L1-016. 查验身份证

    xiaoxiao2021-03-26  8

    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118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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