一、数论中的基本概念与性质
1、整除
定义
若整数\(b\)除以非零整数\(a\),商为整数,且余数为零, 我们就说\(b\)能被\(a\)整除(或说\(a\)能整除\(b\)),表示为\(a \mid b\)
性质
(1)反身性
\(a \mid a\)
证明:
\(\because a \div a=1\)
\(\therefore a \mid a\)
(2)反对称性
\(a \mid b\)且\(b \mid a\)则\(\left\vert a \right\vert=\left\vert b \right\vert\)
证明:
\(\because a \mid b,b \mid a\)
\(\therefore\)设\(b \div a=x,a \div b=y \left(x,y \in Z\right)\)
\(\therefore \begin{cases}b=ax\\a=by\end{cases}\)
\(\therefore b=ax=bxy\)
\(\therefore xy=1\)
\(\therefore \begin{cases} x= \pm 1\\y= \pm 1\end{cases}\)
\(\therefore \left\vert a \right\vert=\left\vert b \right\vert\)
\(\left\vert a \right\vert=\left\vert b \right\vert\)则\(a \mid b\)且\(b \mid a\)
证明:
\(\because \left\vert a \right\vert=\left\vert b \right\vert\)
\(\therefore \begin{cases} a \div b= \pm 1\\b \div a= \pm 1\end{cases}\)
\(\therefore a \mid b\)且\(b \mid a\)
(3)传递性
\(a \mid b\)且\(b \mid c\)则\(a \mid c\)
证明:
\(\because a \mid b,b \mid c\)
\(\therefore\)设\(b \div a=x,c \div b=y \left(x,y \in Z\right)\)
\(\therefore \begin{cases}b=ax\\c=by\end{cases}\)
\(\therefore c=by=axy\)
\(\therefore c \div a=xy\)
\(\therefore a \mid c\)
(4)其他性质
①\(a \mid b\)且\(a \mid c\)且\(a \mid d\)则\(a \mid \left(ka+mb+nc+ld\right)\)
证明:
\(\because a \mid b,a \mid c,a \mid d\)
\(\therefore\)设\(b \div a=x,c \div a=y,d \div a=z \left(x,y,z \in Z\right)\)
\(\therefore \begin{cases}b=ax\\c=ay\\d=az\end{cases}\)
\(\therefore \left(ka+mb+nc+ld\right) \div a=\left(ka+mxa+nya+lza\right) \div a=\left(k +mx+ny+lz\right)\)
\(\therefore a \mid \left(ka+mb+nc+ld\right)\)
②质数\(p \mid ab\)则\(p \mid a\)或\(p \mid b\)
证明:
假设\(p \nmid a\)且\(p \nmid b\)
\(\because p \nmid a,p \nmid b\)
\(\therefore a\)中不含有质因子\(p\),\(b\)中不含有质因子\(p\)
\(\therefore ab\)中不含有质因子\(p\)
\(\therefore p \nmid ab\),与\(p \mid ab\)矛盾
\(\therefore\)假设不成立
\(\therefore p \mid a\)或\(p \mid b\)
③连续\(n\)个整数中恰有一个整数是\(n\)的倍数
证明:
设这\(n\)个数为\(a,a+1,\cdots,a+n-1,a \equiv r \pmod{n},1 \leqslant r \leqslant n\)
\(\therefore \left(a+n-r\right) \equiv \left(r+n-r\right) \equiv n \equiv 0 \pmod{n}\)
又\(\because 0 \leqslant n-r < n\)
\(\therefore n \mid \left(a+n-r\right)\)
\(\therefore\)连续\(n\)个整数中恰有一个整数是\(n\)的倍数
④连续\(n\)个整数的乘积为\(n!\)的倍数
证明:
设这\(n\)个数为\(a,a+1,\cdots,a+n-1\)
\(\because C_{a+n-1}^{n}=\frac{\prod \limits_{i=1}^n \left(a+n-i\right)}{n!}\)为整数
\(\therefore n! \mid \prod \limits_{i=1}^n \left(a+n-i\right)\)
2、质数与合数
定义
(1)质数
一个大于1的自然数,除了1和它自身外,不能整除其他自然数的数叫做质数
(2)合数
合数指自然数中除了能被1和本身整除外,还能被其他数(0除外)整除的数
