How many levels of fantasy realm can mine

1.

There are great differences:

1, Rog maximum player's country series (that is, the loser's eye that netizens ridicule) positioning top DIY customized overclocking

2, tuf special forces series positioning top material 5-year warranty

3, Rog Strix Raptor series positioning middle end user

4, prime Master Series mainstream entry-level user

5, b250m-pixiu belongs to e-commerce special edition, Not in the retail market, B250 should have some functions, but the motherboard is very economical

6, b250m-plus is a retail model, and its materials are much better than Pixiu

The material of b250m-pixiu is close to that of b250m-j, and the actual price is estimated to be around 350

2. 1. You can install a computer manager on your computer, then open the toolbox and find the blackmail virus killing tool
3. After killing the virus, you can recover naturally

3. Because the inter-bank transfer must go through the system of the people's Bank of China (all inter-bank transfers should go through the system of the people's Bank of China), the arrival time of inter-bank transfer is generally 1 to 3 working days (legal holidays and weekends are postponed), please pay attention to check; If you haven't reached the 3 day of work, the funds have not been returned to the remittance account. Please login to the online banking service to select "customer service" - "mail service" to provide your transfer time, transfer amount, remittance account, name of the household name, ID number, and online banking number. The name and contact number of the Bank of deposit can be further verified by the staff
this is a similar question I once answered by a friend. I hope it can help you.

4. The composition and dosage of the vaccine are the same for everyone. There is no problem with this operation·

5.

You need to add rules to the hosts file

1. to run Notepad as an administrator

2. open C: system32, drivers, etc, hosts

3. < P > add rules, 127.0.0.1 the forbidden web address

4. will take effect after being saved finally




hosts is a system file without extension, You can use Notepad and other tools to open it. Its function is to establish a "database" between some commonly used web address domain names and their corresponding IP addresses. When the user enters a web address to log in in the browser, the system will first automatically find the corresponding IP address from the hosts file. Once it is found, the system will immediately open the corresponding web page, Then the system will submit the web address to the DNS domain name resolution server for IP address resolution

6. There is an icon in the tracking taskbar. Right click

7. Pslfngnxwjotzgzjdmbto school to send good to oneskqidkzxeqgnyyltrenhzpk; Play; Stability; Win-&# 1379;<
319 ╅ 928 ╅ 4241
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the file system of hard disk, mobile hard disk or U disk has become the solution of raw format
1. System permission problem 1 (the original system is NTFS format)
right click on the raw format drive letter, select properties, and then select security, delete useless users, add your own user name, and then change permissions. If there is no security option, you can remove the check before "use simple file sharing" in the folder option (open my computer, select Tools menu, and then select folder option)
to right-click on the NTFS disk and display the security option in the Properties menu
2. System permission problem 2 (the original system is in NTFS format)
go to the control panel to find the "management tool", open it, open the "local security policy", open the "security options", and find the "network access: local account sharing and security mode" in the right window, then change the "guest only" to "classic", Exit and go back to my computer, right-click on disk d to find the "security" tag, delete the garbled code, and then add the user of this computer
rxdnkxvkskpyojtalojzrno resist the space event with witcayrqgrygeuzbipny,

