B1031 check ID card

1031 Check your ID card (15 branch )

A legal ID number is from 17 Location 、 Date number and sequence number plus 1 Bit check code composition . Check code calculation rules are as follows ：

First of all, to the front 17 Bit number weighted sum , The weight is assigned to ：{7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2}; Then the sum of the calculated values is compared to the 11 Take the modulus to get the value Z; Finally, according to the following relationship Z Value and check code M Value ：

Z：0 1 2 3 4 5 6 7 8 9 10
M：1 0 X 9 8 7 6 5 4 3 2

Now give me some ID number. , Please verify the validity of the check code , And output the problem number .

Input format ：
Enter the first line to give a positive integer N（≤100） The number of the ID number entered. . And then N That's ok , Each line gives 1 individual 18 I. D. number .

Output format ：
Output each line in the order of input 1 Question Id number . This is not before the test 17 Is it reasonable , Just before checking 17 Are all digits and last 1 Accurate calculation of bit check code . If all the numbers are OK , The output All passed.

sample input

4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X

sample output

12010X198901011234
110108196711301866
37070419881216001X

sample input

2
320124198808240056
110108196711301862

sample output

All passed

Topic analysis ：

1. Be careful Before inspection 17 Whether all bits are numbers
2. Calculation 1 Check whether the bit check code is correct .

The code is as follows ：

#include <bits/stdc++.h>
using namespace std;
int main(){
    int n;
    char m[]={'1','0','X','9','8','7','6','5','4','3','2'};
    int z[]={7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
    cin>>n;
    char s[101][19];
    int sum[101]={0},flag[101]={0},g=0;
    for(int i=0;i<n;i++){
    	scanf("%s",s[i]);
    	s[i][18]='\0';
    	for(int j=0;j<17;j++){
    		if(s[i][j]<='9'&&s[i][j]>='0'){
    			sum[i]+=(s[i][j]-'0')*z[j];
			}else{
				flag[i]=1;
				break;
			}	
		}
		if(flag[i]==0){
			int a=sum[i]%11;
	    	if(m[a]-s[i][17]!=0){
	    		flag[i]=1;
	    	}
		}
		if(flag[i]==1){
			g++;
		}
	}
	if(g==0){
		cout<<"All passed";
	}else{
		for(int i=0;i<n;i++){
			if(flag[i]==1){
				printf("%s\n",s[i]);
			}
		}
	}
	
    return 0;
}