Power of Cryptography
2013-12-29 20:01
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这个题好坑。。本来以为要很麻烦很麻烦的,然后百度了一下求根函数。。结果。。还真有。。于是。。结果如下。。
Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of 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 of only theoretical interest.
This problem involves the efficient computation of integer roots of numbers.
Given an integer
and an integer
you
are to write a program that determines
, the positive
root
of p. In this problem, given such integers n and p, p will always be of the form
for an integer k (this
integer is what your program must find).
The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs
,
and
there exists an integer k,
such that
.
For each integer pair n and p the value
should be printed, i.e., the number k such that
.
Background
Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematicsonce considered to be of only theoretical interest.
This problem involves the efficient computation of integer roots of numbers.
The Problem
Given an integer and an integer
you
are to write a program that determines
, the positive
root
of p. In this problem, given such integers n and p, p will always be of the form
for an integer k (this
integer is what your program must find).
The 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 ,
and
there exists an integer k,
such that
.
The Output
For each integer pair n and p the value should be printed, i.e., the number k such that
.
Sample Input
2 16 3 27 7 4357186184021382204544
Sample Output
4 3 1234
1 2 3 4 5 6 7 8 9 10 11 12
#include <stdio.h> #include <math.h> int main() { int n; double p; while(scanf("%d %lf", &n, &p)!=EOF) { printf("%d\n", (int)(pow(p,1.0/n)+0.5)); } return 0; }
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