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poj 1152

雪黛依梦 — Sun, 29 Aug 2010 03:41:00 GMT

#include
#include
int main ()
{
    int  i;
    char R;
    int r;
    while ( (R = getchar()) != EOF )
    {
            getchar();
          //��?nbsp;R 转化为数�?nbsp;
          if ( '0' <= R && R <= '9')
             r = R - '0';
             else if ( 'A' <= R && R <= 'Z' )
                  r = R - 'A' + 10;
                  else if ( 'a' <= R && R <= 'z' )
                  r = R - 'a' + 36;
             for (i = 2; i <= 62; i++)
             {
                 if ( (i - 1) % r == 0 )
                 {
                      printf ("%d\n", i);
                      break;
                 }
                 else
                     continue;
             }
    }
    //system("pause");
    return 0;
}

雪黛依梦 2010-08-29 11:41 发表评论

poj 1001

雪黛依梦 — Sun, 29 Aug 2010 03:35:00 GMT

#include
#include
#include
#define LENGTH 6  //��数的位敎ͼ�含小数点�Q?nbsp;

//��字�W��{化�ؓ数字
unsigned int change (char s[LENGTH], unsigned int s1[LENGTH - 1])
{
     int ss[LENGTH - 1];  //ss 放未逆置的整�?nbsp;
     memset (ss, 0, sizeof(ss));

     int k = 0 ;
     for (int i = 0; i < LENGTH && s[i]; ++i)  //考虑�Ҏ(gu��)��数据如：0.0001
     {
         if (s[i] != '.')
            ss[k++] = s[i] - '0';
     }
     for (int j = 0;j < LENGTH - 1; j++)
     {
         s1[j] = ss[LENGTH - 2 - j];
     }

     int m = 0;
     while ( (s[m] != '.') && s[m] )
           ++m;
           return LENGTH - 1 - m;   //��数点位�?nbsp;
}

//大数乘法�q�算
//函数�q�回 s2
void mu1 (unsigned int s1[LENGTH - 1],unsigned int s2[130])
{
     int ss[130];
     memset ( ss, 0, sizeof(ss) );

     for ( int i = 0; i < LENGTH - 1; i++)
         for (int j = 0;j < 130; j++)  //隄��:因�ؓ�q�回新的s2之后位数会增�?nbsp;最多时 5* 25 = 125
         ss [i + j] += s1[i] * s2[j];

     //��?nbsp;两个大数�怹�得的�U�ss中进行进位处理后攑ֈ�s2 �?nbsp;
     int c = 0;
     for (int i = 0;i < 130;i++)
     {
         s2[i] = (c + ss[i]) % 10;
         c = (c + ss[i]) / 10;
     }
}

int main()
{
    int n;
    char s[LENGTH];  //要处理的�q?nbsp;R
    unsigned int s1[LENGTH - 1];  //��?nbsp;R 转化成数�?nbsp;
    unsigned int s2[130];

    while(scanf ("%s%d", s, &n) != EOF)
    {
        memset (s1, 0, sizeof (s1));
        memset (s2, 0, sizeof (s2));

        int j = change (s, s1);      //得到��数�Ҏ(gu��)��在位�|?nbsp;
        change (s,s2);              //得到s2 �?nbsp;s1 �q�行�q�运��?nbsp;
        for ( int i = 1; i < n; i ++)
            mu1 (s1,s2);

        //在s2中前面的代表��数位，后面的代表整��C��Q?br>        //所以关键是通过数值关�p�L��到小数点的位�|?br>

         //例：0.1010  * 0.1010 = 0.01020100
        int m = 129;//��L��前导0
        while ( (!s2[m]) && m)
        --m;

        int k = 0; //��L��?
        while ( ( !s2[k] ) && (k < 130))
        ++k;

        //输出整数�?nbsp;
        for (int i = m; i >= n * j; i--)
            printf ("%d",s2[i]);

