1031 查验身份证 (15分)
一个合法的身份证号码由17位地区、日期编号和顺序编号加1位校验码组成。校验码的计算规则如下:
首先对前17位数字加权求和,权重分配为:{7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};然后将计算的和对11取模得到值Z;最后按照以下关系对应Z值与校验码M的值:
Z:0 1 2 3 4 5 6 7 8 9 10
M:1 0 X 9 8 7 6 5 4 3 2
现在给定一些身份证号码,请你验证校验码的有效性,并输出有问题的号码。
输入格式:
输入第一行给出正整数N(≤100)是输入的身份证号码的个数。随后N行,每行给出1个18位身份证号码。
输出格式:
按照输入的顺序每行输出1个有问题的身份证号码。这里并不检验前17位是否合理,只检查前17位是否全为数字且最后1位校验码计算准确。如果所有号码都正常,则输出All passed。
输入样例1:
4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X
输出样例1:
12010X198901011234
110108196711301866
37070419881216001X
输入样例2:
2
320124198808240056
110108196711301862
输出样例2:
All passed
解题思路:按题干要求输出即可
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int num = Integer.parseInt(line);
String[] lines = new String[num];
for (int i = 0; i < num; i++) {
lines[i] = br.readLine();
}
int count = 0;
for (int i = 0; i < num; i++) {
int sum = 0;
boolean flag = true;
for (int j = 0; j < lines[i].length()-1; j++) {
char c = lines[i].charAt(j);
if (c >= '0' && c <= '9') {
switch (j) {
case 0:
sum += (c - 48) * 7;
break;
case 1:
sum += (c - 48) * 9;
break;
case 2:
sum += (c - 48) * 10;
break;
case 3:
sum += (c - 48) * 5;
break;
case 4:
sum += (c - 48) * 8;
break;
case 5:
sum += (c - 48) * 4;
break;
case 6:
sum += (c - 48) * 2;
break;
case 7:
sum += (c - 48) * 1;
break;
case 8:
sum += (c - 48) * 6;
break;
case 9:
sum += (c - 48) * 3;
break;
case 10:
sum += (c - 48) * 7;
break;
case 11:
sum += (c - 48) * 9;
break;
case 12:
sum += (c - 48) * 10;
break;
case 13:
sum += (c - 48) * 5;
break;
//7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2
case 14:
sum += (c - 48) * 8;
break;
case 15:
sum += (c - 48) * 4;
break;
case 16:
sum += (c - 48) * 2;
break;
}
} else {
flag=false;
}
}
if(flag){
char d = lines[i].charAt(17);
char e = ' ';
switch (sum % 11) {
case 0:
e = '1';
break;
case 1:
e = '0';
break;
case 2:
e = 'X';
break;
case 3:
e = '9';
break;
case 4:
e = '8';
break;
case 5:
e = '7';
break;
case 6:
e = '6';
break;
case 7:
e = '5';
break;
case 8:
e = '4';
break;
//Z:0 1 2 3 4 5 6 7 8 9 10
//M:1 0 X 9 8 7 6 5 4 3 2
case 9:
e = '3';
break;
case 10:
e = '2';
break;
}
if (d == e) {
count++;
} else {
System.out.println(lines[i]);
}
}else{
System.out.println(lines[i]);
}
}
if (count == num) {
System.out.println("All passed");
}
}
}