Power
of Cryptography
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 18018 | Accepted: 9090 |
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
【题目来源】
México
and Central America 2004
http://poj.org/problem?id=2109
【题目分析】
这道题目里面 p, n 没有超过 double 范围，所以可以直接算。
一开始看了分类说是贪心，还想了半天
各种数据类型范围：
unsigned int 0～4294967295
int 2147483648～2147483647
unsigned long 0～4294967295
long 2147483648～2147483647
long long的最大值：9223372036854775807
long long的最小值：-9223372036854775808
unsigned long
long的最大值：1844674407370955161
__int64的最大值：9223372036854775807
__int64的最小值：-9223372036854775808
unsigned
__int64的最大值：18446744073709551615
#include<stdio.h>
#include<math.h>
int main()
{
double m,n;
while(scanf("%lf%lf",&m,&n)!=EOF)
{
printf("%.0lf\n",pow(n,1.0/m));
}
return 0;
}
数论 --- 简单计算