复习1

1.欧几里得

#include<cstdio>
using namespace std;
int a,b;
int gcd(int x,int y){
    if(y==0)return x;
    return gcd(y,x%y);
}
int main(){
    scanf("%d%d",&a,&b);
    printf("%d",gcd(a,b));
}

2.扩展欧几里得

#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
#define ll long long
ll a,b,x,y;
ll exgcd(ll a,ll b){
    if(b==0){
        x=1;y=0;
        return a;
    }
    exgcd(b,a%b);
    ll tmp=x;
    x=y;y=tmp-(a/b)*y;
}
int main(){
    scanf("%lld%lld",&a,&b);
    exgcd(a,b);
    while(x<0)x+=b;
    cout<<x;
}

3.欧拉筛

void prepare(){
    for(int i=2;i<=n;i++){
        if(!vis[i])p[++cnt]=i;
        for(int j=1;j<=cnt&&i*p[j]<=n;j++){
            vis[i*p[j]]=1;
            if(i%p[j]==0)break;
        }
    }
}

4.欧拉函数

int euler_phi(int n){//单个欧拉函数 
    int m=(int)sqrt(n+0.5);
    int ans=n;
    for(int i=2;i<=m;i++)if(n%i==0){
        ans=ans/i*(i-1);
        while(n%i==0)n/=i;
    }
    if(n>1)ans=ans/n*(n-1);
}

void prepare(){//函数表
    phi[1]=1;
    for(int i=2;i<=1000;i++){
        if(!vis[i])p[++p[0]]=i,phi[i]=i-1;
        for(int j=1;j<=p[0]&&i*p[j]<=1000;j++){
            vis[i*p[j]]=1;
            if(i%p[j]==0){
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
            else phi[i*p[j]]=phi[i]*(p[j]-1);
        }
    }
}

5.中国剩余定理

/*
    x=p(mod 23)
    x=e(mod 28)
    x=i(mod 33)
    题目就是要求上面同余方程的解，用中国剩余定理求解 
*/
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[5],m[5];
void Exgcd(int a,int b,int &x,int &y){
    if(b==0){x=1;y=0;return;}
    Exgcd(b,a%b,x,y);
    int tmp=x;
    x=y;
    y=tmp-(a/b)*y;
}
int CRT(int a[],int m[],int n){
    int M=1,ans=0;
    for(int i=1;i<=n;i++)M*=m[i];
    for(int i=1;i<=n;i++){
        int x,y,Mi=M/m[i];
        Exgcd(Mi,m[i],x,y);
        ans=(ans+Mi*x*a[i])%M;
    }
    if(ans<0)ans+=M;
    return ans;
}
int main(){
    freopen("Cola.txt","r",stdin);
    int p,e,i,d,Case=0;
    while(1){
        Case++;
        scanf("%d%d%d%d",&p,&e,&i,&d);
        if(p==-1&&e==-1&&i==-1&&d==-1)return 0;
        a[1]=p;a[2]=e;a[3]=i;
        m[1]=23;m[2]=28;m[3]=33;
        int ans=CRT(a,m,3);
        if(ans<=d)ans+=21252;
        printf("Case %d: the next triple peak occurs in %d days.
",Case,ans-d);
    }
}

6.卡特兰数

#include<iostream> //没有取模操作
#include<cstdio>
using namespace std;
long long h[20];
int main(){
    int n;
    scanf("%d",&n);
    h[0]=1;h[1]=1;
    for(int i=2;i<=n;i++)h[i]=h[i-1]*(4*i-2)/(i+1);
    cout<<h[n];
}

/*注意不要用另类递推式，因为里面的除法在取模的时候会出问题*/
#include<cstdio>//有取模操作
using namespace std;
int h[110];
int main(){
    int n,i,j;
    h[0]=1;
    scanf("%d",&n);
    for(i=1;i<=n;++i)
        for(j=0;j<i;++j)
            h[i]=(h[i]+h[j]*h[i-1-j])%100;
    printf("%d
",h[n]);
    return 0;
}

7.逆元（求组合数）

#include<iostream>//O(n)预处理+O(1)查询
#include<cstdio>
#define N 200001
#define mod 1000000007
using namespace std;
long long fac[N]={1,1},inv[N]={1,1},f[N]={1,1};
int n,m;
long long C(int x,int y){
    return fac[x]*inv[y]%mod*inv[x-y]%mod;
}
int main(){
    for(int i=2;i<N;i++){
        fac[i]=fac[i-1]*i%mod;
        f[i]=(mod-mod/i)*f[mod%i]%mod;
        inv[i]=inv[i-1]*f[i]%mod;
    }
    while(scanf("%d%d",&n,&m)!=EOF){
        cout<<C(n+m-4,m-2)<<endl;
    }
    return 0;
}

#include<iostream>//费马小定理
#include<cstdio>
#define mod 1000000007
using namespace std;
long long fac[1000001];
int n,m;
long long Pow(long long a,long long b){
    long long res=1;
    while(b){
        if(b&1)res=res*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return res;
}
long long C(int x,int y){
    return fac[x]*Pow(fac[y],mod-2)%mod*Pow(fac[x-y],mod-2)%mod;
}
int main(){
    fac[1]=1;fac[0]=1;
    for(int i=2;i<=1000000;i++)fac[i]=fac[i-1]*i%mod;
    while(scanf("%d%d",&n,&m)!=EOF){
        cout<<C(n+m-4,m-2)<<endl;
    }
}

8.卢卡斯定理（求组合数取模）

#include<bits/stdc++.h>
#define N 100010
using namespace std;
typedef long long ll;
ll a[N];
int p;
ll pow(ll y,int z,int p){
    ll res=1;y%=p;
    while(z){
        if(z&1)res=res*y%p;
        y=y*y%p;
        z>>=1;
    }
    return res;
}
ll C(ll n,ll m){
    if(m>n)return 0;
    return ((a[n]*pow(a[m],p-2,p))%p*pow(a[n-m],p-2,p)%p);
}
ll Lucas(ll n,ll m){
    if(!m)return 1;
    return C(n%p,m%p)*Lucas(n/p,m/p)%p;
}
int main(){
    int T;scanf("%d",&T);
    int n,m;
    while(T--){
        scanf("%d%d%d",&n,&m,&p);
        a[0]=1;
        for(int i=1;i<=p;i++)a[i]=(a[i-1]*i)%p;
        cout<<Lucas(n+m,n)<<endl;
    }
}