POJ 2402--Palindrome Numbers

Sun, 29 Mar 2009 07:00:00 GMT

  1#include <cstdio>
  2#include <cmath>
  3const int MAX = 20;
  4
  5__int64 arr_bit[MAX]; //记录i个位有几个回文数
  6char strNum[50]; //保存�l�果
  7
  8void Cal()
  9{
10    int i, j;
11    __int64 t;
12
13    arr_bit[1] = 9;
14
15    for ( i = 2; i <= MAX; ++i )
16    {
17        arr_bit[i] = arr_bit[i - 1];
18        t = 9;
19        for ( j = 0; j < (i >> 1) - 1; ++j )
20                t *= 10;
21        if ( i % 2 != 0 ){
22            t *= 10;
23        }
24        arr_bit[i] += t;
25    }
26}
27
28void GetNum(const __int64& n)
29{
30    //对少�?00的数直接输出
31    if ( n <= 9 )
32    {
33        printf("%I64d\n", n);
34    }
35    else if ( n <= 18 )
36    {
37        printf("%I64d%I64d\n", (n - 9), (n - 9));
38    }
39    else {
40        int i, pos, wh, t, k;
41        __int64 del;
42
43        //扑և�当前数有多少个位
44        for ( i = 3; i <= MAX; ++i )
45        {
46            if ( arr_bit[i] >= n )
47            {
48                pos = i;
49                break;
50            }
51        }
52
53        //pos个位的第几个
54        del = n - arr_bit[pos - 1] - 1;
55
56        wh = pos >> 1;
57
58        if ( pos % 2 == 0 ) {
59            wh--;
60        }
61
62        //��定每个�?/span>
63        t = (int)pow(10, wh);
64        strNum[0] = strNum[pos - 1] = (int)(del / t) + 1 + '0';
65        del = del % t;
66
67        for ( i = wh - 1, k = 1; i >= 0; --i, ++k )
68        {
69            t = (int)pow(10, i);
70            strNum[k] = strNum[pos - k - 1] = (int)(del / t) + '0';
71            del = del % t;
72        }
73
74        strNum[pos] = '\0';
75
76        printf("%s\n", strNum);
77
78    }
79
80}
81
82int main()
83{
84    __int64 n;
85
86    Cal();
87
88    while ( true )
89    {
90        scanf("%I64d", &n);
91
92        if ( n == 0 )
93            break;
94
95        GetNum(n);
96    }
97    return 0;
98}
99
100
101
102

�?/a> 2009-03-29 15:00 发表评论

Pku 3258--River Hopscotch(二分枚�D�?

Sat, 07 Mar 2009 08:03:00 GMT

二分�?br>不断枚�D一个��|��然后�Ҏ��题目要求判断�Q�这个��g��正确�{�案的大��关�p�，修改上下界的��|��
无穷��D��正确�{�案�Q�最后就是正��答�?img height=20 src="http://m.shnenglu.com/Emoticons/QQ/noidea.gif" width=20 border=0>

其实我还没有真正了解�Q�通过别�h的代码去做的。。�?br>

1#include <cstdio>
2#include <algorithm>
3
4const int SIZE = 50002 ;
5
6int length , N , M ;
7int distance[SIZE] ;
8
9bool Judge(const int& value)
10{
11    int num , i , pre ;
12    num = 0 ;
13    pre = 0 ;
14
15    for ( i = 0 ; i < N ; ++i )
16    {
17        if ( distance[i] - pre < value )
18            num++ ;
19        else
20            pre = distance[i] ;
21    }
22
23    if ( length - pre < value )
24        num++ ;
25
26    if ( num <= M )
27        return true ;
28    else
29        return false ;
30
31
32}
33
34int BiSearch()
35{
36    int ans = 0 ;
37    int mid , left = 0 , right = length ;
38
39    while ( left <= right )
40    {
41        mid = (left + right) >> 1 ;
42
43        if ( Judge(mid) )
44            left = mid + 1 ;
45        else
46            right = mid - 1 ;
47    }
48
49    ans = left - 1 ;
50
51    return ans ;
52}
53
54int main()
55{
56//    freopen("1.txt", "r", stdin) ;
57
58    int i ;
59    char strNum[12] ;
60
61    scanf("%d %d %d", &length, &N, &M) ;
62
63    getchar() ;
64    for ( i = 0 ; i < N ; ++i )
65    {
66    //    scanf("%d", &distance[i]) ;
67        gets(strNum) ;
68        distance[i] = atoi(strNum) ;
69    }
70
71    std::sort(distance, distance + N) ;
72
73    printf("%d\n", BiSearch()) ;
74
75    return 0 ;
76}

�?/a> 2009-03-07 16:03 发表评论

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POJ 2402--Palindrome Numbers

Pku 3258--River Hopscotch(二分枚�D�?