数论章节模板

 1 --------------------------------------卢卡斯---------------------------
 2 ll mg(ll a,ll b,ll c)
 3 {
 4     ll ans=1;
 5     a=a%c;
 6     while(b) {
 7         if(b&1) ans=ans*a%c;
 8         a=a*a%c;
 9         b>>=1;
10     }
11     return ans;
12 }
13 
14 ll c(ll a,ll b)
15 {
16     if(a<b) return 0;
17     if(a==b) return 1;
18     if(b>a-b) b=a-b;
19 
20     ll A=1,B=1;
21     for(ll i=0; i<b; ++i) {
22         A=(A*(a-i))%mod;
23         B=(B*(b-i))%mod;
24     }
25     return (A*mg(B,mod-2,mod))%mod;
26 
27 }
28 
29 ll lucas(ll n,ll m)
30 {
31     if(!m) return 1;
32     return c(n%mod,m%mod)*lucas(n/mod,m/mod)%mod;
33 }
34 --------------------------------------卢卡斯---------------------------

 1 --------------------------------------扩欧----------------------------
 2 ll exgcd(ll l,ll r,ll &x1,ll &y1)
 3 {
 4     ll d,x2,y2;
 5     if(r==0) {
 6         x1=1;
 7         y1=0;
 8         return l;
 9     }
10     d=exgcd(r,l%r,x2,y2);
11     x1=y2,y1=x2-l/r*y2;
12     return d;
13 }
14 --------------------------------------扩欧----------------------------

1 -------------------------------------扩欧求逆元-----------------------------
2 ll ni(ll a,ll n)
3 {
4     ll x,y;
5     if(exgcd(a,n,x,y)==1) return (x+n)%n;
6     else return -1;
7 }
8 -------------------------------------扩欧求逆元-----------------------------

1 -------------------------------------线性求逆元-----------------------------
2 ll ni()
3 {
4     ni[1]=1;
5     for(int i=2; i<=n; i++)
6         ni[i]=(p-(p/i))*ni[p%i]%p;
7 }
8 -------------------------------------线性求逆元-----------------------------

 1 --------------------------------------素数测试------------------------
 2 ll mod-mul(ll a, ll b,ll n)
 3 {
 4     ll res=0;
 5     while(b) {
 6         if(b & 1) res=(res+a) % n;
 7         a=(a+a)%n;
 8         b>>=1;
 9     }
10     return res;
11 }
12 
13 ll pow-mod(ll a,ll b,ll n)
14 {
15     ll res=1;
16     while(b) {
17         if(b & 1) res=mod-mul(res, a, n);
18         a=mod-mul(a,a,n);
19         b>>=1;
20     }
21     return res;
22 }
23 
24 bool test(ll n,ll a,ll d)
25 {
26     if(n==2) return true;
27     if(n==a) return false;
28     if(!(n&1)) return false;
29     while(!(d&1)) d>>=1;
30     ll t=pow-mod(a,d,n);
31     while(d!=n-1&&t!=n-1&&t!=1) {
32         t=t*t%n;
33         d<<=1;
34     }
35     return t==n-1||(d&1)==1;
36 }
37 
38 bool isprime(ll n)
39 {
40     int a[]= {2,3,5,7};
41     for(int i=0; i<=3; ++i) {
42         if(n==a[i]) return true;
43         if(!test(n,a[i],n-1)) return false;
44     }
45     return true;
46 }
47 --------------------------------------素数测试------------------------

 1 --------------------------------------中国剩余定理--------------------
 2 int China(int B[],int W[],int k)
 3 {
 4     int i,d,x,y,ans=0,Mi,P=1;
 5     for(i=1; i<=k; i++) P*=W[i];
 6     for(i=1; i<=k; i++) {
 7         Mi=P/W[i];
 8         d=exgcd(Mi,W[i],x,y);
 9         ans=(ans+xMiB[i])%P;
10     }
11     if(ans>0)return ans;
12     else return(ans+P);
13 } 
14 --------------------------------------中国剩余定理--------------------

 1 --------------------------------------直接计算欧拉函数----------------
 2 int euler(int x)
 3 {
 4     int i,ans= x, a = x;
 5     for(i=2; i*i<= a; i++) {
 6         if(a%i==0) {
 7             ans=ans-ans/i;
 8             while(a%i == 0) a=a/i;
 9         }
10     }
11     if(a>1) ans=ans-ans/a;
12     return ans;
13 } 
14 --------------------------------------直接计算欧拉函数-----------------

 1 --------------------------------------线性筛---------------------------
 2 void getPhi()
 3 {
 4     int i,j,tot=0;
 5     for(i=2; i<=N; i++) {
 6         if(Mark[i]==false) {
 7             Prime[++tot]=i;
 8             phi[i]=i-1;
 9         }
10         for(j=1; j<=tot && i*Prime[j]<=N; j++) {
11             Mark[i*Prime[ j]]=true;
12             if(i%Prime[ j]==0) {
13                 phi[i*Prime[j]]=phi[i]*Prime[ j];
14                 break;
15             } else phi[i*Prime[j]]=phi[i]*(Prime[j]-1);
16         }
17     }
18 }
19 --------------------------------------线性筛---------------------------

 1 ------------------------------扩展中国剩余定理-------------------
 2 void exchina(ll c[],ll m[],ll n)
 3 {
 4     for (ll i=2; i<=n; ++i) {
 5         ll m1=m[i-1],m2=m[i],c1=c[i-1],c2=c[i];
 6         ll gcd=exgcd(m1,m2,x,y);
 7         if((c2-c1)%gcd) break;
 8         m[i]=m1m2/gcd;
 9         c[i]=ni(m1/gcd,m2/gcd)((c2-c1)/gcd)%(m2/gcd)*m1+c1;
10         c[i]=(c[i]%m[i]+m[i])%m[i];
11     }
12 }
13 ------------------------------扩展中国剩余定理-------------------