水题一道
推一下就是
f[n] = f[n - 1] + f[n - 2]
发现n很大
所以用矩阵快速幂就好了
还有P很大 那就指数循环节
一定要注意
这个条件
#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <list> #include <cmath> #include <bitset> #define rap(i, a, n) for(int i=a; i<=n; i++) #define rep(i, a, n) for(int i=a; i<n; i++) #define lap(i, a, n) for(int i=n; i>=a; i--) #define lep(i, a, n) for(int i=n; i>a; i--) #define rd(a) scanf("%d", &a) #define rlld(a) scanf("%lld", &a) #define rc(a) scanf("%c", &a) #define rs(a) scanf("%s", a) #define rb(a) scanf("%lf", &a) #define rf(a) scanf("%f", &a) #define pd(a) printf("%d ", a) #define plld(a) printf("%lld ", a) #define pc(a) printf("%c ", a) #define ps(a) printf("%s ", a) #define MOD 2018 #define LL long long #define ULL unsigned long long #define Pair pair<int, int> #define mem(a, b) memset(a, b, sizeof(a)) #define _ ios_base::sync_with_stdio(0),cin.tie(0) //freopen("1.txt", "r", stdin); using namespace std; const int maxn = 1100000, INF = 0x7fffffff; int prime[maxn+10], phi[maxn+10]; bool vis[maxn+10]; int ans; LL P; void get_phi() { ans = 0; phi[1] = 1; for(int i=2; i<=maxn; i++) { if(!vis[i]) { prime[++ans] = i; phi[i] = i - 1; } for(int j=1; j<=ans; j++) { if(i * prime[j] > maxn) break; vis[i * prime[j]] = 1; if(i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; } else phi[i * prime[j]] = phi[i] * (prime[j] - 1); } } } LL q_pow(LL a, LL b) { LL res = 1; while(b) { if(b & 1) res = res * a % P; a = a * a % P; b >>= 1; } return res; } int tot; struct Matrix { LL v[110][110]; Matrix() { memset(v, 0, sizeof(v)); } Matrix operator *(const Matrix B) // 重载的速度比写独立的函数慢点。 { int i, j, k; Matrix C; for(i = 0; i <= tot; i ++) for(j = 0; j <= tot; j ++) for(k = 0; k <= tot; k ++) { C.v[i][j] = (C.v[i][j] + v[i][k] * B.v[k][j]) % phi[P]; } return C; } }; Matrix mtPow(Matrix A, LL k) // 用位运算代替递归求 A^k。 { int i; Matrix B; for(i = 0; i <= tot; i ++) { B.v[i][i] = 1; } while(k) { if(k & 1) B = B * A; A = A * A; k >>= 1; } return B; } int main() { get_phi(); int kase = 0; int T; rd(T); while(T--) { tot = 1; LL a, b, n; cin >> a >> b >> P >> n; if(n == 1) printf("Case #%d: %lld ", ++kase, a % P); else if(n == 2) printf("Case #%d: %lld ", ++kase, b % P); else if(n == 3) printf("Case #%d: %lld ", ++kase, a * b % P); else { Matrix A, B; A.v[0][0] = A.v[0][1] = A.v[1][0] = 1; A.v[1][1] = 0; B = mtPow(A, n - 3); LL d1 = B.v[0][0] + B.v[0][1], d2 = B.v[1][0] + B.v[1][1]; if(d2 >= phi[P]) d2 = d2 % phi[P] + phi[P]; if(d1 >= phi[P]) d1 = d1 % phi[P] + phi[P]; a = q_pow(a, d2) % P; b = q_pow(b, d1) % P; printf("Case #%d: ", ++kase); cout << a * b % P << endl; } } return 0; }