性质
质数无穷多
证明(质数无穷多的证明方法有许多,这里只展示1种,其实我知道两种,感兴趣的可以上网查):
假设质数只有\(n\)个
从小到大依次排列为\(p_{1},p_{2}, \cdots ,p_{n}\),设\(N=\prod \limits_{i=1}^n p_{i}\)
\(\because p_{1} \nmid N\),\(p_{2} \nmid N,\cdots,p_{n} \nmid N\)
\(\therefore N\)为质数,与质数只有\(n\)个矛盾
\(\therefore\)假设不成立
\(\therefore\)质数无穷多
算术基本定理
每一个合数都可以以唯一形式被写成质数的乘积
证明:
假设合数可以以多种方式写成多个质数的乘积,设最小的是\(n\)
设\(n=\prod \limits_{i=1}^{r}(p_{i}^{a_{i}})=\prod \limits_{i=1}^{s}(q_{i}^{b_{i}})\)
\(\because p_{1} \mid \prod \limits_{i=1}^{s}(q_{i}^{b_{i}})\)
\(\therefore q_{1}^{b_{1}},q_{2}^{b_{2}},\cdots,q_{s}^{b_{s}}\)中有一个数能被\(p_{1}\)整除
\(\therefore\)不妨设为\(q_{1}\)
又\(\because q_{1}\)也是质数,因此\(q_{1}=p_{1}\)
假设\(a_{1} > b_{1}\)
\(\therefore p_{1}^{a_{1}-b_{1}} \prod \limits_{i=2}^{r}(p_{i}^{a_{i}})=\prod \limits_{i=2}^{s}(q_{i}^{b_{i}})\)
\(\therefore q_{2}^{b_{2}},q_{3}^{b_{3}},\cdots,q_{s}^{b_{s}}\)中有一个数能被\(p_{1}\)整除
又\(\because p_{1}=q_{1} \ne q_{i}(i \ne 1)\)
\(\therefore a_{1} \leqslant b_{1}\)
同理,\(\therefore a_{1} \geqslant b_{1}\)
\(\therefore a_{1} = b_{1}\)
\(\therefore\)存在小于\(n\)的整数\(m=\prod \limits_{i=2}^{r}(p_{i}^{a_{i}})=\prod \limits_{i=2}^{s}(q_{i}^{b_{i}})\)可以用多于一种的方式写成多个质数的乘积,这与\(n\)的最小性矛盾
\(\therefore\) 每一个合数都可以以唯一形式被写成质数的乘积
3、最大公约数和最小公倍数
定义
(1)约数与倍数
如果整数\(a\)能被整数\(b\)整除,\(a\)就叫做\(b\)的倍数,\(b\)就叫做\(a\)的约数
(2)公约数与公倍数
几个整数中公有的约数,叫做这几个整数的公约数;几个整数中公有的倍数,叫做这几个整数的公倍数
(3)最大公约数与最小公倍数
几个整数的公约数中,最大的一个,叫做这几个数的最大公约数;几个整数的公倍数中,最小的一个,叫做这几个数的最小公倍数数
(4)互质
\(\forall a,b \in N\),若(a,b)=1,则称\(a,b\)互质
(5)欧拉函数
\(1\)~\(N\)中与\(N\)互质的数的个数被称为欧拉函数,记为\(\varphi \left(N\right)=N \times \prod \limits_{质数p|N}(1-\frac{1}{p})\)
(6)积性函数
如果当\(a,b\)互质,有\(f \left(ab\right)=f \left(a\right) \times f \left(b\right)\),那么称函数\(f\)为积性函数
性质
①\(\forall a,b \in Z\),\(gcd\left(a,b\right) \times lcm\left(a,b\right)=ab\)
证明:
设\(gcd\left(a,b\right)=d,a=a_{0}d,b=b_{0}d,(a_{0},b_{0})=1\)
\(lcm\left(a,b\right)=lcm\left(a_{0},b_{0}\right) \times d=a_{0}b_{0}d\)
\(\therefore gcd\left(a,b\right) \times lcm\left(a,b\right)=d \times a_{0}b_{0}d=a_{0}b_{0}d^{2}=\left(a_{0}d\right) \times \left(b_{0}d\right)=ab\)
②\(\forall n > 1,1-n\)中与\(n\)互质的数的和为\(\frac{n \times \varphi \left(n\right)}{2}\)
证明:
\(\because gcd \left(n,x\right)=gcd \left(n,n-x\right)\)
\(\therefore\)与\(n\)不互质的数\(x,n-x\)成对出现,平均值为\(\frac{n}{2}\)
\(\therefore1-n\)中与\(n\)互质的数的和为\(\frac{n \times \varphi \left(n\right)}{2}\)
③欧拉函数是积性函数
若\(a,b\)互质,则\(\varphi \left(ab\right)=\varphi \left(a\right) \times \varphi \left(b\right)\)
