算法笔试题_1

1、将一整数逆序后放入一数组中（要求递归实现）

void convert(int *result, int n)

{

     if(n>=10)

         convert(result+1, n/10);

     *result = n%10;   

}

int main(int argc, char* argv[])

{

     int n = 123456789, result[20]={};

     convert(result, n);

     printf("%d:", n);

     for(int i=0; i<9; i++)

         printf("%d", result[i]);

     return getchar();

}

2、求高于平均分的学生学号及成绩（学号和成绩人工输入）

double find(int total, int n)

{

     int number, score, average;

     scanf("%d", &number);

     if(number != 0){

         scanf("%d", &score);

         average = find(total+score, n+1);

         if(score >= average)

              printf("%d:%d/n", number, score);

         return average;

     }else{

         printf("Average=%d/n", total/n);

         return total/n;

     }

}

int main(int argc, char* argv[])

{

     find(0, 0);

     return getchar();

}

3、递归实现回文判断（如：abcdedbca就是回文）

int find(char *str, int n)

{

     if(n<=1) return 1;

     else if(str[0]==str[n-1])   return find(str+1, n-2);

     else     return 0;

}

 

int main(int argc, char* argv[])

{

     char *str = "abcdedcba";

     printf("%s: %s/n", str, find(str,

 strlen(str)) ? "Yes" : "No");

     return getchar();

}

 

4、组合问题（从M个不同字符中任取N个字符的所有组合）

void find(char *source, char *result, int n)

{

     if(n==1){

         while(*source)

            printf("%s%c/n", result, *source++);

     }else{

         int i, j;

         for(i=0; source[i] != 0; i++);

         for(j=0; result[j] != 0; j++);

         for(; i>=n; i--)

         {

              result[j] = *source++;

              result[j+1] = '/0';

              find(source, result, n-1);

         }

     }

}

 

int main(int argc, char* argv[])

{

     int const n = 3;

     char *source = "ABCDE", result[n+1] = {0};

     if(n>0 && strlen(source)>0 && n<=strlen(source))

         find(source, result, 3);

     return getchar();

}

5、分解成质因数(如435234=251*17*17*3*2)

void prim(int m, int n)

{

     if(m>n){

         while(m%n != 0) n++;

         m /= n;

         prim(m, n);

         printf("%d*", n);

     }

}

int main(int argc, char* argv[])

{

     int n = 435234;

     printf("%d=", n);

     prim(n, 2);

     return getchar();

}

 

6、寻找迷宫的一条出路（o：通路； X障碍）

#define MAX_SIZE 8

int H[4] = {0, 1, 0, -1};

int V[4] = {-1, 0, 1, 0};          

char Maze[MAX_SIZE][MAX_SIZE] = {{'X','X','X','X','X','X','X','X'},

                                 {'o','o','o','o','o','X','X','X'},

                                 {'X','o','X','X','o','o','o','X'},

                                 {'X','o','X','X','o','X','X','o'},

                                 {'X','o','X','X','X','X','X','X'},

{'X','o','X','X','o','o','o','X'},

                                {'X','o','o','o','o','X','o','o'},

                                 {'X','X','X','X','X','X','X','X'}};

void FindPath(int X, int Y)

{

    if(X == MAX_SIZE || Y == MAX_SIZE){

         for(int i = 0; i < MAX_SIZE; i++)

for(int j = 0; j < MAX_SIZE; j++)

                  printf("%c%c", Maze[i][j], j < MAX_SIZE-1 ? ' ' : '/n');

}else for(int k = 0; k < 4; k++)

if(X >= 0 && Y >= 0 && Y < MAX_SIZE && X < MAX_SIZE && 'o' == Maze[X][Y]){

                  Maze[X][Y] = ' ';

                  FindPath(X+V[k], Y+H[k]);

                  Maze[X][Y] ='o';

}

}

int main(int argc, char* argv[])

{

    FindPath(1,0);

    return getchar();

}

 

7、随机分配座位，共50个学生，使学号相邻的同学座位不能相邻(早些时候用C#写的，没有用C改写）。

static void Main(string[] args)

{

     int Tmp = 0, Count = 50;            

     int[] Seats = new int[Count];            

     bool[] Students = new bool[Count];

