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]
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
Sum += Stamp[Flag[i]];
printf("%d ", Stamp[Flag[i]]);
}
printf("/tSum=%d/n/n", Sum);
}else for(int i=1; i0; 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
TempTask[i] = Task[i];
}else{
for(int i=0; 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/* 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
if(Board[i-k][j+k]) return 0;
return 1;
}
void Trial(int i, int n)
{
if(i==n){
for(int k=0; k
for(int m=0; m
printf("%d ", Board[k][m]);
printf("/n");
}
printf("/n");
}else{
for(int j=0; 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]);
}
};
