2014届百田校招笔试

          这次百田的笔试共有三道编程题，60分钟完卷。感觉发挥十分槽糕。第二题事后反思自己的思维方式有误，丢弃掉原本数学的知识反而去想一些递归什么的。学习编程和算法要有这理科生的心态，而不是一味死记硬背。好了不多说。下面是3道大题。

    题目1：有n位同学（n是偶数），编号从1到n。现在需要把n位同学分配到2张桌子吃饭，有m组约束，要求a[i]不得与b[i]同桌。a[],b[]的长度都是m。如果有多种解决方案，返回其中一种一桌的编号。如果没有结果，返回空。

           

import java.io.BufferedInputStream;import java.util.Scanner;public class Main {public int[] arrange(int n, int m, int[] a, int[] b) {int[] l1 = new int[n / 2];int[] l2 = new int[n / 2];int l1size = 0;int l2size = 0;for (int i = 0; i < m; i++) {if (i == 0) {l1[0] = a[0];l2[0] = b[0];l1size++;l2size++;} else if (!findInArray(a[i], l1size, l1)&& !findInArray(b[i], l2size, l2)&& !findInArray(a[i], l2size, l2)&& !findInArray(b[i], l1size, l1)) {l1[l1size] = a[i];l2[l2size] = b[i];l1size++;l2size++;} else if (findInArray(a[i], l1size, l1)&& !findInArray(b[i], l2size, l2)&& !findInArray(b[i], l1size, l1)) {l2[l2size] = b[i];l2size++;} else if (!findInArray(a[i], l1size, l1)&& !findInArray(a[i], l2size, l2)&& findInArray(b[i], l1size, l1)) {l2[l2size] = a[i];l2size++;} else if (!findInArray(b[i], l1size, l1)&& !findInArray(b[i], l2size, l2)&& findInArray(a[i], l2size, l2)) {l1[l1size] = b[i];l1size++;} else if (!findInArray(a[i], l1size, l1)&& findInArray(b[i], l2size, l2)&& !findInArray(a[i], l2size, l2)) {l1[l1size] = a[i];l1size++;} else if ((findInArray(a[i], l1size, l1)&& findInArray(b[i], l2size, l2)&& !findInArray(a[i], l2size, l2) && !findInArray(b[i],l1size, l1))|| (findInArray(b[i], l1size, l1)&& findInArray(a[i], l2size, l2)&& !findInArray(b[i], l2size, l2) && !findInArray(a[i], l1size, l1))) {} else if ((findInArray(a[i], l1size, l1) && findInArray(b[i],l1size, l1))|| (findInArray(a[i], l2size, l2) && findInArray(b[i],l2size, l2))) {if ((findInNum(a[i], i, a) + findInNum(a[i], i, b)) >= 2&& (findInNum(b[i], i, a) + findInNum(b[i], i, b)) >= 2)return null;else if ((findInNum(a[i], i, a) + findInNum(a[i], i, b)) == 1&& (findInNum(b[i], i, a) + findInNum(b[i], i, b)) >= 2) {int t = 0;if (findInArray(a[i], i, a)) {for (int j = 0; j < i; j++) {if (a[j] == a[i])t = b[j];}for (int k = 0; k < l1size; k++) {if (a[i] == l1[k]) {l1[k] = t;}}for (int k = 0; k < l2size; k++) {if (t == l2[k]) {l1[k] = a[i];}}} else {for (int j = 0; j < i; j++) {if (b[j] == a[i])t = a[j];}for (int k = 0; k < l1size; k++) {if (a[i] == l1[k]) {l1[k] = t;}}for (int k = 0; k < l2size; k++) {if (t == l2[k]) {l1[k] = a[i];}}}} else if ((findInNum(b[i], i, a) + findInNum(b[i], i, b)) == 1&& (findInNum(a[i], i, a) + findInNum(a[i], i, b)) >= 2) {int t = 0;if (findInArray(b[i], i, a)) {for (int j = 0; j < i; j++) {if (a[j] == b[i])t = a[j];}for (int k = 0; k < l1size; k++) {if (b[i] == l1[k]) {l1[k] = t;}}for (int k = 0; k < l2size; k++) {if (t == l2[k]) {l1[k] = b[i];}}} else {for (int j = 0; j < i; j++) {if (b[j] == b[i])t = a[j];}for (int k = 0; k < l1size; k++) {if (b[i] == l1[k]) {l1[k] = t;}}for (int k = 0; k < l2size; k++) {if (t == l2[k]) {l1[k] = b[i];}}}}}}if (l1size < l1.length || l2size < l2.length) {for (int j = 1; j <= n; j++) {if (!findInArray(j, l1size, l1) && !findInArray(j, l2size, l2)) {if (l1size < l1.length) {l1[l1size] = j;l1size++;} else if (l2size < l2.length) {l2[l2size] = j;l2size++;}}}}for (int i = 0; i < l1.length; i++) {System.out.println(l1[i]);}for (int i = 0; i < l2.length; i++) {System.out.println(l2[i]);}return l1;}private boolean findInArray(int m, int j, int[] a) {for (int i = 0; i < j; i++) {if (a[i] == m) {return true;}}return false;}private int findInNum(int m, int j, int[] a) {int result = 0;for (int i = 0; i < j; i++) {if (a[i] == m) {++result;}}return result;}public static void main(String[] args) {int n = 0, m = 0;Scanner cin = new Scanner(new BufferedInputStream(System.in));if (cin.hasNextInt()) {n = cin.nextInt();} else {System.out.println("输入不对");}if (cin.hasNextInt()) {m = cin.nextInt();} else {System.out.println("输入不对");}int k = 0;int j = 0;int[] result = new int[n / 2];int[] a = new int[m];int[] b = new int[m];while (k < m && cin.hasNextInt()) {a[k] = cin.nextInt();k++;}while (j < m && cin.hasNextInt()) {b[j] = cin.nextInt();j++;}result = new Main().arrange(n, m, a, b);}}

