a+new+problem+has

@笪朗535:翻译 a new problem threatens to earn them -
禄享13444682700…… a new problem threatens to earn them 翻译: 一个新问题(==数据得不到保护)威胁着要击溃他们/它们.

@笪朗535:帮忙翻译一下:a new problem threatens to earn them -
禄享13444682700…… “just as...,a new problem threatens to earn them.” “当...一个新的问题开始威胁他们.”这是原文的翻译.

@笪朗535:There is a new problem;involved in the popularity of private cars that road conditions need -
禄享13444682700…… There is a new problem involved in the popularity of private cars that road conditions need to be improved. involved in the popularity of private cars为前面problem的定语.此句为过去分词做定语. that引导的是a new problem的同位语, road conditions 和后面的部分对problem进行解释说明.

@笪朗535:There is a new problem - -------in the popularity of private cars that road conditions need impro... -
禄享13444682700…… A 试题分析:考查固定搭配.动词involve sb in sth使某人参与某事;转换为be involved in…参与…,涉及…;本题中的形容词短语involved in…对名词a new problem进行解释.句意:有一个涉及到私家车流行的新的问题,就是道路情况需要改善.故A正确. 点评:动词involve的固定搭配为:involve sb in sth使某人参与某事;转换为be involved in…参与…,涉及…;在考查的时候,经常使用其中的involved in…说明主语的情况.

@笪朗535:4.There is a new problem involved in the popularity of private cars - --road conditions need - -
禄享13444682700…… 首先,地点状语修饰的是地名,而这里的主语是problem.其次,由于被修饰的是problem,从句的内容又是problem的具体解释,从句是同位语.by the way, need 后面加to, to be improved 是被动语态.

@笪朗535:C程序 A+B Problem(A+B问题) -
禄享13444682700…… #include <stdio.h> #include <stdlib.h> void main() { int a,b,sum; scanf("%d %d",&a,&b); if(a>=0&&a<=10&&b>=0&&b<=10){ sum=a+b; printf("%d\n",sum); } else printf("error"); }

@笪朗535:关于c语言A+B problem的问题 -
禄享13444682700…… //又是oj上的题目? #includeint main() { int a,b; while(scanf("%d %d",&a,&b)!=EOF) { //EOF代表文件结尾,测试时通过文件输入数据 //当输入到文件尾部时停止处理 //当在控制台输入时按Ctrl+Z代表文件末尾 printf("%d\n",a+b); } return 0; }

@笪朗535:用java编个A+B problem问题
禄享13444682700…… public static void main(String args[]) { BufferedReader br = new BufferedReader( new InputStreamReader(System.in)); this.m_msg = br.readLine(); System.out.println("用户输入:"+this.m_msg); int val_a =0; int val_b =0; int val_max=0; String[]...

@笪朗535:C程序 A+B Problem(A+B问题) -
禄享13444682700…… /*因为a、b没有初始值,所以一进入循环判断的时候就会提示错误......还有,循环的条件应该是 a<0 || a>10 的吧.......*/ #include "stdio.h"main(){ int a=-1,b=-1; while(a<0||a>10) { scanf("%d",&a); } while(b<0||b>10) { scanf("%d",&b); } printf("%d",a+b);}

@笪朗535:如何用Python解决OJ中的A+B problem. -
禄享13444682700…… 描述 Calculate a + b 输入 Two integer a,,b (0 ≤ a,b ≤ 10) 输出 Output a + b 样例输入 1 2 样例输出 3 在python3 中,以这样的格式输入两个数要把这两个数当作一个字符串来处理,找到空格字符,让后两个数相加即可.错误解题方法是

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