poj 2109 Power of Cryptography
Power of Cryptography
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 17912 | Accepted: 9034 |
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).
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
Source
正常算法是二分+高精度。
可是由下表:
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
可以一步算出结果,如果多步会丢失精度:
1 //180K 0MS C++ 195B 2014-05-08 13:47:15 2 #include<stdio.h> 3 #include<math.h> 4 int main(void) 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 }
写了二分法的不过应该是处理不够好,没过就不贴了。