运行结果截图

题目2

       已知  f(n)=f(n-1)+f(n-2),如果知道f（0）的值为a，n的个数和f（n）的值，求f（1）的值。

      分别输入 a,n,f(n)的值输入f(1).

     例如输入  1   3  11 输出 5.

  代码

import java.io.BufferedInputStream;import java.util.Scanner;public class Main {public int findB(int a, int n, int f) {int anum = 0;int bnumpre = 0;int bnum = 1;for (int i = 2; i <= n; i++) {anum = bnum;bnum += bnumpre;bnumpre = anum;}return (f - a * anum) / bnum;}public static void main(String[] args) {Scanner cin = new Scanner(new BufferedInputStream(System.in));int a = 0;int n = 0;int f = 0;if (cin.hasNextInt()) {a = cin.nextInt();}if (cin.hasNextInt()) {n = cin.nextInt();}if (cin.hasNextInt()) {f = cin.nextInt();}System.out.println(new Main().findB(a, n, f));}}

运行结果：

 题目3

      输入16进制字符串如“3A”输出“111010”

     代码

import java.io.BufferedInputStream;import java.util.Scanner;public class Main {public String hxToBin(String s) {StringBuffer result = new StringBuffer();for (int i = 0; i < s.length(); i++) {if (i == 0) {result.append(trimZero(chatToString(s.charAt(i))));} else {result.append(chatToString(s.charAt(i)));}}return result.toString();}private String trimZero(String s) {String result = null;result = s.replaceFirst("^0+", "");return result;}private String chatToString(char a) {switch (a) {case '0':return "0000";case '1':return "0001";case '2':return "0010";case '3':return "0011";case '4':return "0100";case '5':return "0101";case '6':return "0110";case '7':return "0111";case '8':return "1000";case '9':return "1001";case 'A':return "1010";case 'B':return "1011";case 'C':return "1100";case 'D':return "1101";case 'E':return "1110";case 'F':return "1111";default:return null;}}public static void main(String[] args) {Scanner cin = new Scanner(new BufferedInputStream(System.in));String s = null;if (cin.hasNext()) {s = cin.next();}System.out.println(new Main().hxToBin(s));}}

运行结果：