1. Link:
Http://poj.org/problem? Id = 2109
Http://bailian.openjudge.cn/practice/2109/
2. content:
Power of cryptography
Time limit:1000 ms |
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Memory limit:30000 K |
Total submissions:18872 |
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Accepted:9520 |
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 shoshould be printed, I. e., the number k such that k N = P.
Sample Input
2 163 277 4357186184021382204544
Sample output
431234
Source
México and Central America 2004
3. Method:
Derivation process:
K ^ n = P
Logk P = N
Log (P)/log (K) = N (bottom-changing formula)
Log (K) = 1/N * log (P)
K = e ^ (1/n * log (p) = (E ^ log (p) ^ (1/n) = P ^ (1/n) = POW (P, 1/n)
Therefore, using double to read data will result in loss of some precision, but this question seems to be less accurate, so it is feasible
4. Code:
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 5 using namespace std; 6 7 int main() 8 { 9 //freopen("D://input.txt","r",stdin);10 11 double p,n;12 13 while(cin >> n >> p)14 {15 cout << pow(p,1.0/n) << endl;16 }17 18 return 0;19 }
5. Reference:
Poj 2109/openjudge 2109 power of cryptography