a的n次幂 —— POJ 3100 Root of the Problem

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Root of the Problem

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11962 Accepted: 6409

Description

Given positive integers B and N, find an integer A such that A^N is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that A^N may be less than, equal to, or greater than B.

Input

The input consists of one or more pairs of values for B and N. Each pair appears on a single line, delimited by a single space. A line specifying the value zero for both B and N marks the end of the input. The value of B will be in the range 1 to 1,000,000 (inclusive), and the value of N will be in the range 1 to 9 (inclusive).

Output

For each pair B and N in the input, output A as defined above on a line by itself.

Sample Input

4 35 327 3750 51000 52000 53000 51000000 50 0

Sample Output

123444516

题意：

给出B, N （B在[1, 1000000]内， N在[1, 9]内）；求一个整数A， 使A^N最接近B。

思路：

从小到大枚举A，或者对[1, 1000000]进行二分，数据比较小。

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#define Min 1#define Max 1000000int b, n;double My_pow(int a, int n){if(1 == n) return (double)a;if(n & 1){double ans;ans = My_pow(a, (n - 1) / 2);return ans * ans * a;}else{double ans;ans = My_pow(a, n / 2);return ans * ans;}}void Bsearch(int l, int r, int *x, int *y){while(l <= r){int mid = ((l + r) >> 1);double s = My_pow(mid, n);if(s == (double)b){*x = *y = mid;break;}else if(s > (double)b){*y = mid;r = mid - 1;}else{*x = mid;l = mid + 1;}}}int main(){#if 0freopen("in.txt","r",stdin);#endifwhile(scanf("%d%d", &b, &n), b && n){int x = 1, y = 1;Bsearch(Min, Max, &x, &y);if(abs(My_pow(x, n) - b) < abs(My_pow(y, n) - b))printf("%d\n", x);elseprintf("%d\n", y);}return 0;}