算法1-1 模拟与高精度

目录

一 阶乘数码

 二 麦森数

 三 模拟题

一 阶乘数码

本题中n<=1000,1000的阶乘为以下这么大，远超long的范围
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

使用高精度计算，BigInteger可以表示任意大的数，仅受到内存的限制

数码就是一个数0-9

public class P1591 {public static void main(String[] args) {Scanner scanner=new Scanner(System.in);int t = scanner.nextInt();for (int i = 0; i < t; i++) {int n = scanner.nextInt();int a= scanner.nextInt();BigInteger m=new BigInteger("1");for (int j = 1; j <=n ; j++) {m=m.multiply(new BigInteger(j+""));}System.out.println(m);// 求sum中a的个数String string = m.toString();char[] chars = string.toCharArray();int count=0;for (char c : chars) {if (c-'0'==a){count++;}}System.out.println(count);}}
}

使用高精度计算，BigInteger可以表示任意大的数，仅受到内存的限制

数码就是一个数0-9

public class P1591 {public static void main(String[] args) {Scanner scanner=new Scanner(System.in);int t = scanner.nextInt();for (int i = 0; i < t; i++) {int n = scanner.nextInt();int a= scanner.nextInt();BigInteger m=new BigInteger("1");for (int j = 1; j <=n ; j++) {m=m.multiply(new BigInteger(j+""));}System.out.println(m);// 求sum中a的个数String string = m.toString();char[] chars = string.toCharArray();int count=0;for (char c : chars) {if (c-'0'==a){count++;}}System.out.println(count);}}
}

 二 麦森数

求一个数的位数，我们可以根据数学公式直接计算，不需要知道该数具体的值

法一：使用数学公式计算位数

public static void main(String[] args) {// 给一个数字P，求2的p次方-1 有多少位数字和最后500位Scanner scanner=new Scanner(System.in);int p=scanner.nextInt();// 使用数学公式直接计算位数int count= (int) ((Math.floor(p*Math.log10(2)))+1);System.out.println(count);// 直接算需要的后500位BigInteger two=new BigInteger("2");// 需要知道一个数的后n位，就对10的n次方取模BigInteger mod=BigInteger.TEN.pow(500);BigInteger last500=two.modPow(new BigInteger(p+""),mod).subtract(BigInteger.ONE);String str=last500.toString();if (str.length()<500){// 高位补0int diff=500-str.length();StringBuilder stringBuilder=new StringBuilder(0);for (int i = 0; i < diff; i++) {stringBuilder.append("0");}str=stringBuilder+str;}else{str=str.substring(str.length()-500);}// 每行输出50位 输出10行for (int i = 0; i < 10; i++) {System.out.println(str.substring(i*50,i*50+50));}scanner.close();
}

 法二：快速幂计算值，得出该数的位数

public static void main(String[] args) {// 给一个数字P，求2的p次方-1 有多少位数字和最后500位Scanner scanner=new Scanner(System.in);int p=scanner.nextInt();BigInteger result;// 快速幂result= quickMi(p);result=result.subtract(new BigInteger("1"));String str=result.toString();System.out.println(str.length());if (str.length()<500){// 高位补0int diff=500-str.length();StringBuilder stringBuilder=new StringBuilder(0);for (int i = 0; i < diff; i++) {stringBuilder.append("0");}str=stringBuilder+str;}else{str=str.substring(str.length()-500);}// 每行输出50位 输出10行for (int i = 0; i < 10; i++) {System.out.println(str.substring(i*50,i*50+50));}
}private static BigInteger quickMi(int p) {BigInteger base = new BigInteger("2");BigInteger result = BigInteger.ONE;while (p > 0) {if (p % 2 == 1) {result = result.multiply(base);}base = base.multiply(base); // 平方p /= 2; // 指数除以2}return result;
}private static BigInteger quickMi(int p) {if (p==1){return new BigInteger("2");}if (p%2==0){// p位偶数BigInteger sum=quickMi(p/2);return sum.multiply(sum);}else{// p为奇数BigInteger sum=quickMi(p/2);return sum.multiply(sum).multiply(new BigInteger("2"));}}

 三 模拟题

思路：

• 根据给定x，y值找到中心点，然后将2r+1阶矩阵顺时针旋转或者逆时针旋转

1 2 3 8 5 1 3 7 11

5 6 7 顺时针旋转90° -> 9 6 2 逆时针旋转90° -> 2 6 9

8 9 11 11 7 3 1 5 8

顺时针旋转：将某一列的值放到对应的行上(例如将第一列赋值到第一行)，从下到上

逆时针旋转：最后一列元素放到第一行上，从上到下

注：在赋值元素时，如果直接替换容易元素覆盖，可以使用一个临时数组存储原来的元素，然后将对应元素赋值

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner=new Scanner(System.in);int n = scanner.nextInt();int m = scanner.nextInt();int[][] array=new int[n+1][n+1];// 初始化数组int number=1;for (int i = 1; i <= n; i++) {for (int j = 1; j <=n ; j++) {array[i][j]=number++;}}// 改变 m 次矩阵for (int i = 0; i < m; i++) {int x = scanner.nextInt();int y = scanner.nextInt();int r = scanner.nextInt();int z = scanner.nextInt();int[][] temp = new int[n+1][n+1];for (int j = x - r; j <= x + r; j++) {for (int k = y - r; k <= y + r; k++) {temp[j][k] = array[j][k];}}if (z == 0) {int x1 = x + r;int y1 = y - r;// 2.1 顺时针将行内元素反转for (int j = x - r; j <= x + r; j++) {for (int k = y - r; k <= y + r; k++) {// 将一列元素赋值到array对应的一行上array[j][k] = temp[x1][y1];x1--;}x1 = x + r;y1++;}} else {int x1 = x - r;int y1 = y + r;// 逆时针for (int j = x - r; j <= x + r; j++) {for (int k = y - r; k <= y + r; k++) {// 将一列元素赋值到array对应的一行上array[j][k] = temp[x1][y1];x1++;}x1 = x - r;y1--;}}}for (int i = 1; i< array.length; i++) {for (int j = 1; j < array.length; j++) {System.out.print(array[i][j]+" ");}System.out.println();}
}
}