        //输出��数�?br>        if ( j && n * j >= k + 1)
        printf (".");

        for (int i = n*j -1; i >= k; --i)
        printf ("%d", s2[i]);
        printf ("\n");
    }

    return 0;
   // system ("pause");
}

雪黛依梦 2010-08-29 11:35 发表评论

poj 1503 ��L��_�ֺ�计算 �?R �?n �ơ幂

雪黛依梦 — Mon, 09 Aug 2010 05:21:00 GMT

//思�\�Q?.调用函数change() ��小�?s 转化成去掉小数点的整�?s1 �q�且�q�回得到��数点的位置�?br>//      2.调用函数mu1()用大��C��法模拟整数运��，�q�将�怹�的结果放在s2中，�q�回s2和是�q�行下面�?n -1�ơ相�?
//      3.在主函数里面通过已知的小数点的位�|�，利用数值关�p�输�?
//      4.注意��数点前没有前导0�Q�小数点后面没有��?�?/p>

//隄��Q�for ( int i = 1; i < n; i ++) mu1 (s1,s2);

1
2#include <stdio.h>
3#include <stdlib.h>
4#include <string.h>
5#define LENGTH 6  //��数的位敎ͼ�含小数点�Q?nbsp;
6
7//��字�W��{化�ؓ数字
8unsigned int change (char s[LENGTH], unsigned int s1[LENGTH - 1])
9{
10     int ss[LENGTH - 1];  //ss 放未逆置的整�?nbsp;
11     memset (ss, 0, sizeof(ss));
12
13     int k = 0 ;
14     for (int i = 0; i < LENGTH && s[i]; ++i)  //考虑�Ҏ(gu��)��数据如：0.0001
15     {
16         if (s[i] != '.')
17            ss[k++] = s[i] - '0';
18     }
19     for (int j = 0;j < LENGTH - 1; j++)
20     {
21         s1[j] = ss[LENGTH - 2 - j];
22     }
23
24     int m = 0;
25     while ( (s[m] != '.') && s[m] )
26           ++m;
27           return LENGTH - 1 - m;   //��数点位�?nbsp;
28}
29
30//大数乘法�q�算
31//函数�q�回 s2
32void mu1 (unsigned int s1[LENGTH - 1],unsigned int s2[130])
33{
34     int ss[130];
35     memset ( ss, 0, sizeof(ss) );
36
37     for ( int i = 0; i < LENGTH - 1; i++)
38         for (int j = 0;j < 130; j++)  //隄��:因�ؓ�q�回新的s2之后位数会增�?nbsp;最多时 5* 25 = 125
39         ss [i + j] += s1[i] * s2[j];
40
41     //��?nbsp;两个大数�怹�得的�U�ss中进行进位处理后攑ֈ�s2 �?nbsp;
42     int c = 0;
43     for (int i = 0;i < 130;i++)
44     {
45         s2[i] = (c + ss[i]) % 10;
46         c = (c + ss[i]) / 10;
47     }
48}
49
50int main()
51{
52    int n;
53    char s[LENGTH];  //要处理的�q?nbsp;R
54    unsigned int s1[LENGTH - 1];  //��?nbsp;R 转化成数�?nbsp;
55    unsigned int s2[130];
56
57    while(scanf ("%s%d", s, &n) != EOF)
58    {
59        memset (s1, 0, sizeof (s1));
60        memset (s2, 0, sizeof (s2));
61
62        int j = change (s, s1);      //得到��数�Ҏ(gu��)��在位�|?nbsp;
63        change (s,s2);              //得到s2 �?nbsp;s1 �q�行�q�运��?nbsp;
64        for ( int i = 1; i < n; i ++)
65            mu1 (s1,s2);
66
67        //在s2中前面的代表��数位，后面的代表整��C��Q?br>68        //所以关键是通过数值关�p�L��到小数点的位�|?br>69
70
71         //例：0.1010  * 0.1010 = 0.01020100
72        int m = 129;//��L��前导0
73        while ( (!s2[m]) && m)
74        --m;
75
76        int k = 0; //��L��?
77        while ( ( !s2[k] ) && (k < 130))
78        ++k;
79
80        //输出整数�?nbsp;
81        for (int i = m; i >= n * j; i--)
82            printf ("%d",s2[i]);
83
84        //输出��数�?br>85        if ( j && n * j >= k + 1)
86        printf (".");
87
88        for (int i = n*j -1; i >= k; --i)
89        printf ("%d", s2[i]);
90        printf ("\n");
91    }
92
93    return 0;
94   // system ("pause");
95}
96

