PAT(Advance Level)Practice1002

mac2022-06-30  22

 

题目分析:

输入2项 2.4*a^1+3.2*a^02项 1.5*a^2+0.5*a^1输出2.4*a^1+1.5*a^2+0.5*a^1+3.23项 =1.5*a^2+2.9*a+3.2

 

Java pat 1002在实践中遇到的问题:

1.Please be accurate to 1 decimal place.

翻译:请精确到小数点后一位

2.Notice that there must be NO extra space at the end of each line

翻译:每行的某位不能有额外的空格。

3.K is the number of nonzero terms in the polynomial,

翻译:K是多项式中非0项的数目。

4. Ni and a~Ni~ (i=1, 2, ..., K) are the exponents and coefficients,

翻译:Ni和a~Ni(i=1, 2,…,K)是指数和系数

5.java 怎么定义数组并初始化:

一.数组:数组的定义:一维数组的定义:type [ ] arr_name;type arr_name[ ];二维数组的定义:type [ ][ ] arr_name;type arr_name [ ][ ];多维数组的定义:int [ ] arr1;String [ ] arr2;float arr3[ ];String [ ][ ] arr4;

二.数组的初始化:静态初始化:(定义的同时指定数组元素的内容)int [ ] arr1={1,2,3,4,5};String [ ] arr2={"tom","rose","sunny"};String [ ][ ] arr3 ={{"tom","American"},{"jack","england"},{"张三","china"}};动态初始化:(定义的同时通过new关键字开辟指定大小的存储空间)int [ ] arr1=new int [2] ;arr1[0] =10;arr2[1]=20;arr3[2]=30;

6.Java数组内元素由大到小排序:

1 for(int p=0;p<(K2+(K1-num));p++) { 2 for(int q=p+1;q<(K2+(K1-num));q++) { 3 if(a[p]<a[q]) { 4 temp=a[p]; 5 a[p]=a[q]; 6 a[q]=temp; 7 } 8 } 9 } 排序

7Java中的Boolean类型怎么写:

 

转载于:https://www.cnblogs.com/Catherinezhilin/p/9316512.html

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