poj 2109 Power of Cryptography

题目链接：

　　http://poj.org/problem?id=2109

题目描述：

　　给n，p。肯定有k(1<=k<=109)，满足kn=p，1<=n<=200,1<=p<=10101,问k是几？

解题思路：

　　由于p太大，无法对p进行开方，只有枚举k，又因为k太大又知道k的范围，所以很自然就想到了二分枚举，大树乘法加上二分枚举就可以解决。^_^……（奉上代码）！！

代码：　

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #define INF 1000000000
 6 #define N 110
 7 
 8 int a[N], b[N], n, len;
 9 char str[N];
10 int judge (int x);
11 int main ()
12 {
13     int i, low, high, mid;
14     while (scanf ("%d %s", &n, str) != EOF)
15     {
16         memset (a, 0, sizeof(a));
17         len = strlen (str);
18         for (i=0; i<len; i++)
19             a[i] = str[i] - '0';
20 
21         low = 1;
22         high = INF;
23 
24         while (low <= high)
25         {
26             mid = (low + high) / 2;
27             int m = judge (mid);
28             if (m < 0)
29                 low = mid + 1;
30             else if (m > 0)
31                 high = mid - 1;
32             else
33                 break;
34         }
35         printf ("%d\n", mid);
36     }
37     return 0;
38 }
39 
40 int judge (int x)
41 {
42     int m , i, j;
43     m = n * log10(x) + 1;
44     if (m > len)
45         return 1;
46     else if (m < len)
47         return -1;
48     else
49     {
50         memset (b, 0, sizeof(b));
51         b[0] = 1;
52         for (i=0; i<n; i++)
53         {
54             int yu = 0;
55             for (j=0; j<110; j++)
56             {
57                 int s = x * b[j] + yu;
58                 b[j] = s % 10;
59                 yu = s / 10;
60             }
61         }
62         for (i=0; i<len; i++)
63             if (b[len-1-i] > a[i])
64                 return 1;
65             else if (b[len-1-i] < a[i])
66                 return -1;
67         return 0;
68     }
69 }
View Code

 

posted @ 2014-12-16 20:28  罗茜  阅读(134)  评论(0)  编辑  收藏  举报