2017 ACM/ICPC Asia Regional Shenyang Online：number number number hdu 6198【矩陣快速冪】
阿新 • • 發佈：2017-09-23
cpc 相同 exp -128 integer regional test atom online
? F0=0,F1=1;
? Fn=Fn?1+Fn?2 (n≥2).
Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤?≤ak, this positive number is mjf?good. Otherwise, this positive number is mjf?bad.
Now, give you an integer k, you task is to find the minimal positive mjf?bad
The answer may be too large. Please print the answer modulo 998244353.
Each test case includes an integer k which is described above. (1≤k≤109)
Problem Description We define a sequence F:
? F0=0,F1=1;
? Fn=Fn?1+Fn?2 (n≥2).
Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0≤a1≤a2≤?≤ak, this positive number is mjf?good. Otherwise, this positive number is mjf?bad.
Now, give you an integer k, you task is to find the minimal positive mjf?bad
The answer may be too large. Please print the answer modulo 998244353.
Input There are about 500 test cases, end up with EOF.
Each test case includes an integer k which is described above. (1≤k≤109)
Output For each case, output the minimal mjf?bad number mod 998244353.
Sample Input 1
思路：找規律，當k=1時，n=F5-1=4。k=2，n=F7-1=12。k=3，n=F9-1=33。所以大膽推測n=F(2*k+3)-1；再用矩陣快速冪輸出F(2n+3)-1。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <vector> 5 using namespace std; 6 const int mod = 998244353; 7 typedef long long LL;8 LL n; 9 typedef vector<LL>vec; 10 typedef vector<vec>mat; 11 mat mul(mat &A, mat &B) 12 { 13 mat C(A.size(), vec(B[0].size()));///分配大小，A的行，B的列 14 for (int i = 0; i<A.size(); i++) ///矩陣A的行 15 for (int k = 0; k<B.size(); k++) ///矩陣B的行 16 for (int j = 0; j<B[0].size(); j++) ///矩陣B的列 17 C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod + mod) % mod; 18 return C; 19 } 20 ///計算A^n 21 mat pow(mat A, LL n) 22 { 23 mat B(A.size(), vec(A.size()));///和矩陣A的大小相同 24 for (int i = 0; i<A.size(); i++) 25 B[i][i] = 1; 26 while (n>0) 27 { 28 if (n & 1) B = mul(B, A); 29 A = mul(A, A); 30 n >>= 1; 31 } 32 return B; 33 } 34 void solve() 35 { 36 mat A(2, vec(2));///2*2的矩陣 37 A[0][0] = 1; 38 A[0][1] = 1; 39 A[1][0] = 1; 40 A[1][1] = 0; 41 A = pow(A, n); 42 printf("%d\n", (A[1][0] % mod - 1 + mod) % mod); 43 } 44 int main() 45 { 46 while (~scanf("%lld", &n)) 47 { 48 n = 2 * n + 3; 49 solve(); 50 } 51 }
2017 ACM/ICPC Asia Regional Shenyang Online：number number number hdu 6198【矩陣快速冪】