     System.Random RandStudent=new System.Random();

     Students[Seats[0]=RandStudent.Next(0,Count)]=true;

     for(int i = 1; i < Count; )

{

         Tmp=(int)RandStudent.Next(0,Count);

         if((!Students[Tmp])&&(Seats[i-1]-Tmp!=1) && (Seats[i-1] - Tmp) != -1){

              Seats[i++] = Tmp;

Students[Tmp] = true;

         }

     }

     foreach(int Student in Seats)

         System.Console.Write(Student + " ");

     System.Console.Read();

}

 

8、求网格中的黑点分布（有6*7的网格，在某些格子中有黑点，已知各行与各列中有黑点的点数之和）

#define ROWS 6

#define COLS 7

int iPointsR[ROWS] = {2, 0, 4, 3, 4, 0};           // 各行黑点数和的情况

int iPointsC[COLS] = {4, 1, 2, 2, 1, 2, 1};        // 各列黑点数和的情况

int iCount, iFound;

int iSumR[ROWS], iSumC[COLS], Grid[ROWS][COLS];

 

int Set(int iRowNo)

{

if(iRowNo == ROWS){

        for(int iColNo=0; iColNo < COLS && iSumC[iColNo]==iPointsC[iColNo]; iColNo++)

           if(iColNo == COLS-1){

               printf("/nNo.%d:/n", ++iCount);

               for(int i=0; i < ROWS; i++)

                  for(int j=0; j < COLS; j++)

                      printf("%d%c", Grid[i][j], (j+1) % COLS ? ' ' : '/n');

               iFound = 1;                        // iFound = 1，有解

           }

    }else{

        for(int iColNo=0; iColNo < COLS; iColNo++)

        {

            if(iPointsR[iRowNo] == 0){

                Set(iRowNo + 1);

  }else if(Grid[iRowNo][iColNo]==0){

Grid[iRowNo][iColNo] = 1;

iSumR[iRowNo]++; iSumC[iColNo]++;                                  if(iSumR[iRowNo]<iPointsR[iRowNo] && iSumC[iColNo]<=iPointsC[iColNo])

                     Set(iRowNo);

else if(iSumR[iRowNo]==iPointsR[iRowNo] && iRowNo < ROWS)

                     Set(iRowNo + 1);

                Grid[iRowNo][iColNo] = 0;

                iSumR[iRowNo]--; iSumC[iColNo]--;

            }

        }

    }

return iFound;   // 用于判断是否有解

}

int main(int argc, char* argv[])

{

    if(!Set(0))

        printf("Failure!");

    return getchar();

}

9、有4种面值（面值为1, 4, 12, 21）的邮票很多枚，从中最多任取5张进行组合，求邮票最大连续组合值

#define N 5

#define M 5

int k, Found, Flag[N];

int Stamp[M] = {0, 1, 4, 12, 21};

 

// 在剩余张数n中组合出面值和Value

int Combine(int n, int Value)

{

     if(n >= 0 && Value == 0){

         Found = 1;

         int Sum = 0;

         for(int i=0; i<N && Flag[i] != 0; i++){

              Sum += Stamp[Flag[i]];

              printf("%d ", Stamp[Flag[i]]);

         }

         printf("/tSum=%d/n/n", Sum);

     }else for(int i=1; i<M && !Found && n>0; i++)

         if(Value-Stamp[i] >= 0){

              Flag[k++] = i;

              Combine(n-1, Value-Stamp[i]);

              Flag[--k] = 0;

         }

     return Found;

}

 

int main(int argc, char* argv[])

{

     for(int i=1; Combine(N, i); i++, Found=0);

     return getchar();

}

 

10、大整数数相乘的问题。

void Multiple(char A[], char B[], char C[])

{

    int TMP, In=0, LenA=-1, LenB=-1;

    while(A[++LenA] != '/0');

    while(B[++LenB] != '/0');

    int Index, Start = LenA + LenB - 1;

    for(int i=LenB-1; i>=0; i--)

    {

        Index = Start--;

        if(B[i] != '0'){

            for(int In=0, j=LenA-1; j>=0; j--)

