HDU 3037 Saving Beans (number theory, Lucas theorem)

Test instructions: Ask how many ways to put in n trees with no more than M seeds.

Analysis: Test instructions can be converted to X1 + x2 +. + xn = m, how many solutions, and then using the combined knowledge to get the answer is C (N+m, M).

Then we ask for this value directly, so we can use the Lucas theorem to break down the combination number, which is Lucas (N,M,P) =c (n%p,m%p) * Lucas (N/P,M/P,P).

And then we can do it according to the Fermat theorem.

The code is as follows:

The first type:

#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;cons T LL LNF = 0x3f3f3f3f3f3f;const Double PI = ACOs ( -1.0); const double EPS = 1e-8;const int maxn = 10005;const LL mod = 10000 000000007;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", "10 00 "," 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, 29, 31, 30, 31, 0, +, +,, +, +, +,, N, H, H, C, h, 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;}

  Second type:

#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;cons T LL LNF = 0x3f3f3f3f3f3f;const Double PI = ACOs ( -1.0); const double EPS = 1e-8;const int maxn = 10005;const LL mod = 10000 000000007;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", "10 XX "," 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, 29, 31, 30, 31, 0, +, +,, +, +, +, +, +, 31};const int monn[] N, H, H, C, h, 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;}

  Third Type:

#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;cons T LL LNF = 0x3f3f3f3f3f3f;const Double PI = ACOs ( -1.0); const double EPS = 1e-8;const int maxn = 10005;const LL mod = 10000 000000007;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", "10 XX "," 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, 29, 31, 30, 31, 0, +, +,, +, +, +, +, +, 31};const int monn[] N, H, H, C, h, 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 (number theory, Lucas theorem)