证明:
设\(a=\prod \limits_{i=1}^{r}(p_{i}^{a_{i}}),b=\prod \limits_{i=1}^{s}(q_{i}^{b_{i}})\)
\(\therefore \varphi \left(a\right)=a \times \prod \limits_{i=1}^{r}(1-\frac{1}{p_{i}}),\varphi \left(b\right)=b \times \prod \limits_{i=1}^{s}(1-\frac{1}{q_{i}})\)
\(\therefore \varphi \left(ab\right)=ab \times \prod \limits_{i=1}^{r}(1-\frac{1}{p_{i}}) \times \prod \limits_{i=1}^{s}(1-\frac{1}{q_{i}})=\varphi \left(a\right) \times \varphi \left(b\right)\)
4、同余
定义
(1)同余
若整数\(a\)和整数\(b\)除以正整数\(m\)的余数相等,则称\(a,b\)模\(m\)同余,记为\(a \equiv b \pmod{m}\)
(2)同余类
对于\(\forall a \in \left[0,m-1\right]\),集合\(\left\{a+km\right\}(k \in Z)\)的所有数模\(m\)同余,余数都是\(a\),该集合称为一个模\(m\)的同余类,简记为\(\overline{a}\)
(3)完全剩余系
模\(m\)的同余类一共有m个,分别为\(\overline{0} , \overline{1} , \cdots , \overline{m-1}\),它们构成\(m\)的完全剩余系
(4)简化剩余系
\(1-m\)中与\(m\)互质的数代表的同余类共有\(\varphi \left(m\right)\)个,它们构成\(m\)的简化剩余系
(5)数论倒数(乘法逆元)
若整数\(a,x\)满足\(ax \equiv 1 \pmod{b}\),则\(x\)为\(a\)对模\(m\)意义下的数论倒数(乘法逆元)记为\(a^{-1} \pmod{m}\)
性质
(1)同余的充要条件
\(a \equiv b \pmod{m}\)的充要条件是\(m \mid \left(a-b\right)\)
证明:
\(\because m \mid \left(a-b\right)\)
\(\therefore\)存在整数\(t\)使得\(a-b=mt\)
\(\therefore a=b+mt\)
\(\therefore a \equiv b \pmod{m}\)
***
\(\because a \equiv b \pmod{m}\)
\(\therefore\)存在整数\(t\)使得\(a=b+mt\)
\(\therefore a-b=mt\)
\(\therefore m \mid \left(a-b\right)\)
(2)反身性
\(a \equiv a \pmod{m}\)
证明:
\(\because a-a=0,m \ne 0\)
\(\therefore m \mid a-a\)
\(\therefore a \equiv a \pmod{m}\)
(3)对称性
\(a \equiv b \pmod{m}\)则\(b \equiv a \pmod{m}\)
证明:
\(\because a \equiv b \pmod{m}\)
\(\therefore m \mid \left(a-b\right)\)
\(\therefore m \mid \left(b-a\right)\)
\(\therefore b \equiv a \pmod{m}\)
(4)传递性
\(a \equiv b \pmod{m}\)且\(b \equiv c \pmod{m}\)则\(a \equiv c \pmod{m}\)
\(\because a \equiv b \pmod{m}\)
\(\therefore m \mid \left(a-b\right)\)
又\(\because b \equiv c \pmod{m}\)
\(\therefore m \mid \left(b-c\right)\)
\(\therefore m \mid \left[\left(a-b\right)+\left(b-c\right)\right]\)
\(\therefore m \mid \left(a-c\right)\)
\(\therefore a \equiv c \pmod{m}\)
(5)可加性
\(a \equiv b \pmod{m}\)且\(c \equiv d \pmod{m}\)则\(a+c \equiv b+d \pmod{m}\)
证明:
\(\because a \equiv b \pmod{m}\)
\(\therefore m \mid \left(a-b\right)\)
又\(\because c \equiv d \pmod{m}\)
\(\therefore m \mid \left(c-d\right)\)
\(\therefore m \mid \left[\left(a-b\right)+\left(c-d\right)\right]\)
\(\therefore m \mid \left[\left(a+c\right)-\left(b+d\right)\right]\)
\(\therefore a+c \equiv b+d \pmod{m}\)
