri(){ 68 int x = 0, f = 1, c = gc(); 69 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 70 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 71 return x*f; 72 } 73 74 template<class T> 75 inline string toString(T x) { 76 ostringstream sout; 77 sout << x; 78 return sout.str(); 79 } 80 81 inline int toInt(string s) { 82 int v; 83 istringstream sin(s); 84 sin >> v; 85 return v; 86 } 87 88 //min <= aim <= max 89 template<typename T> 90 inline bool BETWEEN(const T aim, const T min, const T max) { 91 return min <= aim && aim <= max; 92 } 93 94 typedef unsigned int uI; 95 typedef long long LL; 96 typedef unsigned long long uLL; 97 typedef vector< int > VI; 98 typedef vector< bool > VB; 99 typedef vector< char > VC; 100 typedef vector< double > VD; 101 typedef vector< string > VS; 102 typedef vector< LL > VL; 103 typedef vector< VI > VVI; 104 typedef vector< VB > VVB; 105 typedef vector< VS > VVS; 106 typedef vector< VL > VVL; 107 typedef vector< VVI > VVVI; 108 typedef vector< VVL > VVVL; 109 typedef pair< int, int > PII; 110 typedef pair< LL, LL > PLL; 111 typedef pair< int, string > PIS; 112 typedef pair< string, int > PSI; 113 typedef pair< string, string > PSS; 114 typedef pair< double, double > PDD; 115 typedef vector< PII > VPII; 116 typedef vector< PLL > VPLL; 117 typedef vector< VPII > VVPII; 118 typedef vector< VPLL > VVPLL; 119 typedef vector< VS > VVS; 120 typedef map< int, int > MII; 121 typedef unordered_map< int, int > uMII; 122 typedef map< LL, LL > MLL; 123 typedef map< string, int > MSI; 124 typedef map< int, string > MIS; 125 typedef set< int > SI; 126 typedef stack< int > SKI; 127 typedef deque< int > DQI; 128 typedef queue< int > QI; 129 typedef priority_queue< int > PQIMax; 130 typedef priority_queue< int, VI, greater< int > > PQIMin; 131 const double EPS = 1e-8; 132 const LL inf = 0x7fffffff; 133 const LL infLL = 0x7fffffffffffffffLL; 134 const LL mod = 1e9 + 7; 135 const int maxN = 1e6 + 7; 136 const LL ONE = 1; 137 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 138 const LL oddBits = 0x5555555555555555; 139 140 struct Matrix{ 141 int row, col; 142 LL MOD; 143 VVL mat; 144 145 Matrix(int r, int c, LL p = mod) : row(r), col(c), MOD(p) { 146 mat.assign(r, VL(c, 0)); 147 } 148 Matrix(const Matrix &x, LL p = mod) : MOD(p){ 149 mat = x.mat; 150 row = x.row; 151 col = x.col; 152 } 153 Matrix(const VVL &A, LL p = mod) : MOD(p){ 154 mat = A; 155 row = A.size(); 156 col = A[0].size(); 157 } 158 159 // x * 单位阵 160 inline void E(int x = 1) { 161 assert(row == col); 162 Rep(i, row) mat[i][i] = x; 163 } 164 165 inline VL& operator[] (int x) { 166 assert(x >= 0 && x < row); 167 return mat[x]; 168 } 169 170 inline Matrix operator= (const VVL &x) { 171 row = x.size(); 172 col = x[0].size(); 173 mat = x; 174 return *this; 175 } 176 177 inline Matrix operator+ (const Matrix &x) { 178 assert(row == x.row && col == x.col); 179 Matrix ret(row, col); 180 Rep(i, row) { 181 Rep(j, col) { 182 ret.mat[i][j] = mat[i][j] + x.mat[i][j]; 183 ret.mat[i][j] %= MOD; 184 } 185 } 186 return ret; 187 } 188 189 inline Matrix operator* (const Matrix &x) { 190 assert(col == x.row); 191 Matrix ret(row, x.col); 192 Rep(k, x.col) { 193 Rep(i, row) { 194 if(mat[i][k] == 0) continue; 195 Rep(j, x.col) { 196 ret.mat[i][j] += mat[i][k] * x.mat[k][j]; 197 ret.mat[i][j] %= MOD; 198 } 199 } 200 } 201 return ret; 202 } 203 204 inline Matrix operator*= (const Matrix &x) { return *this = *this * x; } 205 inline Matrix operator+= (const Matrix &x) { return *this = *this + x; } 206 }; 207 208 // 矩阵快速幂，计算x^y 209 inline Matrix mat_pow_mod(Matrix x, LL y) { 210 Matrix ret(x.row, x.col); 211 ret.E(); 212 while(y){ 213 if(y & 1) ret *= x; 214 x *= x; 215 y >>= 1; 216 } 217 return ret; 218 } 219 220 LL N; 221 Matrix X(VVL({{1, 1, 0}, {0, 0, 1}, {1, 0, 0}})); 222 Matrix ans(VVL({{3, 2, 1}})); 223 224 int main(){ 225 //freopen("MyOutput.txt","w",stdout); 226 //freopen("input.txt","r",stdin); 227 //INIT(); 228 scanf("%lld", &N); 229 ans = ans * mat_pow_mod(X, N - 1); 230 printf("%lld\n", ans.mat[0][2]); 231 return 0; 232 }
