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HDU 3037 Saving Beans (数论,Lucas定理)
题意:问用不超过 m 颗种子放到 n 棵树中,有多少种方法。
析:题意可以转化为 x1 + x2 + .. + xn = m,有多少种解,然后运用组合的知识就能得到答案就是 C(n+m, m)。
然后就求这个值,直接求肯定不好求,所以我们可以运用Lucas定理,来分解这个组合数,也就是Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)。
然后再根据费马小定理就能做了。
代码如下:
第一种:
#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 int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2}; const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; LL exgcd(LL a,LL b,LL &x,LL &y){LL d = a;if(b!=0){d=exgcd(b,a%b,y,x);y-=(a/b)*x;}else{x=1;y=0;}return d;} LL mod_inverse(LL a,LL m){LL x,y;exgcd(a,m,x,y);return (m+x%m)%m; } 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 fact[100005]; LL mod_fact(LL n, LL p, LL &e){ e = 0; if(!n) return 1; LL res = mod_fact(n / p, p, e); e += n / p; if(n / p % 2 != 0) return res * (p - fact[n%p]) % p; return res * fact[n%p] % p; } LL mod_comb(LL n, LL k, LL p){ if(n < 0 || k < 0 || n < k) return 0; LL e1, e2, e3; LL a1 = mod_fact(n, p, e1); LL a2 = mod_fact(k, p, e2); LL a3 = mod_fact(n-k, p, e3); if(e1 > e2+e3) return 0; return a1 * mod_inverse(a2*a3%p, p) % p; } int main(){ fact[0] = 1; int T; cin >> T; while(T--){ LL p, m, n; scanf("%I64d %I64d %I64d", &n, &m, &p); for(int i = 1; i < p; ++i) fact[i] = fact[i-1] * (LL)i % p; printf("%I64d\n", mod_comb(n+m, m, p)); } return 0; }
第二种:
#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 int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2}; const int hc[]= {-1, 1, -2, 2, -2, 2, -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 fact[100005]; LL p; LL quick_pow(LL a, LL b){ LL ans = 1LL; a %= p; while(b){ if(b & 1) ans = ans * a % p; a = a * a % p; b >>= 1; } return ans; } LL C(LL n, LL m){ if(n < m) return 0; return fact[n] * quick_pow(fact[m]*fact[n-m], p-2) % p; } LL lucas(LL n, LL m){ if(!m) return 1LL; return C(n%p, m%p) * lucas(n/p, m/p) % p; } int main(){ fact[0] = 1; int T; cin >> T; while(T--){ LL m, n; scanf("%I64d %I64d %I64d", &n, &m, &p); for(int i = 1; i < p; ++i) fact[i] = fact[i-1] * (LL)i % p; printf("%I64d\n", lucas(n+m, m)); } return 0; }
第三种:
#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 int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2}; const int hc[]= {-1, 1, -2, 2, -2, 2, -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 b){ LL ans = 1LL; a %= p; while(b){ if(b & 1) ans = ans * a % p; a = a * a % p; b >>= 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 1LL; return C(n%p, m%p) * lucas(n/p, m/p) % p; } int main(){ int T; cin >> T; while(T--){ LL m, n; scanf("%I64d %I64d %I64d", &n, &m, &p); printf("%I64d\n", lucas(n+m, m)); } return 0; }
HDU 3037 Saving Beans (数论,Lucas定理)
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