(6)可减性
\(a \equiv b \pmod{m},c \equiv d \pmod{m}\)则\(a-c \equiv b-d \pmod{m}\)
证明:
\(\because a \equiv b \pmod{m}\)
\(\therefore m \mid \left(a-b\right)\)
又\(\because c \equiv d \pmod{m}\)
\(\therefore m \mid \left(c-d\right)\)
\(\therefore m \mid \left[\left(a-b\right)-\left(c-d\right)\right]\)
\(\therefore m \mid \left[\left(a-c\right)-\left(b-d\right)\right]\)
\(\therefore a-c \equiv b-d \pmod{m}\)
(7)可乘性
①\(a \equiv b \pmod{m}\)则\(ac \equiv bc \pmod{m}\)
证明:
\(\because a \equiv b \pmod{m}\)
\(\therefore m \mid \left(a-b\right)\)
\(\therefore m \mid \left(a-b\right)c\)
\(\therefore m \mid \left(ac-bc\right)\)
\(\therefore ac \equiv bc \pmod{m}\)
②\(a \equiv b \pmod{m},c \equiv d \pmod{m}\)则\(ac \equiv bd \pmod{m}\)
证明:
\(\because a \equiv b \pmod{m}\)
\(\therefore ac \equiv bc \pmod{m}\)
又\(\because c \equiv d \pmod{m}\)
\(\therefore bc \equiv bd \pmod{m}\)
\(\therefore ac \equiv bd \pmod{m}\)
(8)有关同余的其他性质
①\(ac \equiv bc \pmod{m}\)且\(\left(m,c\right)=1\)则\(a \equiv b \pmod{m}\)
证明:
\(\because ac \equiv bc \pmod{m}\)
\(\therefore m \mid \left(ac-bc\right)\)
\(\therefore m \mid \left(a-b\right)c\)
又\(\because \left(m,c\right)=1\)
\(\therefore m \mid \left(a-b\right)\)
\(\therefore a \equiv b \pmod{m}\)
②\(ac \equiv bc \pmod{mc}\)则\(a \equiv b \pmod{m}\)
证明:
\(\because ac \equiv bc \pmod{mc}\)
\(\therefore mc \mid \left(ac-bc\right)\)
\(\therefore m \mid \left(a-b\right)\)
\(\therefore a \equiv b \pmod{m}\)
(9)有关完全剩余系的性质
\(gcd\left(m,a\right)=1\)且\(\left\{\overline{b_{i}}\right\}\left(i\in \left[1,m\right]\right)\)是模\(m\)的一个完全剩余系,则\(\left\{\overline{ab_{i}}\right\}\left(i\in \left[1,m\right]\right)\)也是模\(m\)的一个完全剩余系
证明:
假设存在两个整数\(a \times b_{i} \equiv a \times b_{j} \pmod{m}\)
\(b_{i} \equiv b_{j} \pmod{m}\),与\(\left\{\overline{b_{i}}\right\}\left(i\in \left[1,m\right]\right)\)是模\(m\)的一个完全剩余系矛盾
\(\therefore\)假设不成立
\(\therefore \left\{\overline{ab_{i}}\right\}\left(i\in \left[1,m\right]\right)\)是模\(m\)的一个完全剩余系
(10)有关简化剩余系的性质
\(gcd\left(m,a\right)=1\)且\(\left\{\overline{b_{i}}\right\}\left(i\in \left[1,\varphi \left(m\right)\right]\right)\)是模\(m\)的一个简化剩余系,则\(\left\{\overline{ab_{i}}\right\}\left(i\in \left[1,\varphi \left(m\right)\right]\right)\)也是模\(m\)的一个完全剩余系
证明:
假设存在两个整数\(a \times b_{i} \equiv a \times b_{j} \pmod{m}\)
\(b_{i} \equiv b_{j} \pmod{m}\),与\(\left\{\overline{b_{i}}\right\}\left(i\in \left[1,m\right]\right)\)是模\(m\)的一个简化剩余系矛盾
\(\therefore\)假设不成立
