数论入门

1.求乘法逆元

扩展欧几里得算法

欧几里得算法：任意a,b∈N,b≠0,gcd(a,b) = gcd(b,a mod b)

定理1: 设a和b不全为0,则存在整数x和y,使得ax + by = gcd(a,b)。

当b = 0时,ax + by = gcd(a,b) 有一组解为x = 1,y = 0。

当b ≠ 0时,bx' + (a%b)y' = gcd(b,a%b) = gcd(a,b)

bx' + (a - a / b × b)y' = gcd(a,b) （此时 / 为整除）

ay' + b(x' - (a / b) y') = gcd(a,b)

令x = y',y = x' - (a / b)y',则ax + by = gcd(a,b)

void exgcd(int a,int b,long long  &x,long long  &y){
    if(!b)x = 1,y = 0;
    else exgcd(b,a%b,y,x),y -= a/b*x;
}

exgcd(i,p,x,y);

(x % b + b) % b // 求最小正整数解

 

还有线性算法的，给个大佬博客：传送门

2.中国剩余定理

转自 中国剩余定理 && 扩展中国剩余定理

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 tp=x;
    x=y; y=tp-a/b*y;
}

int china()
{
    int ans=0,lcm=1,x,y;
    for(int i=1;i<=k;++i) lcm*=b[i];
    for(int i=1;i<=k;++i)
    {
        int tp=lcm/b[i];
        exgcd(tp,b[i],x,y);
        x=(x%b[i]+b[i])%b[i];//x要为最小非负整数解
        ans=(ans+tp*x*a[i])%lcm;
    }
    return (ans+lcm)%lcm;
}

 

 eg 1:

Biorhythms POJ - 1006 

while(~scanf("%d%d%d%d",&p,&e,&i,&d))
{
    if(d == -1)break;
    x=(5544*p+14421*e+1288*i-d+21252)%21252;
    if(x==0)
        x=21252;
    printf("Case %d: the next triple peak occurs in %d days.\n", ++icase, x);
}

 3.扩展欧几里得算法

 

lt exgcd(lt a,lt b,lt &x,lt &y)
{
    if(b==0){x=1;y=0;return a;}
    lt gcd=exgcd(b,a%b,x,y);
    lt tp=x;
    x=y; y=tp-a/b*y;
    return gcd;
}

lt excrt()
{
    lt x,y,k;
    lt M=bi[1],ans=ai[1];
    for(int i=2;i<=n;i++)
    {
        lt a=M,b=bi[i],c=(ai[i]-ans%b+b)%b;
        lt gcd=exgcd(a,b,x,y),bg=b/gcd;
        if(c%gcd!=0) return -1; 
        
        x=mul(x,c/gcd,bg);
        ans+=x*M;
        M*=bg;
        ans=(ans%M+M)%M;
    }
    return (ans%M+M)%M;
}