            {

                TMP = (C[Index]-'0') + (A[j]-'0') * (B[i] - '0') + In;

                C[Index--] = TMP % 10 + '0';

                In = TMP / 10;

            }

            C[Index] = In + '0';

        }

    }

}

 

int main(int argc, char* argv[])

{

    char A[] = "21839244444444448880088888889";

    char B[] = "38888888888899999999999999988";

char C[sizeof(A) + sizeof(B) - 1];

 

    for(int k=0; k<sizeof(C); k++)

        C[k] = '0';

    C[sizeof(C)-1] = '/0';

 

    Multiple(A, B, C);

    for(int i=0; C[i] != '/0'; i++)

        printf("%c", C[i]);

    return getchar();

}

 

11、求最大连续递增数字串（如“ads3sl456789DF3456ld345AA”中的“456789”）

int GetSubString(char *strSource, char *strResult)

{

    int iTmp=0, iHead=0, iMax=0;

    for(int Index=0, iLen=0; strSource[Index]; Index++)

    {

        if(strSource[Index] >= '0' && strSource[Index] <= '9'

&& strSource[Index-1] > '0' && strSource[Index] == strSource[Index-1]+1)

{

            iLen++;                    // 连续数字的长度增1

        }else{                          // 出现字符或不连续数字

            if(iLen > iMax)

            {

            iMax = iLen;

iHead = iTmp;

            }       

           

            // 该字符是数字，但数字不连续

            if(strSource[Index] >= '0' && strSource[Index] <= '9'){

                iTmp = Index;

iLen = 1;

            }

        }   

    }

       

    for(iTmp=0 ; iTmp < iMax; iTmp++) // 将原字符串中最长的连续数字串赋值给结果串

        strResult[iTmp] = strSource[iHead++];

    strResult[iTmp]='/0';

    return iMax;                      // 返回连续数字的最大长度

}

int main(int argc, char* argv[])

{

    char strSource[]="ads3sl456789DF3456ld345AA", char strResult[sizeof(strSource)];

printf("Len=%d, strResult=%s /nstrSource=%s/n", GetSubString(strSource, strResult),

strResult, strSource);

    return getchar();

}

 

12、四个工人，四个任务，每个人做不同的任务需要的时间不同，求任务分配的最优方案。（2005年5月29日全国计算机软件资格水平考试——软件设计师的算法题）。

#include "stdafx.h"

#define N 4

int Cost[N][N] = { {2, 12, 5, 32},       // 行号：任务序号，列号：工人序号

                    {8, 15, 7, 11},       // 每行元素值表示这个任务由不同工人完成所需要的时间

                    {24, 18, 9, 6},

                    {21, 1, 8, 28}};

int MinCost=1000;

int Task[N], TempTask[N], Worker[N];

void Assign(int k, int cost)

{

     if(k==N)

     {

         MinCost = cost;   

         for(int i=0; i<N; i++)

              TempTask[i] = Task[i];

     }else{

         for(int i=0; i<N; i++){

              if(Worker[i]==0 && cost+Cost[k][i] < MinCost)

              {

                   Worker[i] = 1;     Task[k] = i;

                   Assign(k+1, cost+Cost[k][i]);

                   Worker[i] = 0; Task[k] = 0;

              }

         }

     }

}

 

int main(int argc, char* argv[])

{

     Assign(0, 0);

     printf("最佳方案总费用=%d/n", MinCost);

     for(int i=0; i<N; i++)      /* 输出最佳方案 */

         printf("/t任务%d由工人%d来做：%d/n", i, TempTask[i], Cost[i][TempTask[i]]);

     return getchar();     

}

 

13、八皇后问题（输出所有情况，不过有些结果只是旋转了90度而已）。哈哈：）回溯算法的典型例题

#define N 8

int Board[N][N];

int Valid(int i, int j)     // 所下棋子有效性的严正

{

     int k = 1;

     for(k=1; i>=k && j>=k;k++)

         if(Board[i-k][j-k])    return 0;

     for(k=1; i>=k;k++)

         if(Board[i-k][j])      return 0;

     for(k=1; i>=k && j+k<N;k++)

         if(Board[i-k][j+k])    return 0;

     return 1;

}

 

void Trial(int i, int n)

