roof(开N次方以及上下取整)

开N次方

pow()函数
头文件：#include
形式：pow(x,y)
格式：

1. 表示对x开y次方
2. x，y，函数返回值都为double型
3. y不可以为小数
4. x,y不可同为0

上下取整函数

向上取整
形式：floor(x)
含义：floor(x)返回的是小于或等于x的最大整数,floor(x)是向负无穷大舍入
返回形式：double型(c语言)
eg：floor(-2.5)=-3

向下取整
形式：ceil(y)
含义：ceil(x)返回的是大于x的最小整数，ceil()是向正无穷大舍入
返回形式：double型(c语言)
eg：ceil(-2.5)=-2

应用：

Root of the Problem

Description
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.

Example Input:
4 3
5 3
27 3
750 5
1000 5
2000 5
3000 5
1000000 5
0 0
Example Output:
1
2
3
4
4
4
5
16
AC代码(c语言)如下

#include
#include
#include
#include
int main()
{int N,B;double Z;int x,y,A;double a,b;while(scanf("%d%d",&B,&N)&&B&&N){//pow(x,y)表示x的y次方//先对B开N次方Z=pow(B*1.0,1.0/N);//向上取整x=floor(Z);//向下取整y=ceil(Z);//上下取整的值更接近B的就是A，所以直接比较结果就行a=pow(x,N);//对向上取整的结果开N次方b=pow(y,N);//对向下取整的结果开N次方//上差值(B大)：B-a//下差值(B小)：b-Bif(B-a>b-B)A=y;elseA=x;printf("%lld\n",A);}return 0;
}

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