Question

acm运行时显示RUNTIME_ERROR [ACCESS_VIOLATION，怎么解决

#include<stdio.h>
int main()
{
int F[]={7,11};
int n,i;
while(scanf("%d",&n)){
if(n>=2) {
for(i=2;i<=n;i++){
F[i]=F[i-1]+F[i-2];
}
if(F[n]%3==0)
printf("yes\n");
else
printf("no\n");
}
else{
printf("no\n");
}
}
return 0;
}

Answer 1

runtime error （运行时错误）就是程序运行到一半，程序就崩溃了。
比如说：
①除以零
②数组越界：int a[3]; a[10000000]=10;
③指针越界：int * p; p=(int *)malloc(5 * sizeof(int)); *(p+1000000)=10;
④使用已经释放的空间：int * p; p=(int *)malloc(5 * sizeof(int));free(p); *p=10;
⑤数组开得太大，超出了栈的范围，造成栈溢出：int a[100000000];

Answer 2

①除以零

②数组越界：int a[3]; a[10000000]=10;

③指针越界：int * p; p=(int *)malloc(5 * sizeof(int)); *(p+1000000)=10;

④使用已经释放的空间：int * p; p=(int *)malloc(5 * sizeof(int));free(p); *p=10;

⑤数组开得太大，超出了栈的范围，造成栈溢出：int a[100000000];

打开/etc/mysql/mysql.conf.d/mysqld.cnf 文件,命令如下

$ sudo vi /etc/mysql/mysql.conf.d/mysqld.cnf

2.找到[mysqld]段，并加入一行“skip-grant-tables”,如下图，

3.重启mysql服务,用空密码进入mysql管理命令行，切换到mysql库,操作命令如下，

$ mysql

Welcome to the MySQL monitor.  Commands end with ; or \g.

mysql> use mysql

Reading table information for completion of table and column names

You can turn off this feature to get a quicker startup with -A

Database changed

mysql> update mysql.user set authentication_string=password('newpass') where user='root' and Host ='localhost';

Query OK, 1 row affected, 1 warning (0.00 sec)

Rows matched: 1  Changed: 1  Warnings: 1

mysql> update user set plugin="mysql_native_password";

Query OK, 0 rows affected (0.00 sec)

Rows matched: 3  Changed: 0  Warnings: 0

mysql> flush privileges;

Query OK, 0 rows affected (0.01 sec)

mysql> quit;

Bye

4.回到sudo vi /etc/mysql/mysql.conf.d/mysqld.cnf，把刚才加入的那一行“skip-grant-tables”注释或删除掉。

5.再次重启mysql服务sudo service mysql restart，使用新的密码登陆，修改成功。

$ mysql -u root -p new_pass 

Welcome to the MySQL monitor.  Commands end with ; or \g. 

mysql> 

6.至此，问题解决

Answer 3

F 数组访问越界