雪黛依梦 2010-08-09 13:21 发表评论

poj 1503 多个大数相加求和

雪黛依梦 — Mon, 09 Aug 2010 05:12:00 GMT

1
2#include<stdio.h>
3#include<stdlib.h>
4#include<string.h>
5#define MAXSIZE 101
6int main()
7{
8    char line[MAXSIZE];
9    int sum[MAXSIZE + 1];
10
11    memset (sum, 0, sizeof(sum));
12
13    while (  scanf ("%s",line) && strcmp (line, "0") )  //求和�l�止条�g
14    {
15    //处理负数情况
16      while ( line[0] == '-')
17             return 0;
18
19      //��字�W�数转化为数�?nbsp;�Q��ƈ且相加存到sum【】数�l�中
20      int j = strlen(line);
21      for (int i = j - 1; i >= 0; i--)
22      {
23          sum[j-1-i] += (line[i] - '0');
24      }
25    }
26
27    //对sum【】进行进位的处理
28    for ( int i = 0; i <= MAXSIZE; i++ )
29    {
30        if ( sum[i] >= 10 )
31        {
32             sum[i+1] += (sum[i] / 10);
33             sum[i] = sum[i] % 10;
34        }
35    }
36    /*int c = 0;
37    for (int i = 0; i < MAXSIZE; i++)
38    {
39        c += sum[i];
40        sum[i] = c % 10;
41        c = c / 10;
42    }
43
44    for (int i = MAXSIZE; i >= 0; i--)
45    {
46        if (sum[i] != 0)
47        printf ("%d", sum[i]);
48    }*/
49
50    //�q�行输出处理
51    bool target = false;
52    for ( int i = MAXSIZE; i >= 0; i--)
53    {
54        if (target)
55           printf ("%d", sum[i]);
56           else if ( sum[i] )
57           {
58                printf ("%d", sum[i]);
59                target = true;
60           }
61    }
62    printf ("\n");
63    system("pause");
64    return 0;
65}
66

雪黛依梦 2010-08-09 13:12 发表评论

poj grids 2737 大数除法

雪黛依梦 — Mon, 09 Aug 2010 05:11:00 GMT

摘要: //思�\: 大数问题的处理，通常都是以字�W�串的�Ş式读入，再将字符转化为数字进行处�?/因�ؓ除法�q�算的实质是�Q�被除数能够减去除数的多��倍；�?546 / 23 ��Z��// 开始时�Q?546 - 23 �?100倍可以减 3 ��?�{�于 646 �Q�所以商增加 300// &nb... 阅读全文

