题目地址：http://poj.org/problem?id=2109
1 /*
2 题意：k ^ n = p，求k
3 1. double + pow：因为double装得下p，k = pow (p, 1 / n);
4 基础知识: 类型 长度 (bit) 有效数字 绝对值范围
5 float 32 6~7 10^(-37) ~ 10^38
6 double 64 15~16 10^(-307) ~ 10^308
7 long double 128 18~19 10^(-4931) ~ 10 ^ 4932
8 2. 二分查找：和1类似
9 3. 取对数：n*ln(k)=ln(p) ln(k)=ln(p)/n k=exp(ln(p)/n)
10 */
11 #include <cstdio>
12 #include <iostream>
13 #include <algorithm>
14 #include <cstring>
15 #include <cmath>
16 #include <string>
17 #include <map>
18 #include <queue>
19 #include <vector>
20 using namespace std;
21
22 const int MAXN = 1e6 + 10;
23 const int INF = 0x3f3f3f3f;
24
25 int main(void) //POJ 2109 Power of Cryptography
26 {
27 //freopen ("D.in", "r", stdin);
28 double n, p, k;
29
30 while (~scanf ("%lf%lf", &n, &p))
31 {
32 printf ("%.0f
", pow (p, 1 / n));
33 }
34
35 return 0;
36 }
37
38 /*
39 #include <cstdio>
40 #include <cmath>
41 #include <algorithm>
42 #include <iostream>
43 using namespace std;
44
45 void BinarySearch(int l, int r, double n, double p)
46 {
47 int mid;
48
49 while (l <= r)
50 {
51 mid = l + (r - l) / 2;
52 double tmp = pow (mid, n);
53 if (tmp == p)
54 {
55 printf ("%d
", mid); return ;
56 }
57 else if (tmp < p) l = mid + 1;
58 else r = mid - 1;
59 }
60 }
61
62 int main(void)
63 {
64 //freopen ("D.in", "r", stdin);
65
66 double n, p;
67
68 while (~scanf ("%lf%lf", &n, &p))
69 {
70 BinarySearch (1, 1e9, n, p);
71 }
72
73 return 0;
74 }
75 */
76
77 /*
78 #include <cstdio>
79 #include <cmath>
80 #include <algorithm>
81 #include <iostream>
82 using namespace std;
83
84 int main(void) //POJ 2109 Power of Cryptography
85 {
86 //freopen ("D.in", "r", stdin);
87
88 double n, p;
89
90 while (~scanf ("%lf%lf", &n, &p))
91 {
92 printf ("%.0f
", exp (log (p) / n));
93 }
94
95 return 0;
96 }
97 */