Power of Cryptography

//只用一行核心代码就可以过的天坑题目............= =

题目：

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<10 ¹⁰¹ and there exists an integer k, 1<=k<=10 ⁹ such 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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

 

哎~不多说了，代码如下：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    double n,p;
    while(cin>>n>>p)
    {
        cout<<pow(p,1/n)<<endl;
    }
    return 0;
}

 正解：二分+高精

代码：

#include <stdio.h>
#include <string.h>

// 交换字符串函数
void swap_str(char str[]) {
    int len = strlen(str);
    for (int i=0; i<len/2; i++) {
        int tmp = str[i];
        str[i] = str[len-i-1];
        str[len-i-1] = tmp;
    }
}

// 大数与整型相乘函数（大数以字符串形式给出）
void my_mul(char str[], int x) {
    int len = strlen(str);
    int cp = 0, i, tmp;
    swap_str(str);
    for (i=0; i<len; i++) {
        tmp = (str[i]-'0')*x + cp;
        str[i] = (tmp%10) + '0';
        cp = tmp / 10;
    }
    while (cp) {
        str[i++] = (cp%10) + '0';
        cp /= 10;
    }
    while ('0'==str[i-1] && i>1)
        i--;
    str[i] = '\0';
    swap_str(str);
}
// 比较两个大数的大小（大数前没有0）
int my_numCmp(char str1[], char str2[]) {
    int len1, len2;
    len1 = strlen(str1);
    len2 = strlen(str2);
    if (len1 > len2)
        return 1;
    if (len1 < len2)
        return -1;
    return strcmp(str1, str2);
}

// 字符串存储开方结果
void my_pow(char str[], int k, int n) {
    str[0] = '1', str[1] = '\0';
    while (n--) {
        my_mul(str, k);
    }
}

// 二分查找正确答案
int my_binary_search(int n, char str[]) {
    int high = 1e9, low = 0;
    int mid;
    char tot[2005];

    while (low < high) {
        mid = low + (high-low)/2;
        my_pow(tot, mid, n);
        int tmp = my_numCmp(tot, str);
        if (0 == tmp)
            return mid;
        if (tmp < 0)
            low = mid + 1;
        else
            high = mid;
    }
    return mid;
}

int main() {
    char str[105];
    int n;
    while (scanf("%d%s", &n, str) != EOF) {
        printf("%d\n", my_binary_search(n, str));
    }
    return 0;
}

代码来源：http://blog.csdn.net/zcube/article/details/8545523