LHL的代码：

#include<iostream>
#include<string>
#include<cmath>
using namespace std;
int main()
{
	int pj[17]={7,9,10,5,8,4,2,1,6,3,7,9,10,5,8, 4 , 2 };
	int z[11]={ 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 };
	char m[11]={'1','0','X','9','8','7','6','5','4','3','2'};
	int n,i,j,k,l;
	int t,x,y;
	int s[100]={0};
	cin>>n;
	char a[102][18];
	int f[1000]={0};
	int q[1000]={0};
	getchar();
	for(i=0;i<n;i++)
	{
		scanf("%s",a[i]);
	}
	
	for(i=0;i<n;i++)
	{
		for(j=0;j<17;j++)
		{
			if(!(a[i][j]>=48&&a[i][j]<=9+48))
			{
				f[i]++;
				break;
			}
			s[i]=s[i]+((int)(a[i][j]-'0'))*pj[j];
		}
		q[i]=s[i]%11;
			if(a[i][17]!=m[q[i]])
			{
				f[i]++;
			}
	}
	int e=0;
	for(i=0;i<n;i++)
	{
		if(f[i]!=0)
		{
			e++;
		for(j=0;j<18;j++)
		printf("%c",a[i][j]);	
		printf("\n");
		}
		
		
	}
	if(e==0)
	printf("All passed");
	return 0;
}

标准答案

#include<iostream>
#include<string>
using namespace std;
int main()
{
    int n;
    cin>>n;
    string* str = new string[n];
    for(int i = 0;i<n;i++)
    {
        cin>>str[i];
        
    }
    char a[n];
    int sum[n];
    int shaobing[n];
    for(int i = 0;i<n;i++)
    {
        a[i]=str[i].at(17);
        sum[i]=0;
        shaobing[i]=1;
     } 
     char M[11]={'1','0','X','9','8','7','6','5','4','3','2'};
     int N[17]={7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
     for(int j = 0;j<n;j++)
     {
        for(int i=0;i<17;i++)
     {
         if(str[j].at(i)>='0'&&str[j].at(i)<='9')
         {
             sum[j]+=(str[j].at(i)-'0')*N[i];
     
            }
         else
         {
             shaobing[j]=0;
             break;
            }
     }
     }
         int flag = 1;
     for(int i=0;i<n;i++)
     {
        if(shaobing[i]==0||M[sum[i]%11]!=a[i])
        {
            cout<<str[i]<<endl;
            flag = 0;
         }
     }
     if(flag == 1)
     {
        cout<<"All passed"<<endl;
     }
    return 0;
}