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hdu 1048

雪黛依梦 — Tue, 07 Sep 2010 03:02:00 GMT

一水题也被我WA 了好几次哦！��是开头的那个必须是START
所以必��d��加一个strcmp ( first, ''START'')

#include
#include
using namespace std;

int main ()
{
    char first[11];
    char last[8];
    char str[201];

    while (  scanf ("%s", &first) && strcmp( first, "EDNOFINPUT") && !strcmp (first, "START") )
    {
         getchar ();
         int len = 0;
         while ( (str[len] = getchar ()) != '\n')
         len ++;

         for ( int i = 0; i < len; i ++ )
         {
             if ( str[i] >= 'A' && str[i] <= 'E')
             {
                  printf ("%c", str[i] + 21);
             }
             else if ( str[i] >= 'F' && str[i] <= 'Z')
             {
                  printf ("%c", str[i] - 5);
             }
             else
             printf ("%c", str[i]);
         }
         printf ("\n");
         scanf ("%s", last);
    }
   //system ("pause");
   return 0;
}

雪黛依梦 2010-09-07 11:02 发表评论

//hdu 1020

雪黛依梦 — Fri, 03 Sep 2010 03:50:00 GMT

#include
#include
#include
#include
int main ()
{
    int n;
    char str[10000];
    while ( scanf ("%d", &n) != EOF && n )
    {
          getchar ();
          for ( int i = 0; i < n; i ++ )
          {
              scanf ("%s", &str);
              int len = 0;
              while ( str[len] != '\0')   //错点
                    len ++;

              int index = 0;
              while ( index < len )
              {
                  int count = 1;
                  if ( str[index] != str[index + 1] )
                  {
                       printf ("%c", str[index]);
                       index ++;
                  }
                  else
                  {
                      count ++;
                      while (str[index] == str[ index + count])  //错点�Q�是 ==
                      count ++;

                      printf ("%d%c", count, str[index]);

                      index += count;
                  }
              }
              printf ("\n");
          }
    }

   return 0;
}

雪黛依梦 2010-09-03 11:50 发表评论

雪黛依梦 — Sat, 28 Aug 2010 14:46:00 GMT

1、先计算Gcd(a,b)�Q�若n不能�?/strong>Gcd(a,b)整除�Q�则方程无整�?/strong>解；否则�Q�在方程两边同时除以Gcd(a,b)�Q�得到新的不定方�E?/strong>a' * x + b' * y = n'�Q�此�?/strong>Gcd(a',b')=1;
2、利用上面所说的�Ƨ几里�d��法求出方程a' * x + b' * y = 1的一�l�整数解x0,y0�Q�则n' * x0,n' * y0是方�E?/strong>a' * x + b' * y = n'的一�l�整数解�Q?
3、根据数��Z��的相兛_��理，可得方程a' * x + b' * y = n'�?/strong>所有整数解为：
x = n' * x0 + b' * t
y = n' * y0 - a' * t (t为整�?/strong>)
上面的解也就�?/strong>a * x + b * y = n 的全部整数解�?nbsp;

此时方程的所有解为：x=c*k1-b*t�?
x的最��的可能值是0
§�?/span>x=0�Q�可求出�?/span>x最��时�?/span>t的取��|��但由�?/span>x=0是可能的最��取��|��实际上可�?/span>x�Ҏ��取不�?/span>0�Q�那么由计算机的取整�?/span>法可知：�?t=c*k1/b��出�?/span>t�Q�代�?/span>x=c*k1-b*t中�?§求出�?/span>x可能会小�?/span>0�Q�此时��ot=t+1�Q�求出的x必大�?/span>0�Q�如果代回后x仍是大于�{�于0的，那么不需要再做修正�?

//满��关系�Q?x + m * s) - (y + n *s) = k*l  (k= 0 1 2 )卻I��(n - m)*s + l*k = x-y;
//利用拓展的欧几里得解出可能的s
#include
#include

//求最大公�U�数
__int64 gcd (__int64 a, __int64 b)
{
    if (b == 0)
       return a;
       else
       gcd (b, a % b);
}

//求得满�� a*x + b*y = d�Q�的x  y
__int64 ex_Gcd (__int64 a, __int64 b, __int64 &x1, __int64 &y1)
{
    if ( b == 0 )
    {
         x1 = 1;
         y1= 0;
         return a;
    }

    int r = ex_Gcd( b, a%b, x1, y1);

    int temp = x1;
    x1 = y1;
    y1 = temp - a/b * y1;
}

int main ()
{
    __int64  x, y, m, n, l;
    int a, b, product, d;//a b 的最大公�U�数 product
    int s, k;
    __int64 s1, k1, s2, k2;

    while ( scanf ("%I64d%I64d%I64d%I64d%I64d", &x, &y, &m, &n, &l ) != EOF )
    {
          a = n - m;
          b = l;
          d = x-y;
          product = gcd (a,b);

          if ( d % product != 0 )
             printf ("Impossible\n");
             else
             {
                 a = a / product;
                 b = b / product;
                 d = d / product;
                 ex_Gcd (a, b, s1, k1);               //得到(n-m)/product * s + l/product * k = 1;�?nbsp;s k的解
                 s2 = d * s1;                     //得到(n-m)/product * s + l/product * k = d;�?nbsp;s k的解
                 k2 = d * k1;

                 int t;
                 //s = s2 - b * t;    用下面的�Ҏ��处理满��条�g的解
                 //k = k2 - a * t;

                 t = s2 / b;
                 s = s2 - b * t;
                 if ( s <= 0)
                 {
                      s += b;
                 }
                 printf ("%d\n", s);
             }
    }
    //system ("pause");
    return 0 ;
}

