int Combination(int n, int m) { const int M = 10007; int ans = 1; for(int i=n; i>=(n-m+1); --i) ans *= i; while(m) ans /= m--; return ans % M; }
方案2
const int M = 10007; const int MAXN = 1000; int C[MAXN+1][MAXN+1]; void Initial() { int i,j; for(i=0; i<=MAXN; ++i) { C[0][i] = 0; C[i][0] = 1; } for(i=1; i<=MAXN; ++i) { for(j=1; j<=MAXN; ++j) C[i][j] = (C[i-1][j] + C[i-1][j-1]) % M; } } int Combination(int n, int m) { return C[n][m]; }
方案3
//用筛法生成素数 const int MAXN = 1000000; bool arr[MAXN+1] = {false}; vector<int> produce_prim_number() { vector<int> prim; prim.push_back(2); int i,j; for(i=3; i*i<=MAXN; i+=2) { if(!arr[i]) { prim.push_back(i); for(j=i*i; j<=MAXN; j+=i) arr[j] = true; } } while(i<=MAXN) { if(!arr[i]) prim.push_back(i); i+=2; } return prim; } //计算n!中素因子p的指数 int Cal(int x, int p) { int ans = 0; long long rec = p; while(x>=rec) { ans += x/rec; rec *= p; } return ans; } //计算n的k次方对M取模，二分法 int Pow(long long n, int k, int M) { long long ans = 1; while(k) { if(k&1) { ans = (ans * n) % M; } n = (n * n) % M; k >>= 1; } return ans; } //计算C(n,m) int Combination(int n, int m) { const int M = 10007; vector<int> prim = produce_prim_number(); long long ans = 1; int num; for(int i=0; i<prim.size() && prim[i]<=n; ++i) { num = Cal(n, prim[i]) - Cal(m, prim[i]) - Cal(n-m, prim[i]); ans = (ans * Pow(prim[i], num, M)) % M; } return ans; }
方案4
# include <stdio.h> const int M = 10007; int ff[M+5]; //打表，记录n!，避免重复计算 //求最大公因数 int gcd(int a,int b) { if(b==0) return a; else return gcd(b,a%b); } //解线性同余方程，扩展欧几里德定理 int x,y; void Extended_gcd(int a,int b) { if(b==0) { x=1; y=0; } else { Extended_gcd(b,a%b); long t=x; x=y; y=t-(a/b)*y; } } //计算不大的C(n,m) int C(int a,int b) { if(b>a) return 0; b=(ff[a-b]*ff[b])%M; a=ff[a]; int c=gcd(a,b); a/=c; b/=c; Extended_gcd(b,M); x=(x+M)%M; x=(x*a)%M; return x; } //Lucas定理 int Combination(int n, int m) { int ans=1; int a,b; while(m||n) { a=n%M; b=m%M; n/=M; m/=M; ans=(ans*C(a,b))%M; } return ans; } int main(void) { int i,m,n; ff[0]=1; for(i=1;i<=M;i++) //预计算n! ff[i]=(ff[i-1]*i)%M; scanf("%d%d",&n, &m); printf("%d\n",func(n,m)); return 0; }
LL solve (LL n, LL r) { vector<int> p; for (int i=2; i*i<=n; ++i) if (n % i == 0) { p.push_back (i); while (n % i == 0) n /= i; } if (n > 1) p.push_back (n);
LL sum = 0; for (int msk=1; msk<(1<<p.size()); ++msk) { LL mult = 1, bits = 0; for (int i=0; i<(LL)p.size(); ++i) if (msk & (1<<i)) { ++bits; mult *= p[i]; }
LL cur = r / mult; if (bits % 2 == 1) sum += cur; else sum -= cur; }
int mobius(int n){ int m = 1; for(int i=2;i*i<=n;i++){ if(n%i==0){ m*=-1; int k = 0; n/=i; if(n%i==0) {m=0;break;}//某个素因子的幂大于1 } } if(n>1) m *= -1; return m; }
/*******************************************/ 线性筛法预处理mobius函数 int prime[N],kp; int Is_or[N],mu[N];
