浙大ACM 2818题答案

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本文介绍了一个算法，该算法用于寻找整数A，使得A的N次方尽可能接近给定的整数B。文章提供了一段C语言代码实现，通过迭代逼近的方法找到最接近目标值B的A值，并考虑了A的N次方可能小于、等于或大于B的情况。

题目连接

ZOJ Problem Set - 2818

Root of the Problem

Time Limit: 2 Seconds Memory Limit: 65536 KB

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.

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

#include<stdio.h>
#include<math.h>

int t;
int pow(int a,int n){
	int i=1;
	t=a;
	for(i;i<n;i++){
		t *= a;
	}
	return t;
}

int main(){
	int B,N,A;
	int y;
	while(scanf("%d%d",&B,&N),B||N){         //自己摸索的，呵呵，当输入0 0 时结束
		for(A=1;(B-pow(A,N))>0;A++){
		}
		y=pow(A,N)-B;
		(B-pow(A-1,N)>=y)?(printf("%d\n",A)):(printf("%d\n",A-1));
	}
}