\(\therefore \left\{\overline{ab_{i}}\right\}\left(i\in \left[1,m\right]\right)\)是模\(m\)的一个简化剩余系
(11)费马小定理
若\(p\)是质数,则对任意与\(p\)互质的整数\(a\),有\(a^{p-1} \equiv 1 \pmod{p}\)
证明:
设\(p\)的完全剩余系为\(\left\{\overline{i}\right\}\left(i \in\left[1,p-1\right]\right)\)
\(\because gcd\left(a,p\right)=1\)
\(\therefore \{\overline{ai}\}(i\in\left[1,p-1\right])\)也是\(p\)的一个完全剩余系
\(\therefore\)对于每一个\(i(i\in\left[1,p-1\right])\),总存在一个\(j\)使得\(i \equiv a \times j \pmod{p}\)
\(\therefore \left(p-1\right)! \equiv \left(p-1\right)! \times a^{p-1} \pmod{p}\)
又\(\because \left(p,\left(p-1\right)!\right)=1\)
\(\therefore a^{p-1} \equiv 1 \pmod{p}\)
(12)欧拉定理
\(gcd\left(a,n\right)=1,a \in Z^{+}\),则\(a^{\varphi \left(n\right)} \equiv 1 \pmod{n}\)
证明:
设\(n\)的简化剩余系为\(\left\{\overline{a_{i}}\right\}\left(i \in\left[1,\varphi\left(n\right)\right]\right)\)
\(\because gcd\left(a,n\right)=1\)
\(\therefore \{\overline{aa_{i}}\}(i\in\left[1,\varphi \left(n\right)\right])\)也是\(p\)的一个化简剩余系
\(\therefore\)对于每一个\(a_{i}(i\in\left[1,\varphi \left(n\right)\right])\),总存在一个\(j\)使得\(i \equiv a \times a_{j} \pmod{p}\)
\(\therefore \prod \limits_{i=1}^{\varphi \left(n\right)} a_{i} \equiv \prod \limits_{i=1}^{\varphi \left(n\right)}a_{i} \times a^{\varphi \left(n\right)} \pmod{n}\)
\(\therefore a^{\varphi \left(n\right)} \equiv 1 \pmod{n}\)
(13)欧拉定理推论
\(gcd\left(a,n\right)=1,a \in Z^{+}\),则对于任意的正整数\(b\),有\(a^b \equiv a^{b \% \varphi \left(n\right)} \pmod{n}\)
证明:
设\(b=q \times \varphi \left(n\right)+r,0 \leqslant r < \varphi \left(n\right)\)
\(a^{b} \equiv a^{q \times \varphi \left(n\right)+r} \equiv \left(a^{\varphi \left(n\right)}\right)^{q} \times a^{r} \equiv 1^{q} \times a^{r} \equiv a^{r} \equiv a^{b \% \varphi \left(n\right)} \pmod{n}\)
5、不定方程(丢番图方程)
定义
不定方程是指未知数的个数多于方程个数,且未知数受到某些限制(如要求是有理数、整数或正整数等等)的方程或方程组
性质
(1)裴蜀Bézout定理
对于任意正整数\(a,b\),存在一对整数\(x,y\),满足\(ax+by=gcd\left(a,b\right)\)
证明:
设\(gcd\left(a,b\right)=d,a=a_{0}d,b=b_{0}d\)
\(aa_{0}+bb_{0}=1\)
\(\therefore \left(a_{0},b_{0}\right)\)是方程\(ax+by=\left(a,b\right)\)的一组解
(2)所有解与特解的关系
\(\left(x_{0},y_{0}\right)\)是方程\(ax+by=c\)的一组整数解,则方程的所有解为\(\begin{cases}x=x_{0}+\frac{b}{gcd\left(a,b\right)}t\\y=y_{0}-\frac{a}{gcd\left(a,b\right)}t\end{cases},t \in Z\)
证明:
设\(ax_{1}+by_{1}=c,gcd\left(a,b\right)=c_{0},a=a_{0}c_{0},b=b_{0}c_{0},gcd\left(a_{0},b_{0}\right)=1\)
\(a_{0}x_{0}+b_{0}y_{0}=c_{0},a_{0}x_{1}+b_{0}y_{1}=c_{0}\)
\(\therefore a_{0}x_{0}+b_{0}y_{0}=a_{0}x_{1}+b_{0}y_{1}\)