8. A total of 24 characters, you only have 20, can't figure out

9. #include <stdio.h>
#include <malloc.h>
#include<string.h>
struct int_char
{
short biao;/*标志是否运算数和运算符，0运算数1运算符*/
long su;/*存储运算数*/
char yunsuanhu;/*存储运算符*/
struct int_char *next;
};
int main(void)
{
int fujianche(char *pstring);
struct int_char *jieshuanshi(char *string);
void yunsuan(struct int_char *suansi);
char sr[100]={0};
struct int_char *p;
int jcsr,off=1;
printf("***The positive integer four mixed operation***\n");
do
{
printf("Please enter the formula:");
gets(sr);
jcsr=fujianche(sr);/*检测输入是否合法*/
if(jcsr==0)break;
if(jcsr==2)
{
printf("除数不能为0\n");
continue;
}
if(jcsr==1)
{
printf("非法的算式\n");
continue;
}
}
while(1);
p=jieshuanshi(sr);/*将字符串转换成链表*/
if(p==0)
{
printf("未输入算式");/*如果链表为空就表示在没有输入任何字符的情况下按下了回车*/
return(0);
}
yunsuan(p); /*运算并输出解题步骤*/
return(0);
}
int fujianche(char *pstring)/*字符串输入合法性检测函数*/
{
char suozuanfan(char fu); /*用于确定当前字符是否数字*/
int i,kuohao1=0,kuohao2=0;/*kuohao1,huohao2,用于检测括号的对称性*/
for(i=0;pstring[i]!='\0';i++)
{
/*算式的开头只能出现数字，'(' 结尾只能出现数字，')'*/
if(i==0&&pstring[i]!=suozuanfan(pstring[i])&&pstring[i]!='(')
{return(1);break;}
if(pstring[i+1]=='\0'&&pstring[i]!=suozuanfan(pstring[i])&&pstring[i]!=')')
if(i==0&&(pstring[i+1]==')'||pstring[i+2]==')'||pstring[i+3]==')'))
{return(1);break;}
if(pstring[i+1]=='\0'&&(pstring[i-1]=='('||pstring[i-2]=='('||pstring[i-3]=='('))
{return(1);break;}
/*检测运算符出现位置的合法性*/
if(i!=0&&pstring[i+1]!='\0')/*判断字符是否在字符串的第一位和倒数第二位*/
{
if(pstring[i]=='+'||pstring[i]=='-'||pstring[i]=='*'||pstring[i]=='/')
{
if(pstring[i-1]!=')'&&pstring[i-1]!=suozuanfan(pstring[i-1]))
{return(1);break;}
if(pstring[i+1]!='('&&pstring[i+1]!=suozuanfan(pstring[i+1]))
{return(1);break;}
}
if(pstring[i]=='(')
{
if(pstring[i-1]!='('&&pstring[i-1]!='+'&&pstring[i-1]!='-'&&pstring[i-1]!='*'&&pstring[i-1]!='/')
{return(1);break;}
if(pstring[i+1]!='('&&pstring[i+1]!=suozuanfan(pstring[i+1]))

{return(1);break;}
}
if(pstring[i]==')'&&pstring[i+1]!='\0')
{
if(pstring[i-1]!=')'&&pstring[i-1]!=suozuanfan(pstring[i-1]))
{return(1);break;}
if(pstring[i+1]!=')'&&pstring[i+1]!='+'&&pstring[i+1]!='-'&&pstring[i+1]!='*'&&pstring[i+1]!='/')
{return(1);break;}
}
}

/*算式中不应出现除数字，加减乘除，()以外的其他字符*/
switch(pstring[i])
{
case '1' :break;
case '2' :break;
case '3' :break;
case '4' :break;
case '5' :break;
case '6' :break;
case '7' :break;
case '8' :break;
case '9' :break;
case '0' :break;
case '+' :break;
case '-' :break;
case '*' :break;
case '/' :{if(pstring[i+1]=='0')return(2);break;}
case '(' :kuohao1++;break;/*检测括号的对称性假设：如果'('和')'数量相同则视为括号对称这里对'('和')'进行统计在循环结束时对其对比.......*/
case ')' :kuohao2++;break;
default :return(-1);
}
}
if(kuohao1!=kuohao2)/*对比括号是否对称*/
return(1);
return(0);
}
char suozuanfan(char fu)/*检查fu是否数字*/
{
switch(fu)
{
case'1':{return('1');break;}
case'2':{return('2');break;}
case'3':{return('3');break;}
case'4':{return('4');break;}

