Accepted ALL lve

int gcd(int a,int b) 求解a和b的最大公约数int gcd(int a,int b){ if(b==0) return a; return gcd(b,a%b);}这也是俗称的欧几里得算法。如果a和b的最大公约数是gcd，那么一定存在x和y使得a*x + b*y = gcd，int e_gcd(int a,int b,int &x,int &y){

素性测试bool is_prime(int n){ for(int i=2;i*i<=n;i++) { if(n%i==0) return false; } return n!=1; //1不是素数 }约数枚举vector divisor(int n){ vector res; for(int i=1;i*i<=n;i++) { if(n%i==0) {

typedef long longll mod_pow(ll x,ll n,ll mod){ ll res=1; while(n>0) { if(n&1) res=res*x%mod; x=x*x%mod; n>>=1; } return res;}x的n次幂取余mod

Nikolay and Asya investigate integers together in their spare time. Nikolay thinks an integer is interesting if it is a prime number. However, Asya thinks an integer is interesting if the amount of it

You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on

先给出计算组合数的公式：#includeusing namespace std;int count_=0;//求组合数C(m,k)个数,k>=1int comb1(int m,int k)//(C(m,k)){ int i; for (i=m;i>=k;i--) { if (k>1) {