HDU 5446 Unknown Treasure (卢卡斯+CRT
代码：
1 #include"bits/stdc++.h" 2 #define db double 3 #define ll long long 4 #define vec vector<ll> 5 #define Mt vector<vec> 6 #define ci(x) scanf("%d",&x) 7 #define cd(x) scanf("%lf",&x) 8 #define cl(x) scanf("%lld",&x) 9 #define pi(x) printf("%d\n",x) 10 #define pd(x) printf("%f\n",x) 11 #define pl(x) printf("%lld\n",x) 12 const int N = 1e5+5; 13 using namespace std; 14 typedef pair<ll, ll> pll; 15 ll F[N], Finv[N], inv[N];//F是阶乘，Finv是逆元的阶乘 16 ll mul(ll a, ll b, ll mod) { 17 ll ret = 0; 18 while(b) { 19 if(b & 1) ret = (ret + a) % mod; 20 a = (a + a) % mod; 21 b >>= 1; 22 } 23 return ret; 24 } 25 void init(int mod){ 26 inv[1] = 1; 27 for(int i = 2; i < N; i ++){ 28 inv[i] = (mod - mod / i) * 1ll * inv[mod % i] % mod; 29 } 30 F[0] = Finv[0] = 1; 31 for(int i = 1; i < N; i ++){ 32 F[i] = F[i-1] * 1ll * i % mod; 33 Finv[i] = Finv[i-1] * 1ll* inv[i] % mod; 34 } 35 } 36 ll C(ll n, ll m,int mod){ //C(n, m)就是C(n, m) 37 if(m < 0 || m > n) return 0; 38 return F[n] * 1ll * Finv[n - m] % mod * Finv[m] % mod; 39 } 40 ll Lucas(ll n, ll m, int p){ 41 return m ? mul(Lucas(n/p, m/p, p),C(n%p, m%p, p),p) : 1; 42 } 43 ll exgcd(ll a, ll b, ll &x, ll &y)//ax+by=gcd(a,b)的整数解 44 { 45 ll d; 46 //if (a == 0 && b == 0) return -1; 47 if (b == 0) 48 { 49 x = 1,y = 0; 50 return a; 51 } 52 d = exgcd(b, a%b, y, x); 53 y -= a / b * x; 54 return d; 55 } 56 ll INV(ll a,ll p) 57 { 58 ll d,x,y; 59 d = exgcd(a,p,x,y); 60 return d==1?(x%p+p)%p:-1; 61 } 62 63 ll china(int n, ll *a, ll *m){ 64 ll M = 1, ret = 0; 65 for(int i = 0; i < n; i ++) M *= m[i]; 66 for(int i = 0; i < n; i ++){ 67 ll w = M / m[i]; 68 ret = (ret + mul(mul(w, INV(w, m[i]),M),a[i],M)) % M; 69 } 70 return (ret + M) % M; 71 } 72 ll n,m; 73 int k; 74 ll P[N],R[N]; 75 int main(){ 76 int T; 77 scanf("%d",&T); 78 while(T--){ 79 scanf("%lld%lld%d",&n,&m,&k); 80 for(int i=0;i<k;i++){ 81 cl(P[i]); 82 init((int)P[i]); 83 R[i]=Lucas(n,m,P[i]); 84 } 85 ll ans=china(k,R,P); 86 pl(ans); 87 } 88 return 0; 89 }