case'5':{return('5');break;}
case'6':{return('6');break;}
case'7':{return('7');break;}
case'8':{return('8');break;}
case'9':{return('9');break;}
case'0':{return('0');break;}
default :return(-1);
}
}
struct int_char *jieshuanshi(char *string)/*这个函数将字符串转换成一个含算式的链表它用到了suozuanfan，suozuanfan1，yunsuanfutiqu三个函数，和struct int_char结构体*/
{
char suozuanfan(char fu);
int suozuanfan1(char fu);
char yunsuanfutiqu(char yunsuanfu);
long yunsuansu=0;
int i,j,x,y,jisuqi,weisuqi;/*jisuqi用于转换后的整数在yunsuansu数组中存储的位置，weisuqi用于数字转换位数的标志*/
int n=0;/*n 位链表长度，为0时表示建立第一节点，非0时表示建立的不是第一节点*/
struct int_char *head=0;
struct int_char *p1,*p2;
if(strlen(string)==0)return(0);
for(i=0,jisuqi=0;string[i]!='\0';i++,jisuqi++)/*遍历字符串*/
{
if(string[i]==suozuanfan(string[i]))/*检查串中的最高位数字*/
{
x=i;/*定位最高数字在串中的位置*/
for(j=i;string[j]==suozuanfan(string[j]);j++);/*检查个位数字的位置*/
y=j-1;/*定位个位数字在串中的位置*/
i=j-1;/*将遍历字符串的计数器定位在最后被检查的字符的位置*/
for(weisuqi=1,yunsuansu=0;y>=x;y--,weisuqi*=10)/*数字转换*/
yunsuansu+=suozuanfan1(string[y])*weisuqi;
if(n==0)/*建立表头及将运算数接入链表*/
{
p1=p2=head=(struct int_char *)malloc(sizeof(struct int_char));
n++;
p1->biao=0;
p1->su=yunsuansu;
p1->yunsuanhu=0;
p1->next=0;
}
else/*将运算数接入链表*/
{
p1=(struct int_char *)malloc(sizeof(struct int_char));
n++;
p2->next=p1;
p2=p2->next;
p1->biao=0;
p1->su=yunsuansu;
p1->yunsuanhu=0;

p1->next=0;
}
}
if(string[i]==yunsuanfutiqu(string[i]))/*检测是否运算符*/
{
if(n==0)/*建立表头及将运算符接入链表*/
{
p1=p2=head=(struct int_char *)malloc(sizeof(struct int_char));
n++;
p1->biao=1;
p1->su=0;
p1->yunsuanhu=string[i];
p1->next=0;
}
else/*将运算符接入链表*/
{
p1=(struct int_char *)malloc(sizeof(struct int_char));
n++;
p2->next=p1;
p2=p2->next;
p1->biao=1;
p1->su=0;
p1->yunsuanhu=string[i];
p1->next=0;
}
}

}
return (head);
}

char yunsuanfutiqu(char yunsuanfu)/*检测yunsuanfu是否运算符*/
{
switch(yunsuanfu)
{
case '(':{return('(');break;}
case ')':{return(')');break;}
case '+':{return('+');break;}
case '-':{return('-');break;}
case '*':{return('*');break;}
case '/':{return('/');break;}
default:return(0);
}
}
void yunsuan(struct int_char *suansi)/*运算函数有两个子函数jjcc用于进行右括号的运算但括号没只有一个运算符时消除括号*/
{ /*wkjjcc用于进行无括号的运算，jjcc,wkjjcc只进行单步运算*/
void jjcc(struct int_char *up,struct int_char *np);
void wkjjcc(struct int_char *up,struct int_char *np);
struct int_char *p1,*p2;/*p1,当前节点p2,上一个节点*/
struct int_char *up,*nP;/*up(的位置，np)的位置*/
short off;
p1=suansi;
if(p1->next==0)/*如果连表中只有一个节点*/
{
printf("=%d\n",p1->su);/*就输出这个接点并结束*/
return;
}
for(;p1!=0;p1=p1->next)
if(p1->biao&&p1->yunsuanhu==')')/*找到')'的位置*/
{
nP=p1;
for(p2=suansi;p2!=nP;p2=p2->next)/*找到'('的位置*/
if(p2->biao&&p2->yunsuanhu=='(')
up=p2;

jjcc(up,nP);
printf("=");
for(p2=suansi;p2!=0;p2=p2->next)/*输出步骤*/
if(p2->biao)
printf("%c",p2->yunsuanhu);
else
printf("%d",p2->su);
printf("\n");
p1=suansi;
}

for(off=1;off;)
{
for(p1=suansi;p1->next;p1=p1->next);
up=suansi;
nP=p1;
if(up==nP)
off=0;
else
{
wkjjcc(up,nP);
printf("=");
for(p2=suansi;p2!=0;p2=p2->next)/*输出步骤*/
if(

10. So the function is not written right.