本题要求:
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:
4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X
输出样例:
12010X198901011234
110108196711301866
37070419881216001X
解题思路 :
注意格式就好
代码 :
#include<iostream>
using namespace std;
int main() {
int i, j;
int n;
char id[
19];
cin >> n;
bool isAll =
true;
for (i =
0; i < n; i++) {
bool isTrue =
true;
int temp =
0;
cin >> id;
for (j =
0; id[j +
1] !=
'\0'; j++) {
if (id[j] <
'0' || id[j] >
'9') {
isTrue =
false;
break;
}
int num =
0;
switch(j +
1) {
case 1:
num =
7;
break;
case 2:
num =
9;
break;
case 3:
num =
10;
break;
case 4:
num =
5;
break;
case 5:
num =
8;
break;
case 6:
num =
4;
break;
case 7:
num =
2;
break;
case 8:
num =
1;
break;
case 9:
num =
6;
break;
case 10:
num =
3;
break;
case 11:
num =
7;
break;
case 12:
num =
9;
break;
case 13:
num =
10;
break;
case 14:
num =
5;
break;
case 15:
num =
8;
break;
case 16:
num =
4;
break;
case 17:
num =
2;
break;
}
temp += (id[j] -
'0') * num;
}
if (!isTrue || j !=
17) {
isAll =
false;
cout << id << endl;
continue;
}
switch(temp %
11) {
case 0:
temp =
'1';
break;
case 1:
temp =
'0';
break;
case 2:
temp =
'X';
break;
case 3:
temp =
'9';
break;
case 4:
temp =
'8';
break;
case 5:
temp =
'7';
break;
case 6:
temp =
'6';
break;
case 7:
temp =
'5';
break;
case 8:
temp =
'4';
break;
case 9:
temp =
'3';
break;
case 10:
temp =
'2';
break;
}
if (temp != id[
17]) {
isAll =
false;
cout << id << endl;
}
}
if (isAll) {
cout <<
"All passed" << endl;
}
return 0;
}
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