Problem Description:
1. The question involves knowledge points.
Round-up and round-down of floats.
Math.ceil(key)Round up 5.6 –> 6
Math.floor(key)round down 5.6 –> 5
2. Solve it yourself.
- Get one decimal place
key*10%10.
package com.chaoxiong.niuke.huawei;
import java.util.Scanner;
/**
* Create by tianchaoxiong on 18-4-8.
*/
public class HuaWei_7 {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
float in = scanner.nextFloat();
int key = (int) (in*10)%10;
if(key<5)
System.out.println((int)Math.floor(in));
else
System.out.println((int)Math.ceil(in));
}
}
}3. Quality answers.
null
4. Summary of this question.
Math.ceil(key)Round up 5.6 –> 6
Math.floor(key)Round down 5.6 –> 5
public static double abs(double a), --------------abs方法求绝对值
public static native double acos(double a) -------------acos求反余弦函数
public static native double asin(double a) -------------asin求反正切函数
public static native double atan(double a) -------------atan求反正切函数
public static native double ceil(double a) -------------ceil返回值最小的大于a的整数
public static native double cos(double a) -------------cos求余弦函数
public static native double exp(double a) -------------exp求e的a次幂
public static native double floor(double a) -------------floor返回最大的小于a的数
public static native double log(double a) -------------log返回lna
public static native double pow(double a，double b)------pow求a的b次幂
public static native double sin(double a) -------------sin求正弦函数
public static native double sqrt(double a) -------------sqrt求a的开平方
public static native double tan(double a) -------------tan求正切函数
public static synchronized double random() -------------返回0~1的随机数