pat的一道题
提交之后有两个测试点过不去
我的代码如下
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.next();
double fu = 0,ou = 0,a_2 = 0;
int i = 0;
if(s.charAt(0)==45){
//如果是负数
fu = 0.5;
}
while(i<s.length()){
//判断2 的个数
if((s.charAt(i)-48)==2){
a_2++;
}
i++;
}
if(s.charAt(s.length()-1)%2==0){
//如果是偶数
ou = 1.0;
}
double fan2 = 0;
if(fu==0){//正数
fan2 = a_2/((double)(s.length()));
}else{//负数
fan2 = a_2/(s.length()-1)*(ou+fu);
}
//测试2的个数/总字符串长度的值
System.out.println(fan2);
System.out.printf("%.2f%%",fan2*2*100);
}
}
请问这是哪点没有通过呢?
The questioner did not understand the meaning of the question clearly. Negative numbers increase by 0.5 times and must be multiplied by 1.5. Even numbers increase by 1 and must be multiplied by 2.
fan2 = a_2/(s.length()-1)*(ou+fu ); // Are you multiplying the scale factor correctly?
Also positive numbers can also be even numbers. I don’t know how you passed the other cases
In addition, as @zjupure said, I feel that the questioner’s understanding of the question is biased, and I also understood it according to @zjupure.
But the strange thing is that the questioner’s code also passed 66.67% of the tests...
In addition, some tests may not run because the boundary conditions and input parameter verification are not judged. For example, the input is not a number, or a value like +0, -0. The subject can also add these ( Although I think it’s fucked up to write these codes as algorithm questions, but in order to pass the test...).
Additional point: If the first digit is a negative sign, then you need to start from the second digit to calculate whether it is 2, and the number of digits in the number