\(\therefore a_{0}\left(x_{0}-x_{1}\right)=b_{0}\left(y_{1}-y_{0}\right)\)
\(\therefore b_{0} \mid a_{0}(x_{0}-x_{1})\)
\(\therefore b_{0} \mid \left(x_{0}-x_{1}\right)\)
\(\therefore x_{0} \equiv x_{1} \pmod{b_{0}}\)
同理,\(\therefore y_{0} \equiv y_{1} \pmod{b_{0}}\)
\(\therefore \begin{cases}x=x_{0}+b_{0}t\\y=y_{0}-a_{0}t\end{cases}\)
\(\therefore \begin{cases}x=x_{0}+\frac{b}{gcd\left(a,b\right)}t\\y=y_{0}-\frac{a}{gcd\left(a,b\right)}t\end{cases}\)
(3)其他性质
①\(gcd\left(a,b\right)=1,a,b \in Z^{+}\),则方程\(ax+by=ab-a-b\)没有非负整数解
证明:
\(\because ax+by=ab-a-b\)
\(\therefore a \left(x+1\right)+b \left(y+1\right)=ab\)
又\(\because a \mid ab\)
\(\therefore a \mid \left(a \left(x+1\right)+b \left(y+1\right)\right)\)
\(\therefore a \mid b \left(y+1\right)\)
又\(\because gcd\left(a,b\right)=1\)
\(\therefore a \mid \left(y+1\right)\)
②\(gcd\left(a,b\right)=1,c>ab-a-b,a,b,c \in Z^{+}\),则方程\(ax+by=c\)有非负整数解
证明:
\(\because gcd\left(a,b\right)=1\)
\(\therefore\)设\(ax_{0}+by_{0}=c,0 \leqslant x_{0} \leqslant b-1\)
\(\therefore y_{0}=\frac{c-ax_{0}}{b}>\frac{ab-a-b-ax_{0}}{b} \geqslant \frac{ab-a-b-a\left(b-1\right)}{b}=-1\)
\(\therefore y_{0} \geqslant 0\)
\(\therefore x_{0},y_{0}\)为非负整数
\(\therefore\)方程\(ax+by=c\)有非负整数解
③\(gcd\left(a,b\right)=1,a,b \in Z^{+},0 \leqslant c \leqslant ab-a-b\),则恰有\(\frac{\left(a-1\right)\left(b-1\right)}{2}\)个整数\(c\)不能表示成\(ax+by\)的形式,\(x,y \in N\)
6、同余方程
定义
设\(f(x)=\sum \limits_{i=0}^{n}a_{i}x^{i}\)是整系数多项式,称\(f(x) \equiv 0 \pmod{m}\)是\(x\)模\(m\)的同余方程
7、高斯函数
定义
(1)取整函数
不超过实数\(x\)的最大整数称为\(x\)的整数部分,记作\(\left[x\right]\)
(2)取小函数
实数\(x\)的非负纯小数部分,记作\(\left\{x\right\}\)
性质
①\(x=\left[x\right]+\left\{x\right\}\)
②\(x-1<\left[x\right] \leqslant x\)
③\(0 \leqslant \left\{x\right\}<1\)
④若\(x \leqslant y\),则\(\left[x\right] \leqslant \left[y\right]\)
证明:
\(\because \left[x\right] \leqslant x \leqslant y<\left[y\right]+1\)
\(\therefore \left[x\right] \leqslant \left[y\right]\)
⑤\(\forall a \in Z^{*},b \in Z\),则\(b=a \left[\frac{b}{a}\right]+a\left\{\frac{b}{a}\right\},a\left\{\frac{b}{a}\right\} \in Z\)
证明:
\(\because \frac{b}{a}=\left[\frac{b}{a}\right]+\left\{\frac{b}{a}\right\}\)
\(\therefore b=a \left[\frac{b}{a}\right]+a\left\{\frac{b}{a}\right\}\)
\(\therefore a\left\{\frac{b}{a}\right\}=b-a \left[\frac{b}{a}\right]\)
\(\therefore a\left\{\frac{b}{a}\right\} \in Z\)
⑥\(\left[x\right]+\left[y\right]=\left[x+y\right]\)或\(\left[x+y\right]=\left[x\right]+\left[y\right]+1\)