雪黛依梦 2010-08-28 22:46 发表评论

hdu 1004

雪黛依梦 — Tue, 24 Aug 2010 08:46:00 GMT
开始时�q�道题目想了��Z��的，也走了一点弯路，其实主要是C 语言的二�l�字�W�数�l�没理解好，都是一些基��知识没有掌握�Q�注意打�U�色的部分就好了

1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4char code[1001][17];
5int num[1001];
6int main ()
7{
8  int n;
9  while (scanf ("%d", &n) != EOF && n != 0)
10  {
11   getchar ();
12   memset (code, 0, sizeof (code));
13   memset (num, 0 , sizeof (num));
14
15   for (int i = 0; i < n; i ++)    //输入处理
16   {
17    int j = 0;
18    while ( (code[i][j] = getchar ()) !=  '\n' )     i 代表�q�是�W�几个串
19    j ++;
20   }
21
22
23   for (int m = 0; m < n; m  ++)     //匚w��
24   {
25    for (int k = m + 1; k < n; k ++)
26    {
27     if ( !strcmp (code[m], code[k]) )  //相等�q�回 0
28      num[m]++;
29    }
30   }
31
32   int max = -1;   int index = 0;
33   for ( int k = 0; k < n; k ++)  //扑ֈ�出现�ơ数最多的下标
34   {
35       if (max < num[k])
36       {
37         max = num[k];
38         index = k;
39       }
40   }
41   printf ("%s", code[index]);
42 }
43 //system ("pause");
44 return 0;
45}
46

雪黛依梦 2010-08-24 16:46 发表评论

poj Grids 2744

雪黛依梦 — Fri, 06 Aug 2010 03:50:00 GMT

其实�q�个题的思�\和简单：扑ֈ�最短的一个串�Q�然后再扑ֈ�他所有的子串和子串的逆串和所�l�的串进行匹配�?br>但是如果不直接调用string.h的函数实现这些功能，我自己就不知道做�Q?br>感觉String.h 有好多功能自�׃��了解�Q�像strstr() strncpy() strrev()可是poj�pȝ��不支持这个函�?br>

1#include <stdio.h>
2#include <string.h>
3char str[100][101];
4
5int t, n;
6void strrev (char *revString)
7{
8    int revLen = strlen (revString);
9    char temp[101];
10
11    int i;
12    for ( i = 0; i < revLen; i ++)
13    {
14        temp[i] = revString[revLen - i - 1];
15    }
16    temp [i] = '\0';
17    strcpy (revString, temp);
18}
19
20int searchSubString (char *source)
21{
22    char minSubString[101];
23    char minrevSubString[101];
24
25    int  minSubStringLen = strlen (source);
26    int sourceLen = strlen (source);
27
28    while (minSubStringLen)
29    {
30        for (int i = 0; i <= sourceLen - minSubStringLen; i ++)
31        {
32            bool tag = true;
33            strncpy (minSubString, source + i, minSubStringLen);
34            strncpy (minrevSubString, source + i, minSubStringLen);
35
36            minSubString[minSubStringLen] = minrevSubString[minSubStringLen] = '\0';
37
38            strrev (minrevSubString);
39
40            for (int i = 0; i < n; i ++)
41            {
42                if ( strstr (str[i], minSubString) == NULL && strstr (str[i], minrevSubString) == NULL )
43                {
44                    tag = false;
45                    break;
46                }
47            }
48
49            if (tag)
50                return minSubStringLen;
51        }
52        minSubStringLen --;
53    }
54
55    return 0;
56}
57
58int main ()
59{
60
61    char minStr[101];
62
63    scanf ("%d", &t);
64    while ( t != 0 )
65    {
66        scanf ("%d", &n);
67
68        int minStrlen = 100;
69        for (int i = 0; i < n; i ++)
70        {
71            scanf ("%s", str[i]);
72            if ( ( strlen(str[i]) ) < minStrlen )
73            {
74                strcpy(minStr, str[i]);
75                minStrlen = strlen (str[i]);
76            }
77        }
78
79        int minSubLen = searchSubString (minStr);
80        printf ("%d\n", minSubLen);
81
82        t --;
83    }
84    return 0;
85}

雪黛依梦 2010-08-06 11:50 发表评论

poj 1298 字符串的处理问题

雪黛依梦 — Thu, 05 Aug 2010 08:00:00 GMT

我晕了！�q�么水的题，我居然WA了N �ơ，�Q�，��是自己没有理解好题意最后还对ENDOFINPUT�q�行了输出处�?br>

1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4int main ()
5{
6    char start[11];
7    char end[4];
8    char cipher[201];
9    int len;
10
11    scanf ("%s", start);
12    while ( strcmp (start, "ENDOFINPUT") )
13    {
14        getchar ();
15
16        gets (cipher);
17
18        getchar ();
19
20        scanf ("%s", end);
21
22        //对密文进行解�?nbsp;
23        len = strlen (cipher);
24        for (int i = 0; i < len; i ++)
25        {
26            if ( ('A' <= cipher[i]) && ( cipher[i] <= 'Z') )
27            {
28                if ( 70 <= cipher[i] && cipher[i]  <= 90)
29                    printf ("%c", cipher[i] - 5);
30                else
31                    printf ("%c", cipher[i] + 21);
32            }
33            else
34                printf ("%c", cipher[i]);
35        }
36        printf ("\n");
37        scanf ("%s", start);
38    }
39    //printf ("ENDOFINPUT\n");   画蛇添��
40    //system ("pause");
41    return 0;
42}
43

雪黛依梦 2010-08-05 16:00 发表评论