算法练习

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1回文词，镜像词判断。（输入舞空白字符和0，可用scanf）

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <ctype.h>const char* rev = "A   3  HIL JM O   2TUVWXY51SE Z  8 ";     //镜像词字符串。mirror charconst char* msg[] = {"not a palindrome","a regular palindrome","a mirrored string","a mirrored palindrome"};//输出数组 output arraychar r( char ch)  {    if(isalpha(ch)) return rev[ch - 'A'];   //judge if alpha,return it's value in the letter table.    return rev[ch - '0' + 25];              //judge and return the number.}int main(){    char s[60];    while( scanf("%s",s) == 1){        int len = strlen(s);        int p = 1, m = 1;        for(int i = 0; i < (len+1)/2; i++){            if(s[i] != s[len-1-i])  p = 0;            if(r(s[i])!=s[len-1-i]) m = 0;        }        printf("%s -- is %s.\n\n",s,msg[m*2+p]);    }    return 0;}

WERTYU，如果手在键盘上放错位置，想输入qwertyu时则输入了wertyui。修改此错误。

#include <stdio.h>#include <stdlib.h>char s[] = "`1234567890-=qwertyuiop[]\\asdfghjkl;'zxcvbnm,./";int main(){    int i,c;    while((c = getchar()) != EOF){        for(i=1;s[i]&& s[i] !=c;i++);        if (s[i]) putchar(s[i-1]) ;        else putchar (c);    }    return 0;}

蛇形填字。

#include <stdio.h>#include <stdlib.h>#include <string.h>#define maxn 20int a[maxn][maxn];  // named a two dimension array respectively represent the line and rowint main(){    int n,x,y,tot = 0;    scanf("%d",&n);    memset(a,0,sizeof(a));    tot = a[x=0][y=n-1] = 1;    while(tot < n*n)    {        while((x+1) < n && !a[x+1][y]) a[++x][y] = ++tot; //if line exist,and the next line is 0. the number next line add 1.        while((y-1>=0) && !a[x][y-1])  a[x][--y] = ++tot; //go back to add 1 on the same line in different column.        while((x-1>=0)&& !a[x-1][y])   a[--x][y] = ++tot; //then add 1 on the same column in different line;        while(y+1 < n && !a[x][y+1])   a[x][++y] = ++tot; //add 1 on the same line in different column from left to right.    }    for(x = 0; x < n; x++)    {        for(y = 0 ; y < n; y++) printf("%3d",a[x][y]);  //%3d,printf an int number take 3 blocks        printf("\n");                                   //printf number in column needn't change line.    }    return 0;}

输出结果：

input：9

25 26 27 28 29 30 31 32 1

24 51 52 53 54 55 56 33 2

23 50 69 70 71 72 57 34 3

22 49 68 79 80 73 58 35 4

21 48 67 78 81 74 59 36 5

20 47 66 77 76 75 60 37 6

19 46 65 64 63 62 61 38 7

18 45 44 43 42 41 40 39 8

17 16 15 14 13 12 11 10 9

开灯问题，n盏灯，第一人把所有灯打开。第2人按下所有2的倍数，第三人按下所有三的倍数。开关再按开的灯会关闭，一共k个人，问最后亮几盏灯。

//n盏灯，第一个人把所有灯打开，第2个人按下所有编号为2的倍数开关，第三个人按3的倍数开关。共k个人，k<=n<=1000.问最后几盏灯亮着#include <stdio.h>#include <stdlib.h>#include <string.h>#define maxn 1010int a[maxn];      //using a[n] represent the light situation.int main(){    int n,k,first = 1;    memset(a,0,sizeof(a));       scanf("%d%d",&n,&k);    for(int i = 1;i<=k; i++)        for(int j = 1; j<=n;j++)        if(j%i==0) a[j]=!a[j];        //i means number of people. if j%i==0,then change the light situation.    for(int i = 1; i<=n; i++)        if(a[i])    {   if(first) first = 0;else printf(" ");printf("%d",i);    }    return 0;}

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