证明:
\(\because x+y=\left[x\right]+\left[y\right]+\left\{x\right\}+\left\{y\right\}\)
\(\therefore\)当\(0 \leqslant \left\{x\right\}+\left\{y\right\}<1\)时,\(\left[x\right]+\left[y\right]=\left[x+y\right]\)
\(\ \ \;\)当\(1 \leqslant \left\{x\right\}+\left\{y\right\}<2\)时,\(\left[x\right]+\left[y\right]+1=\left[x+y\right]\)
二、数论中相关的数和方程的求法
1、埃拉托斯特尼筛法(埃氏筛法)
内容
要得到自然数\(n\)以内的全部素数,必须把不大于\(\sqrt{n}\)的所有素数的倍数剔除,剩下的就是素数
程序实现
void primes(int n)
{
memset(v,0,sizeof(v));
for(int i=2;i<=n;i++){
if(v[i]) continue;
cout<<i<<endl;
for(int j=i;j<=n/i;j++) v[i*j]=1;
}
}
时间复杂度
\(\Theta \left(n \times \log \left(\log \left(n\right)\right)\right)\)
时间复杂度证明
\(\Theta \left(\log \left(n\right)\right)=\Theta \left(\ln \left(n\right)\right)=\Theta \left(\int_{1}^{n+1} \frac{dx}{x}\right)=\Theta \left(\sum \limits_{i=1}^{n} \int_{i}^{i+1} \frac{dx}{x}\right) < \Theta \left(\sum \limits_{i=1}^{n} \frac{1}{i}\right) \leqslant \Theta \left(\prod \limits_{质数p \leqslant n} \left(1+\frac{1}{p}\right) \times \sum \limits_{k=1}^{n} \frac{1}{k^{2}}\right) < \Theta \left(\prod \limits_{质数p \leqslant n} \left(1+\frac{1}{p}\right) \times \left(1+\sum \limits_{k=2}^{n} \left(\frac{1}{k-\frac{1}{2}}-\frac{1}{k+\frac{1}{2}} \right) \right) \right) = \Theta \left(\prod \limits_{质数p \leqslant n} \left(1+\frac{1}{p}\right) \times \left(1+\frac{2}{3}+\frac{1}{n+\frac{1}{2}}\right)\right) < \Theta \left(\frac{5}{3}\prod \limits_{质数p \leqslant n} \left(1+\frac{1}{p}\right)\right) < \Theta \left(\frac{5}{3} \prod \limits_{质数p \leqslant n} \exp \left(\frac{1}{p}\right)\right) = \Theta \left(\frac{5}{3} \exp \left(\sum \limits_{质数p \leqslant n} \frac{1}{p}\right)\right)\)
\(\therefore \Theta \left(n \times \log \left(\log \left(n\right)\right)\right) = \Theta \left(n \times \ln \left(\ln \left(n\right)\right)\right) < \Theta \left(n \times \ln \left(\frac{5}{3} \exp \left(\sum \limits_{质数p \leqslant n} \frac{1}{p}\right)\right)\right) = \Theta \left(n \times \sum \limits_{质数p \leqslant n} \frac{1}{p}\right)\)
2、欧拉筛法
内容
在埃氏筛法的基础上,让每个合数只被它的最小质因子筛选一次,以达到不重复的目的
程序实现
void primes(int n)
{
memset(v,0,sizeof(v));
m=0;
for(int i=2;i<=n;i++){
if(v[i]==0){
v[i]=i;
prime[++m]=i;
}
for(int j=1;j<=m;j++){
if(prime[j]>v[i]||prime[j]>n/i) break;
v[i*prime[j]]=prime[j];
}
}
for(int i=1;i<=m;i++) cout<<prime[i]<<endl;
}
时间复杂度
\(\Theta \left(n\right)\)
3、质因数分解
内容
\(\forall N \in Z^{+} \setminus \left\{1\right\}\),把\(N\)化为\(\prod \limits^{m}_{i=1} p^{c_{i}}_{i}\)的过程叫质因数分解
