哎，这些思路在考试时都有了，可是当时就是写不出来。本菜鸟答案仅供参考。 第一题： 三个数组A[], B[], C[] 求A第3大的数 + B第4小的数 + C第5大的数结果, 不存在则为0 long long GetResult(vector<int> A, vector<int> B, vector<int> C) { } import java.util.Arrays; public class GetResult {     public static void main(String[] arg)     {         int[] A = new int[]{8,2,4,9,3,6};         int[] B = new int[]{8,2,4,9,3,6};         int[] C = new int[]{8,2,4,9,3,6};         System.out.println(getResult(A,B,C));     }     public static int getResult(int[] A,int[] B,int[] C)     {         if(A==null||B==null||C==null||A.length==0||B.length==0||C.length==0)         {             return 0;         }         Arrays.sort(A);         Arrays.sort(B);         Arrays.sort(C);         int sum =0;         int n = A.length,m = B.length,k = C.length;         sum = A[n-3]+B[3]+C[n-5];         return sum;     } } 第二题， 数组A[n] 设数组A的子集为连续m个数，即A[0:m-1] , A[1:m] … A[n-m+1:n] 子集的和为这m个数相加，求和最大的子集(有多个则输出多个) vector<vector<int>> GetResult(vector<int> A, int n, int m) { } 第二题虫取法 尺取就像虫虫一样。往前爬。比如[1,m]到[2,m+1]，求出[1,m]的和后，只要删除1,加上m+1，就是[2,m+1]的和。 import java.util.*; public class GetResult2 {     public static void main(String[] arg)     {         int[] A = new int[]{8,2,4,9,3,6};         System.out.println(getResult2(A,6,2));     }     public static ArrayList<ArrayList<Integer>>getResult2(int[] A, int n , int m) {         ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();         ArrayList<Integer> list1 = new ArrayList<Integer>();         ArrayList<Integer> list2 = new ArrayList<Integer>();         int i = 0, j = m - 1;         int sum = 0;         while (j <= n - 1) {             for (int z = i; z <= j; z++) {                 sum += A[z];             }             System.out.println(sum);             list1.add(sum);             sum = 0;//忘记将sum重新置为0了。             i++;             j++;         }         int z = 0, k = m - 1;//忘记重新更新目录         int max =Collections.max(list1);         System.out.println(max);         while (k <= n - 1) {             for (int h = z; h <= k; h++) {                 sum += A[h];             }             if(sum==max)             {                 for(int g =k-m+1;g<=k;g++)                 list2.add(A[g]);                 System.out.println(list2);                 int size=list2.size()/m;                 while (size>0)                 {                     ret.add(list2);                     size--;                 }             }             sum = 0;//忘记将sum重新置为0了。             z++;             k++;         }         return ret;     } } 第三题 求n * (n+10086) * (n-10086) string GetResult( long long n) { } 大数运算 package com.linbin.struts; import java.math.BigDecimal; public class GetResult3 {     public static void main(String[] arg)     {         System.out.println(GetResult(100000));     }     public static String GetResult( long n) {        BigDecimal b1 = new BigDecimal(10086);        BigDecimal b2 = new BigDecimal(n);        BigDecimal b3=b2.pow(2);        BigDecimal b4 = b1.pow(2);        BigDecimal b5 =b3.subtract(b4);        return  b2.multiply(b5).toString();     } }