zoj 2818 Root of the Problem
2017-03-10 11:53
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Root of the ProblemTime Limit: 2 Seconds Memory Limit: 65536 KB
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.
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.
#include <iostream> #include <cmath> #include <cstdlib> using namespace std; int main(){ double b, n; while(cin >> b >> n){ if(b == 0 && n == 0) break; //pow:double pow(double, double) int a = (int)pow(b, 1/ n);//直接取整,不是四舍五入 ,pow(a, n) < b if(b - pow(a, n) < pow(a + 1, n) - b) cout << a << endl; else cout << a + 1 << endl; } return 0; }
Example Input: | Example Output: |
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 |
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