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FZU 2020 组合 (Lucas定理)
题意:中文题。
析:直接运用Lucas定理即可。但是FZU好奇怪啊,我开个常数都CE,弄的工CE了十几次,在vj上还不显示。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <unordered_map> //#include <tr1/unordered_map> //#define freopenr freopen("in.txt", "r", stdin) //#define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; //const double inf = 0x3f3f3f3f3f3f; //const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 10005; //const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL p; LL quick_pow(LL a, LL n){ LL ans = 1; a %= p; while(n){ if(n & 1) ans = ans * a % p; a = a * a % p; n >>= 1; } return ans; } LL C(LL n, LL m){ if(n < m) return 0; LL a = 1, b = 1; while(m){ a = a * n % p; b = b * m % p; --m; --n; } return a * quick_pow(b, p-2) % p; } LL Lucas(LL n, LL m){ if(!m) return 1; return C(n%p, m%p) * Lucas(n/p, m/p); } int main(){ int T; cin >> T; while(T--){ LL n, m; scanf("%I64d %I64d %I64d", &n, &m, &p); printf("%I64d\n", Lucas(n, m)); } return 0; }
FZU 2020 组合 (Lucas定理)
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