zoukankan      html  css  js  c++  java
  • poj2109 Power of Cryptography【坑~泪目】【水过】【刷题计划】

    Power of Cryptography
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 26249   Accepted: 13121

    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

    Source

    题意:输入n,p找到满足k^n = p的k。

    数据范围给的这么大,吓死我了,以前没有写过这种类型的题,一点思路都没有,没忍住找了一下题解,发现double都能过哎。

    #include<stdio.h>
    #include<math.h>
    
    int main()
    {
        double n,p;
        while(scanf("%lf%lf",&n,&p)!=EOF)
        {
            printf("%.0lf
    ",pow(p,1/n));
        }
        return 0;
    }
  • 相关阅读:
    linkedLoop
    loopqueue
    expect 切换用户
    二叉树的实现
    栈的链表实现, 底层使用链表
    栈的数组实现
    RSA加密算法
    输入一个链表,反转链表后,输出链表的所有元素
    输入一个链表,输出该链表中倒数第k个结点
    ansible中include_tasks和import_tasks
  • 原文地址:https://www.cnblogs.com/hellocheng/p/7843851.html
Copyright © 2011-2022 走看看