//只用一行核心代碼就可以過的天坑題目............= =

題目：

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~不多說了，代碼如下： #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 代碼： 1 #include <stdio.h> 2 #include <string.h> 3 4 // 交換字符串函數 5 void swap_str(char str[]) { 6 int len = strlen(str); 7 for (int i=0; i<len/2; i++) { 8 int tmp = str[i]; 9 str[i] = str[len-i-1]; 10 str[len-i-1] = tmp; 11 } 12 } 13 14 // 大數與整型相乘函數（大數以字符串形式給出） 15 void my_mul(char str[], int x) { 16 int len = strlen(str); 17 int cp = 0, i, tmp; 18 swap_str(str); 19 for (i=0; i<len; i++) { 20 tmp = (str[i]-'0')*x + cp; 21 str[i] = (tmp%10) + '0'; 22 cp = tmp / 10; 23 } 24 while (cp) { 25 str[i++] = (cp%10) + '0'; 26 cp /= 10; 27 } 28 while ('0'==str[i-1] && i>1) 29 i--; 30 str[i] = '\0'; 31 swap_str(str); 32 } 33 // 比較兩個大數的大小（大數前沒有0） 34 int my_numCmp(char str1[], char str2[]) { 35 int len1, len2; 36 len1 = strlen(str1); 37 len2 = strlen(str2); 38 if (len1 > len2) 39 return 1; 40 if (len1 < len2) 41 return -1; 42 return strcmp(str1, str2); 43 } 44 45 // 字符串存儲開方結果 46 void my_pow(char str[], int k, int n) { 47 str[0] = '1', str[1] = '\0'; 48 while (n--) { 49 my_mul(str, k); 50 } 51 } 52 53 // 二分查找正確答案 54 int my_binary_search(int n, char str[]) { 55 int high = 1e9, low = 0; 56 int mid; 57 char tot[2005]; 58 59 while (low < high) { 60 mid = low + (high-low)/2; 61 my_pow(tot, mid, n); 62 int tmp = my_numCmp(tot, str); 63 if (0 == tmp) 64 return mid; 65 if (tmp < 0) 66 low = mid + 1; 67 else 68 high = mid; 69 } 70 return mid; 71 } 72 73 int main() { 74 char str[105]; 75 int n; 76 while (scanf("%d%s", &n, str) != EOF) { 77 printf("%d\n", my_binary_search(n, str)); 78 } 79 return 0; 80 }

代碼來源：http://blog.csdn.net/zcube/article/details/8545523

轉載于:https://www.cnblogs.com/teilawll/p/3204786.html

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