雪黛依梦 2010-08-09 13:11 发表评论

雪黛依梦 — Mon, 09 Aug 2010 05:08:00 GMT

//此题重在理解两个数相乘的��法,

4#include <stdio.h>
5#include <stdlib.h>
6#include <string.h>
7#define MAXSIZE 201
8int main()
9{
10    char line1[MAXSIZE];
11    char line2[MAXSIZE];
12    int a1[MAXSIZE];
13    int a2[MAXSIZE];
14    int product[MAXSIZE];
15
16    while ( scanf ("%s%s", line1, line2) != EOF )
17    {
18          int len1, len2;
19          len1 = strlen (line1);
20          len2 = strlen (line2);
21
22          memset (a1, 0, sizeof (a1));
23          memset (a2, 0, sizeof (a2));
24          memset (product, 0, sizeof (product));
25
26          //对负数的处理
27          if ( line1[0] == '-' || line2[0] == '-' )
28                return 0;
29          //�?nbsp;0 �?nbsp;其它数相乘的处理
30          int mark = 0;
31          if ( ( !strcmp (line1,"0") ) || ( !strcmp (line2, "0") ) )
32          {
33               printf ("%d", 0);
34          }
35
36          //��字�W�数转化为数�?br>37          int j = 0;
38          for (int i = len1 - 1; i >= 0; i--)
39          {
40              a1[j++] = line1[i] - '0';
41          }
42          int k = 0;
43          for (int i = len2 - 1;i >= 0; i--)
44          {
45              a2[k++] = line2[i] - '0';
46          }
47
48          //乘法��法
49          for (int i = 0; i < len2; i++)
50             for (int j = 0; j < len1; j++)
51             {
52                product[i + j] += a1[j] * a2[i];
53             }
54
55          //处理�q�位
56          for (int i = 0; i < MAXSIZE; i++ )
57          {
58              if ( product[i] >= 10 )
59              {
60                   product [i + 1] += product [i] / 10;
61                   product [i] = product[i] % 10;
62              }
63          }
64
65          //输出处理
66          bool target = false;
67          for (int i = MAXSIZE - 1; i >= 0; i--)
68          {
69             if (target)
70                printf ("%d", product[i]);
71                else if (product[i])
72                {
73                     printf ("%d", product[i]);
74                     target = true;
75                }
76          }
77          printf ("\n");
78    }
79    //system ("pause");
80    return 0;
81}
82

以及�Ҏ(gu��)��数据的考虑 �Q�如含有 0 �?负数�?

雪黛依梦 2010-08-09 13:08 发表评论

雪黛依梦 — Mon, 09 Aug 2010 05:07:00 GMT

2//1.用字�W�数�l�输�?nbsp;
3//2.字符转化为数字进行计��?nbsp;
4//3.数组中下标相同的相加�q�存入到sum[]�?br> 5//4.�q�位处理
6//5.存在前导  0  �q�行输出处理  一定要注意最开�?nbsp;0  的存储位�|?nbsp;i = 2*MAXSIZE
7#include<stdio.h>
8#include<stdlib.h>
9#include<string.h>
10#define MAXSIZE 100
11int main()
12{
13    char line[MAXSIZE];
14    int a[MAXSIZE];
15    int sum[MAXSIZE + MAXSIZE];
16
17    memset (sum, 0, sizeof(sum));
18    while ( scanf ("%s",line) && strcmp (line, "0") )  //求和�l�止条�g
19    {
20        while ( line[0] == '-')
21            return 0;
22
23        memset (a, 0, sizeof(a));
24
25        //��字�W�数转化为数�?nbsp;
26        int j = strlen(line);
27        for (int i = j - 1; i >= 0; i--)
28        {
29            a[j-1-i] = line[i] - '0';
30        }
31
32        //��各�l�数据相加到sum数组�?/span>
33        for (int i = 0; i <= MAXSIZE; i++)
34        {
35            sum[i] += a[i];
36        }
37    }
38
39    //对sum【】进行进位的处理
40    for ( int i = 0; i <= 2*MAXSIZE; i++ )
41    {
42        if ( sum[i] >= 10 )
43        {
44            sum[i+1] += (sum[i] / 10);
45            sum[i] = sum[i] % 10;
46
47        }
48    }
49
50    //�q�行输出处理
51    bool target = false;
52    for ( int i = 2*MAXSIZE ; i >= 0; i--)
53    {
54        if (target)
55            printf ("%d", sum[i]);
56        else if ( sum[i] )
57        {
58            printf ("%d", sum[i]);
59            target = true;
60        }
61    }
62    printf ("\n");
63    //system("pause");
64    return 0;
65}
66