Power
of Cryptography
Description
Current
work in cryptography involves (among other things) large prime numbers and
computing powers of numbers among these primes. Work in this area has resulted
in the practical use of results from number theory and other branches of
mathematics once considered to be only of theoretical
interest.
This problem
involves the efficient computation of integer roots of
numbers.
Given an
integer n>=1 and an integer p>= 1 you have to write a program that
determines the n th positive root of p. In this problem, given such integers n
and p, p will always be of the form k to the nth. power, for an
integer k (this integer is what your program must find).
Input
The input
consists of a sequence of integer pairs n and p with each integer on a line by
itself. For all such pairs 1<=n<= 200,
1<=p<10101 and there exists an integer k,
1<=k<=109 such that kn = p.
Output
For each
integer pair n and p the value k should be printed, i.e., the number k such that
k n =p.
Sample
Input
2 16
3 27
7 4357186184021382204544
Sample
Output
4
3
1234
题目大意：输入两个数，n，p。求一个整数k，使得k满足kn=p。
大牛代码，膜拜！
float：2^23 =
8388608，一共七位，这意味着最多能有7位有效数字，但绝对能保证的为6位，也即float的精度为6~7位有效数字；
double：2^52 =
4503599627370496，一共16位，同理，double的精度为15~16位。
首先，题目中的数据强度并不弱，这一点确实如题目中所说：“For all
such pairs 1<=n<= 200,
1<=p<10101，所以，double型是不能精确地表示出所给数据，但是却能表示出一个近似值。
当向double型变量中存入
4357186184021382204544
然后再输出，得到的是
4357186184021382000000
后六位的值变为了0，这一点和int型变量是有很大区别的。也就是说当存入double型变量的值超出了它的精度表示范围时，将低位的数据截断。
在本题中，如果测试数据为：
7
4357186184021382204544
实际上所处理数据是：
7
4357186184021382000000
拿4357186184021382000000开7次方的结果自然就是1234。
为什么不是1233或者1235呢？
12337=4332529576639313702577
12347=4357186184021382204544
12357=4381962969567270546875
可以看出在double型所能表示的精度范围内，它们三个值已经不同了。
1 #include<cstdio>
2 #include<cmath>
3
4 int main()
5 {
6 int n;
7 double p;
8 while(scanf("%d%lf",&n,&p) != EOF)
9 {
10 printf("%.0lf\n",pow(p,1.0/n));
11 }
12 return 0;
13 }
（转）poj2109-Power of Cryptography