题目链接：
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2818
题目描述：
Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN 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.
4 3
5 3
27 3
750 5
1000 5
2000 5
3000 5
1000000 5
0 0
1
2
3
4
4
4
5
16
1 /*问题 输入正整数b和n，找到一个正整数a满足a^n最靠近b 2 解题思路 a^n=b,那么a等于b的算数n次方根，即b的1/n次方等于a, 比较a^n和(a+1)^n谁更接近b */ 3 #include<cstdio> 4 #include<cmath> 5 int main() 6 { 7 double b,n; 8 int a; 9 while(scanf("%lf%lf",&b,&n),b+n != 0) 10 { 11 a=(int)pow(b,1/n);//强制类型转换时直接截取整数部分 12 if(b-pow(a,n) < pow(a+1,n)-b) 13 printf("%d ",a); 14 else 15 printf("%d ",a+1); 16 } 17 return 0; 18 }