贪心 POJ 2109 Power of Cryptography

题目地址：http://poj.org/problem?id=2109

 /*
     题意：k ^ n = p，求k
     1. double + pow：因为double装得下p，k = pow (p, 1 / n);
     基础知识: 类型    长度 (bit)    有效数字    绝对值范围
             float    32             6~7         10^(-37) ~ 10^38
             double    64            15~16        10^(-307) ~ 10^308
        long double    128              18~19         10^(-4931) ~ 10 ^ 4932
     2. 二分查找：和1类似
     3. 取对数：n*ln(k)=ln(p)    ln(k)=ln(p)/n    k=exp(ln(p)/n)
 */
 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <map>
 #include <queue>
 #include <vector>
 using namespace std;

 const int MAXN = 1e6 + ;
 const int INF = 0x3f3f3f3f;

 int main(void)        //POJ 2109 Power of Cryptography
 {
     //freopen ("D.in", "r", stdin);
     double n, p, k;

     while (~scanf ("%lf%lf", &n, &p))
     {
         printf ("%.0f\n", pow (p,  / n));
     }

     return ;
 }

 /*
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <iostream>
 using namespace std;

 void BinarySearch(int l, int r, double n, double p)
 {
     int mid;

     while (l <= r)
     {
         mid = l + (r - l) / 2;
         double tmp = pow (mid, n);
         if (tmp == p)
         {
             printf ("%d\n", mid);    return ;
         }
         else if (tmp < p)    l = mid + 1;
         else    r = mid - 1;
     }
 }

 int main(void)
 {
     //freopen ("D.in", "r", stdin);

     double n, p;

     while (~scanf ("%lf%lf", &n, &p))
     {
         BinarySearch (1, 1e9, n, p);
     }

     return 0;
 }
 */

 /*
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <iostream>
 using namespace std;

 int main(void)        //POJ 2109 Power of Cryptography
 {
     //freopen ("D.in", "r", stdin);

     double n, p;

     while (~scanf ("%lf%lf", &n, &p))
     {
         printf ("%.0f\n", exp (log (p) / n));
     }

     return 0;
 }
 */