1031 查验身份证（附详细注释，逻辑分析）-CFANZ编程社区

写在前面

实现思路

证件号前17位加权和计算，取模
校验位对比

题目较简单，细节处理耗费时间。例如，校验位下标从0开始

大体10分钟a题，尽量掌握调试（虽至今不太习惯，大多靠猜）

测试用例

input:
4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X
output:
12010X198901011234
110108196711301866
37070419881216001X

input:
2
320124198808240056
110108196711301862
output:

ac代码

#include <iostream>
using namespace std;
const int weight[17] = {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
const char ch[11] = {'1','0','X','9','8','7','6','5','4','3','2'};
int check_digit(string p)
{
    int result = 0;
    for(int i=0; i<17; i++) result += (p[i]-'0')*weight[i];
    return result%11;
}
int main()
{
    int N, check=0, counter=0;
    scanf("%d", &N);
    char id[18];
    for(int i=0; i<N; i++)
    {
        scanf("%s", id);
        check = check_digit(id);
        if(ch[check]==id[17]) counter++;
        else printf("%s\n", id);
    }
    if(counter==N) printf("All passed");
    return 0;
}

代码优化后

#include <iostream>
using namespace std;
const int weight[17] = {7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
const char ch[11] = {'1','0','X','9','8','7','6','5','4','3','2'};
bool check_digit(string p)
{
    int result = 0;
    for(int i=0; i<17; i++) result += (p[i]-'0')*weight[i];
    return ch[result%11]==p[17];
}
int main()
{
    int N,counter=0;
    scanf("%d", &N);
    char id[18];
    for(int i=0; i<N; i++)
    {
        scanf("%s", id);
        if(check_digit(id)) counter++;
        else printf("%s\n", id);
    }
    if(counter==N) printf("All passed");
    return 0;
}

学习代码

1031. 查验身份证(15)-PAT乙级真题

自定义函数功能更彻底！

#include <iostream>
using namespace std;
int a[17] = {7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2};
int b[11] = {1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2};
string s;
bool isTrue() {
    int sum = 0;
    for (int i = 0; i < 17; i++) {
        if (s[i] < '0' || s[i] > '9') return false;
        sum += (s[i] - '0') * a[i];
    }
    int temp = (s[17] == 'X') ? 10 : (s[17] - '0');
    return b[sum%11] == temp;
}
int main() {
    int n, flag = 0;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> s;
        if (!isTrue()) {
            cout << s << endl;
            flag = 1;
        }
    }
    if (flag == 0) cout << "All passed";
    return 0;
}