{

     if(i==n){

          for(int k=0; k<n; k++){

              for(int m=0; m<n; m++)

                   printf("%d ", Board[k][m]);

              printf("/n");

         }

         printf("/n");

     }else{

         for(int j=0; j<n; j++){

              Board[i][j] = 1;

              if(Valid(i,j))

                   Trial(i+1, n);

              Board[i][j] = 0;

         }

     }

}

 

int main(int argc, char* argv[])

{

     Trial(0, N);

     return getchar();

}

14、实现strstr功能（寻找子串在父串中首次出现的位置）

char * strstring(char *ParentString, char *SubString)

{

     char *pSubString, *pPareString;

     for(char *pTmp=ParentString; *pTmp; pTmp++)

     {

         pSubString = SubString;

         pPareString = pTmp;   

         while(*pSubString == *pPareString && *pSubString != '/0')

         {

              pSubString++;

              pPareString++;

         }

         if(*pSubString == '/0')     return pTmp;

     }

     return NULL;

}

 

int main(int argc, char* argv[])

{

     char *ParentString = "happy birthday to you!";

     char *SubString = "birthday";

     printf("%s",strstring(ParentString, SubString));

     return getchar();

}}

 

1,写出一个函数，比较两个字符串，返回最大公串，例如abacdaccbadc和cedaccbe返回daccb;

2，有100个数字，其中有正数也有负数，找出连续三个相加之和最大部分；
要求：尽量不要使用库函数！

两道一起来：支持搜索中文，

import java.util.Random;

public class Test
{
  private static int maxSubStart = 0;
  private static int maxSubLength = 0;
  private static char[] c1, c2;

  private static boolean isSame(int i, int j)
  {
    return c1[i] == c2[j];
  }

  private static void setMaxSub(int start1, int start2)
  {
    int i = start1, j = start2;
    int maxStart = 0;
    int maxLength = 0;
    for (; i < c1.length && j < c2.length; i++,j++)
    {
      if (isSame(i, j))
      {
        maxLength++;
      }
      else
        break;
    }
    if (maxLength > maxSubLength)
    {
      maxSubLength = maxLength;
      maxSubStart  = start1;
    }
  }
  private static String getMaxCommonString(String s1, String s2)
  {
    c1 = s1.toCharArray();
    c2 = s2.toCharArray();
    if (c1.length > c2.length)  // swap s1, s2 so s1.length < s2.length
    {
      char[] c = c1;
      c1 = c2;
      c2 = c;
    }
    int minLength = c1.length;
    int maxLength = c2.length;
    for (int i = 0; i < minLength; i++)
    {
      char ch = c1[i];
      for (int j = 0; j < maxLength; j++)
      {
        if (ch == c2[j])
        {
          setMaxSub(i, j);
        }
      }
    }
    return new String(c1, maxSubStart, maxSubLength);
  }

  private static int[] getRandomInt(int length)
  {
    Random r = new Random();
    int[] res = new int[length];
    for (int i = 0; i < length; i++)
    {
      res[i] = r.nextInt(200) - 100;
    }
    return res;
  }
  private static int count(int[] num, int i)
  {
    return num[i]+num[i+1]+num[i+2];
  }
  private static int getMaxThreeStart(int[] num)
  {
    int end = num.length - 3;
    int max = 0;
    int start = 0;
    for (int i = 0; i < end; i++)
    {
      int c = count(num, i);
      if (c > max)
      {
        max = c;
        start = i;
      }
    }
    return start;
  }
  public static void main(String[] args)
  {
    //abacdaccbadc和cedaccbe
    String s1 = "abacd测试汉字 accbadc", s2 = "ced测试汉字 accbe";
    System.out.println(s1 + "   " + s2 + " 的最大公串为: " + getMaxCommonString(s1, s2));
    int[] num = getRandomInt(100);
    int start = getMaxThreeStart(num);
    for (int i = 0; i < 100; i++)
    {
      System.out.print(num[i] + "  ");
      if (i%10 == 9)
      {
        System.out.println();
      }
    }
    System.out.println("三个连续数字之和最大的三个数子为: " + num[start] + "  " + num[start+1] + "  " + num[start+2]);
  }
};