程序实现
void divide(int n){
m=0;
for(int i=2;i*i<=n;i++) if(n%i==0){
p[++m]=i,c[m]=0;
while(n%i==0) n/=i,c[m]++;
}
if(n>1) p[++m]=n,c[m]=1;
for(int i=1;i<=m;i++) cout<<p[i]<<'^'<<c[i]<<endl;
}
时间复杂度
\(\Theta \left(\sqrt N\right)\)
4、欧几里得算法(辗转相除法)
内容
\(\forall a,b \in N,b \ne 0,gcd(a,b)=gcd(b,a \% b)\)
证明
若\(a<b\),则\(gcd(b,a \% b)=gcd(b,a)=gcd(a,b)\),命题成立
若\(a \geqslant b\),设\(a=q \times b+r,0 \leqslant r<b,a,b\)的一个公约数为\(d\)
\(d \mid a,d \mid qb\)
\(\therefore d \mid \left(a-qb\right)\)
\(\therefore d \mid r\)
\(\therefore d\)也是\(b,r\)的公约数
\(\therefore a,b\)的公约数集合与\(b,r\)的公约数集合相同
\(\therefore a,b\)和\(b,r\)的最大公约数相等
程序实现
递归写法
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
非递归写法
int gcd(int a,int b)
{
while(b!=0){
int temp=a;
a=b;
b=temp%a;
}
return a;
}
时间复杂度
\(\Theta \left(\log \left(a+b\right)\right)\)
5、扩展欧几里得算法
内容
求方程\(ax+by=gcd\left(a,b\right),a,b,x,y \in Z\)的解\(x,y\)
当\(b=0\)时,\(x=1,y=0\)为方程的解
当\(b>0\)时,\(gcd \left(a,b\right)=gcd \left(b,a\%b\right)\)
设\(bx'+\left(a\%b\right)y'=gcd\left(b,a\%b\right),x',y' \in Z\)
\(\therefore bx'+\left(a\%b\right)y'=bx'+\left(a-b \left \lfloor \frac{a}{b} \right \rfloor\right)y'=ay'+b \left(x'-\left \lfloor \frac{a}{b} \right \rfloor y'\right)\)
\(\therefore \begin{cases} x=y' \\y=x'-\left \lfloor \frac{a}{b} \right \rfloor y \end{cases}\)
程序实现
int exgcd(int a,int b,int &x,int &y)
{
if(!b){x=1,y=0;return a;}
int d=exgcd(b,a%b,x,y),z=x;
x=y,y=z-y*(a/b);
return d;
}
6、线性同余方程组的解法
内容
设\(m_{1},m_{2},\cdots \cdots,m_{n} \in Z^{+}\),对于任意\(n\)个整数\(a_{1},a_{2},\cdots \cdots,a_{n}\),求方程组\(\begin{cases} x \equiv a_{1} \pmod{m_{1}}\\x \equiv a_{2} \pmod{m_{2}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\x \equiv a_{n} \pmod{m_{n}} \end{cases}\)的解
设前\(k-1\)个方程的解为\(x,m= \sum \limits^{t-1}_{i=1}m_{i},tm \equiv a_{k}-x \pmod{m_{k}}\)
\(\therefore x+tm \equiv a^{k} \pmod{m^{k}}\)
\(\therefore x'=x+tm\)为前\(k\)个方程的解
程序实现
long long n,m[20],a[20];
long long exgcd(long long a,long long b,long long &x,long long &y)
{
if(!b){x=1,y=0;return a;}
long long d=exgcd(b,a%b,x,y),z=x;
x=y,y=z-y*(a/b);
return d;
}
long long xxtyfcz()
{
long long M,A,d,x,y;
M=m[1];
A=a[1];
for(int i=2;i<=n;i++){
d=exgcd(M,m[i],x,y);
if((a[i]-A)%d!=0) break;
long long temp=abs(m[i]/d);
x=x*((a[i]-A)/d);
x=(x%temp+temp)%temp;
A=M*x+A;
M=M*m[i]/d;
}
return (A%M+M)%M;
}
参考材料:
1、《算法竞赛进阶指南》李煜东 著
3、《简明数论》潘承洞、潘承彪 著
2、《数论初步》周春荔 著
ps:写得比较匆忙,有错误请在评论区指出,我将及时更改,谢谢!