数论

欧几里得算法求最大公约数

import java.util.*;
class Main
{public static void main(String[] args){Scanner sc = new Scanner(System.in);int a=sc.nextInt();int b=sc.nextInt();System.out.print(gcd(a,b));}public static int gcd(int a,int b){return b!=0 ? gcd(b,a%b) : a;}
}

线性筛法求质数

import java.util.*;
class Main
{public static int N=1000000;public static int[] primes = new int[N];  //记录素数public static boolean[] st = new boolean[N];  //有没有被筛过public static int[] minp = new int[N];  //最小质因子public static void main(String[] args){Scanner sc = new Scanner(System.in);int n=sc.nextInt();get_primes(n);for(int i=0;i<=n;i++){if(primes[i]==0)break;System.out.print(primes[i]+" ");}}public static void get_primes(int n)  //O(n){int cnt=0;for(int i=2;i<=n;i++){if(!st[i]){minp[i]=i;primes[cnt++]=i;}for(int j=0;primes[j]*i<=n;j++){st[primes[j]*i]=true;minp[primes[j]*i]=primes[j];if(i%primes[j]==0)  break;}}}
}

约数个数与和

$N=p_1^{{\alpha }_1}\cdot p_2^{{\alpha }_2}\cdot \cdot \cdot p_k^{{\alpha }_k}$
约数个数：$({\alpha }_1+1)({\alpha }_2+1)\cdot \cdot \cdot ({\alpha }_k+1)$
约数之和：$(1+p_1+p_1^2+\cdot \cdot \cdot +p_1^{{\alpha }_1})(1+p_2+p_2^2+\cdot \cdot \cdot +p_2^{{\alpha }_2})\cdot \cdot \cdot (1+p_k+p_k^2+\cdot \cdot \cdot +p_k^{{\alpha }_k})$

裴蜀定理

$ax+by=gcd(a,b)$
扩展欧几里得算法

import java.util.*;
class Int  //代替指针
{int v;public Int(){}public Int(int v){this.v=v;}
}
class Main
{public static void main(String[] args){Scanner sc = new Scanner(System.in);int a=sc.nextInt();int b=sc.nextInt();Int x = new Int();Int y = new Int();int d=exgcd(a,b,x,y);System.out.printf("%d * %d + %d * %d = %d",a,x.v,b,y.v,d);}public static int exgcd(int a,int b,Int x,Int y){if(b==0){x.v=1;y.v=0;return a;}int d=exgcd(b,a%b,y,x);y.v-=(a/b)*x.v;return d;}
}

其他数学知识

如果 $a,b$，均是正整数且互质，那么由 $ax+by(x\ge 0,y\ge 0)$ 不能凑出的最大数是 $ab-a-b.$
也即 $a$ 与 $b$ 的最大公因数 $(a,b)=1$，最大不能凑出来的数是 $(a-1)(b-1)-1$