UVA（POJ 2109）中密码学力量如何体现？
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本文共计598个文字，预计阅读时间需要3分钟。
《密码学之力》简介：密码学领域当前研究包括（如：大素数和这些素数中的数幂计算）等。该领域的研究成果已广泛应用于数论领域。
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 n th. 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<10 101and there exists an integer k, 1<=k<=10 9such that k n= 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
2 163 277 4357186184021382204544
Sample Output
431234
Source
México and Central America 2004
#include<stdio.h>
#include<math.h>
#define N 1000000
int main()
{
double n,m;
double t;
int i;
while(scanf("%lf%lf",&m,&n)!=EOF)
{
for(i=1;i<N;i++)
{
t = pow(i,m);
if(t==n)
{
break;
}
}
printf("%d\n",i);
}
return 0;
}
2
#include<stdio.h>
#include<math.h>
int main()
{
double p,n;
while(scanf("%lf%lf",&n,&p)!=EOF)
{
printf("%.lf\n",pow(p,1/n));
}
}
Time Limit:1000MS
Memory Limit:30000K
Total Submissions:18102
Accepted:9135
本文共计598个文字，预计阅读时间需要3分钟。
《密码学之力》简介：密码学领域当前研究包括（如：大素数和这些素数中的数幂计算）等。该领域的研究成果已广泛应用于数论领域。
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 n th. 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<10 101and there exists an integer k, 1<=k<=10 9such that k n= 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
2 163 277 4357186184021382204544
Sample Output
431234
Source
México and Central America 2004
#include<stdio.h>
#include<math.h>
#define N 1000000
int main()
{
double n,m;
double t;
int i;
while(scanf("%lf%lf",&m,&n)!=EOF)
{
for(i=1;i<N;i++)
{
t = pow(i,m);
if(t==n)
{
break;
}
}
printf("%d\n",i);
}
return 0;
}
2
#include<stdio.h>
#include<math.h>
int main()
{
double p,n;
while(scanf("%lf%lf",&n,&p)!=EOF)
{
printf("%.lf\n",pow(p,1/n));
}
}
Time Limit:1000MS
Memory Limit:30000K
Total Submissions:18102
Accepted:9135