PAT (Basic Level) Practise (中文) 1031. 查验身份证(15)
来源:互联网 发布:c语言经典实例 编辑:程序博客网 时间:2024/05/17 02:43
1031. 查验身份证(15)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
CHEN, Yue
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1:432012419880824005612010X19890101123411010819671130186637070419881216001X输出样例1:
12010X19890101123411010819671130186637070419881216001X输入样例2:
2320124198808240056110108196711301862输出样例2:
All passed
//acimport java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;public class Main { public static void main(String[] args) throws NumberFormatException, IOException { int[] quanzhong = { 7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2 }; BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); int n; String result=""; n = Integer.parseInt(bf.readLine()); String[] arr = new String[n]; for (int i = 0; i < arr.length; i++) { arr[i] = bf.readLine(); } for (int i = 0; i < arr.length; i++) { int sum = 0; String temp1 = arr[i].substring(0, 6); String temp2 = arr[i].substring(6, 12); String temp3 = arr[i].substring(12, 17); try { Integer.parseInt(temp1); Integer.parseInt(temp2); Integer.parseInt(temp3); } catch (Exception e) { result+=arr[i] +"\n"; continue; } for (int j = 0; j < arr[i].length()-1; j++) { sum += quanzhong[j] * Integer.parseInt(arr[i].charAt(j) + ""); } int key = sum % 11; switch (key) { case 0: if (!arr[i].substring(17,18).equals("1")) { result+=arr[i] +"\n"; } continue; case 1: if (!arr[i].substring(17,18).equals("0")) { result+=arr[i] +"\n"; } continue; case 2: if (!arr[i].substring(17,18).equals("X")) { result+=arr[i] +"\n"; } continue; case 3: if (!arr[i].substring(17,18).equals("9")) { result+=arr[i] +"\n"; } continue; case 4: if (!arr[i].substring(17,18).equals("8")) { result+=arr[i] +"\n"; } continue; case 5: if (!arr[i].substring(17,18).equals("7")) { result+=arr[i] +"\n"; } continue; case 6: if (!arr[i].substring(17,18).equals("6")) { result+=arr[i] +"\n"; } continue; case 7: if (!arr[i].substring(17,18).equals("5")) { result+=arr[i] +"\n"; } continue; case 8: if (!arr[i].substring(17,18).equals("4")) { result+=arr[i] +"\n"; } continue; case 9: if (!arr[i].substring(17,18).equals("3")) { result+=arr[i] +"\n"; } continue; case 10: if (!arr[i].substring(17,18).equals("2")) { result+=arr[i] +"\n"; } continue; } } // if (result.equals("")||result.isEmpty()) {// System.out.println("All passed");// }else {// result=result.substring(0,result.length()-1);// System.out.println(result);// } System.out.println(result.equals("")||result.isEmpty()?"All passed":result.substring(0,result.length()-1)); }}
阅读全文
0 0
- PAT (Basic Level) Practise (中文)1031. 查验身份证(15)
- PAT (Basic Level) Practise (中文) 1031. 查验身份证(15)
- PAT (Basic Level) Practise (中文)1031. 查验身份证(15)
- PAT BASIC LEVEL 1031. 查验身份证(15)
- PAT (Basic Level)1031. 查验身份证
- PAT(basic level) 1031 查验身份证(15)
- PAT BASIC PRACTICE: 1031. 查验身份证(15)
- Pat(Basic Level)Practice--1031(查验身份证)
- PAT(Basic Level)_1031_查验身份证
- PAT Basic 1031. 查验身份证(15)(C语言实现)
- PAT (Basic Level) Practise (中文)
- PAT (Basic Level) Practise (中文)
- PAT (Basic Level) Practise (中文)--1001
- PAT (Basic Level) Practise (中文)
- PAT (Basic Level) Practise (中文)1001
- PAT (Basic Level) Practise (中文)1002
- PAT (Basic Level) Practise (中文) 1007
- PAT (Basic Level) Practise (中文)1021. 个位数统计 (15)
- 阿里云服务器漏洞phpmyadmin CVE-2016-6617 SQL注入漏洞 解决方法
- js根据不同的方式进行查找结点
- 作业管理
- fpga-静态时序分析
- 517. Super Washing Machines
- PAT (Basic Level) Practise (中文) 1031. 查验身份证(15)
- 电话面试
- 关于h5的所有标签
- 响应式布局
- 浅谈:html5和html的区别
- HDU
- 负margin技术的应用
- shell之判断语句结构
- ACM 2037 今年暑假不AC