题目链接：https://www.luogu.org/problemnew/show/P1962
题目大意：
略
分析：
由于数据规模很大，需要用矩阵快速幂来解。
代码如下：
1 #pragma GCC optimize("Ofast") 2 #include <bits/stdc++.h> 3 using namespace std; 4 5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 6 #define Rep(i,n) for (int i = 0; i < (n); ++i) 7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 13 14 #define inI(x) scanf("%d", &x) 15 #define inII(x, y) scanf("%d%d", &x, &y) 16 #define inIII(x, y, z) scanf("%d%d", &x, &y, &z) 17 #define inL(x) scanf("%lld", &x) 18 #define outI(x) printf("%d", &x) 19 #define outI_S(x) printf("%d ", &x) 20 #define outL(x) printf("%lld", &x) 21 #define outL_S(x) printf("%lld ", &x) 22 #define outD(x) printf("%f", &x) 23 #define outD_S(x) printf("%f ", &x) 24 #define EL() printf("\n") 25 26 #define pr(x) cout << #x << " = " << x << " " 27 #define pr(x) cout << #x << " = " << x << " " 28 #define pr(x) cout << #x << " = " << x << " " 29 #define prln(x) cout << #x << " = " << x << endl 30 31 #define LOWBIT(x) ((x)&(-x)) 32 33 #define ALL(x) x.begin(),x.end() 34 #define INS(x) inserter(x,x.begin()) 35 36 #define ms0(a) memset(a,0,sizeof(a)) 37 #define msI(a) memset(a,inf,sizeof(a)) 38 #define msM(a) memset(a,-1,sizeof(a)) 39 40 #define MP make_pair 41 #define PB push_back 42 #define ft first 43 #define sd second 44 45 template<typename T1, typename T2> 46 istream &operator>>(istream &in, pair<T1, T2> &p) { 47 in >> p.first >> p.second; 48 return in; 49 } 50 51 template<typename T> 52 istream &operator>>(istream &in, vector<T> &v) { 53 for (auto &x: v) 54 in >> x; 55 return in; 56 } 57 58 template<typename T1, typename T2> 59 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 60 out << "[" << p.first << ", " << p.second << "]" << "\n"; 61 return out; 62 } 63 64 typedef long long LL; 65 typedef unsigned long long uLL; 66 typedef pair< double, double > PDD; 67 typedef pair< int, int > PII; 68 typedef set< int > SI; 69 typedef vector< int > VI; 70 typedef map< int, int > MII; 71 const double EPS = 1e-10; 72 const int inf = 1e9 + 9; 73 const LL mod = 1e9 + 7; 74 const int maxN = 1e5 + 7; 75 const LL ONE = 1; 76 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 77 const LL oddBits = 0x5555555555555555; 78 79 LL n, MOD = mod; 80 81 struct Matrix{ 82 static const int sz = 2; 83 LL mat[sz][sz]; 84 85 Matrix(){ ms0(mat); } 86 Matrix(LL A[sz][sz]){ 87 Rep(i, sz) { 88 Rep(j, sz) { 89 mat[i][j] = A[i][j]; 90 } 91 } 92 } 93 Matrix(vector< vector< LL > > &A){ 94 Rep(i, sz) { 95 Rep(j, sz) { 96 mat[i][j] = A[i][j]; 97 } 98 } 99 } 100 Matrix(const initializer_list< initializer_list< LL > > &A){ 101 int i = 0, j = 0; 102 for (auto R : A) { 103 j = 0; 104 for (auto x : R) { 105 mat[i][j] = x; 106 ++j; 107 } 108 ++i; 109 } 110 } 111 112 // x * 单位阵 113 inline void E(int x = 1) { 114 ms0(mat); 115 Rep(i, sz) mat[i][i] = x; 116 } 117 118 inline Matrix operator = (const initializer_list< initializer_list< LL > > &A) const { 119 Matrix ret; 120 int i = 0, j = 0; 121 for (auto R : A) { 122 j = 0; 123 for (auto x : R) { 124 ret.mat[i][j] = x; 125 ++j; 126 } 127 ++i; 128 } 129 return ret; 130 } 131 132 inline Matrix operator = (const Matrix x) { 133 Rep(i, sz) { 134 Rep(j, sz) { 135 mat[i][j] = x.mat[i][j]; 136 } 137 } 138 return *this; 139 } 140 141 inline Matrix operator + (const Matrix &x) { 142 Matrix ret; 143 Rep(i, sz) { 144 Rep(j, sz) { 145 ret.mat[i][j] = mat[i][j] + x.mat[i][j]; 146 ret.mat[i][j] %= MOD; 147 } 148 } 149 return ret; 150 } 151 152 inline Matrix operator - (const Matrix &x) { 153 Matrix ret; 154 Rep(i, sz) { 155 Rep(j, sz) { 156 ret.mat[i][j] = mat[i][j] - x.mat[i][j] + MOD; 157 ret.mat[i][j] %= MOD; 158 } 159 } 160 return ret; 161 } 162 163 inline Matrix operator * (const Matrix &x) { 164 Matrix ret; 165 Rep(i, sz) { 166 Rep(j, sz) { 167 Rep(k, sz) { 168 ret.mat[i][j] += mat[i][k] * x.mat[k][j]; 169 ret.mat[i][j] %= MOD; 170 } 171 } 172 } 173 return ret; 174 } 175 176 inline Matrix operator * (const LL &x) { 177 Matrix ret; 178 Rep(i, sz) { 179 Rep(j, sz) { 180 ret.mat[i][j] = mat[i][j] * x; 181 ret.mat[i][j] %= MOD; 182 } 183 } 184 return ret; 185 } 186 187 inline Matrix operator += (const Matrix &x) { return *this = *this + x; } 188 inline Matrix operator -= (const Matrix &x) { return *this = *this - x; } 189 inline Matrix operator *= (const Matrix &x) { return *this = *this * x; } 190 inline Matrix operator *= (const LL &x) { return *this = *this * x; } 191 192 inline void print() { 193 Rep(i, sz) { 194 Rep(j, sz) { 195 cout << mat[i][j] << " "; 196 } 197 cout << endl; 198 } 199 } 200 }; 201 202 // 矩阵快速幂，计算x^y 203 inline Matrix mat_pow_mod(Matrix x, LL y) { 204 Matrix ret; 205 ret.E(); 206 while(y){ 207 if(y & 1) ret *= x; 208 209 x *= x; 210 y >>= 1; 211 } 212 return ret; 213 } 214 215 Matrix X = {{1, 1}, {1, 0}}; 216 Matrix ans; 217 218 int main(){ 219 cin >> n; 220 ans = mat_pow_mod(X, n); 221 if(n == 0) cout << 0 << endl; 222 else cout << ans.mat[0][1] << endl; 223 return 0; 224 }