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电脑系统崩溃怎么重新安装系统(电脑系统崩溃怎么重装win)
2025年1月三星i809(nokia 610)本文为大家介绍电脑系统崩溃Zěn么重新安装系统(电脑系统崩溃怎么重装win)，下Miàn和小编一起看看详细Nèi容吧。
电脑系统崩溃LiǎoZěn么办？
我用电脑N多年，最烦的Shì开机系统崩溃——，Kàn到一连串不知名的蓝屏，于是电脑专业英语词Huì闪烁甚至全黑屏————头好大！
有些事情还是需要自己解决。这会教你一招BànZhāo，可以作为临时救援。
nokia 610java代码大全（Java源代码
package?.zhidao;import?java.util.ArrayList;import?java.util.List;import?java.util.Scanner;import?static?java.lang.System.out;import?static?java.lang.System.in;/**?*?Created?by?tracy?on?//.?*/public?class?LR?{????static?int?numbers;????static?List《Domain》?domains?=?new?ArrayList《》();????public?static?void?main(String...asd){????????int?m,n;????????out.print(“请输入数字个数N：“);????????n?=?input(,Integer.MAX_VALUE);????????out.print(“请输入计算行数M：“);????????m?=?input(,);????????out.println(“请输Rù“+n+“个数字,每个数字介于跟之间，用空格隔开:“);????????numbers?=?inputNumbers(n);????????for(int?i?=?;?i《m;?i++){????????????domains.add(inputDomain(n,i+));????????}????????for(Domain?domain:domains){????????????domain.output();????????}????}????/**?????*?输入一个介于min跟max之间De整数?????*?param?min?????*?param?max?????*?return?????*/????private?static?int?input(int?min,int?max){????????Scanner?scanner?=?new?Scanner(in);????????try?{????????????int?i?=??scanner.nextInt();????????????if(i《min?||?i》max){????????????????out.print(“输入数字大小不符合要求，请重新输入”“);????????????????return?input(min,max);????????????}????????????return?i;????????}catch?(Exception?e){????????????out.print(“您输入的不是数字，请重新输入““);????????????return?input(min,max);????????}????}????/**?????*?输入N个介Yú到之间的整数?????*?param?n?????*?return?????*/????private?static?int?inputNumbers(int?n){????????Scanner?scanner?=?new?Scanner(in);????????String?numStr?=?scanner.nextLine();????????String?nums?=?numStr.trim().split(“\s+“);????????if(nums.length?!=?n){????????????out.println(“输入数Zì个数不对，请重新输入：“);????????????return?inputNumbers(n);????????}????????int?numbers?=?new?int[n];????????for(int?i=;i《n;i++){????????????try?{????????????????int?number?=?Integer.valueOf(nums[i]);????????????????if(number《?||?number?》){????????????????????out.print(“输入数字大小不符合Yào求，请重新输入”“);????????????????????numbers[i]?=?input(,);????????????????}else{????????????????????numbers[i]?=?number;????????????????}????????????}catch?(NumberFormatException?e){????????????????out.print(String.format(“您输入的%s不是数字，请重新输入：“,nums[i]));????????????????numbers[i]?=?input(,);????????????}????????}????????return?numbers;????}????private?static?Domai
n?inputDomain(int?n,int2025年1月三星i809(nokia 610)?index){????????out.println(“请输入第“+index+“行的个数字L,R(《=L《=R《=%d)用空格隔开：“);????????Scanner?scanner?=?new?Scanner(in);????????String?numStr?=?scanner.nextLine();????????String?nums?=?numStr.trim().split(“\s+“);????????if(nums.length?!=?){????????????out.print(String.format(“输入数字个数不对，请重新输入!“,n));????????????return?inputDomain(n,index);????????}????????String?ls?=?nums;????????String?rs?=?nums;????????try{????????????int?l?=?Integer.valueOf(ls);????????????int?r?=?Integer.valueOf(rs);????????????if(l《?||?r《l?||?r》n){????????????????out.print(String.format(“输入数字大小不符合，请重新输入!“,n));????????????????return?inputDomain(n,index);????????????}????????????return?new?Domain(l,r,index);????????}catch?(NumberFormatException?e){????????????out.println(String.format(“输入数字格式错误，请重新输入个数字L,R(《=L《=R《=%d)用空格隔开：“,n));????????????retur
n?inputDomain(n,index);????????}????}????private?static?class?Domain{????????int?l;????????int?r;????????int?index;????????Domain(int?l,?int?r,?int?index)?{????????????this.l?=?l;????????????this.r?=?r;????????????this.index?=?index;????????}????????void?output(){????????????int?result?=?;????????????for(int?i?=?l-;?i《r;?i++){????????????????result+=?numbers[i];????????????}????????????out.println(String.format(“第%d行：第%d到第%d个数之和Wèi：%d“,index,l,r,result));????????}????}}
publilassStringUtils{ publicstaticvoidmain(Stringargs){ System.out.println(leftPad(““,,’#’)); } publicstaticStringleftPad(Stringstr,intlength,charch){ charchs=newchar[length];//定义一个长度的char数组 Arrays.fill(chs,ch);//将指定的ch（“#”字符分配到char数组的每个序列Zhōng charsrc=str.toCharArray();//将str中的每个元素分配到一个新的char数组中 System.arraycopy(src,,chs,length-src.length,src.length);//将源数组src从第个元素开始数两个长度，拷贝到目标数组chs中，起始的位置Wèilength-src.length() returnnewString(chs);//将新的chs数组返回 }}
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