身份证号校验公式

原理：身份证号的最后一位是根据前 17 位数字计算出来的，具有唯一性。计算方式：将身份证号的 { [第 1 个数字 (2^17/11) 的余数] + [第 2 个数字 (2^16/11)的余数] + …+ [第 17 个数字 * (2^1/11)的余数] } ，将所得的数除以 11 后，得到的余数按照余数转化后01102X39485766758493102依次对应。最后的结果就是第 18 位身份证号校验位。公式代码：Excel公式=LOOKUP(MOD(MID(B3,1,1)*MOD(2^17,11)+MID(B3,2,1)*MOD(2^16,11)+MID(B3,3,1)*MOD(2^15,11)+MID(B3,4,1)*MOD(2^14,11)+MID(B3,5,1)*MOD(2^13,11)+MID(B3,6,1)*MOD(2^12,11)+MID(B3,7,1)*MOD(2^11,11)+MID(B3,8,1)*MOD(2^10,11)+MID(B3,9,1)*MOD(2^9,11)+MID(B3,10,1)*MOD(2^8,11)+MID(B3,11,1)*MOD(2^7,11)+MID(B3,12,1)*MOD(2^6,11)+MID(B3,13,1)*MOD(2^5,11)+MID(B3,14,1)*MOD(2^4,11)+MID(B3,15,1)*MOD(2^3,11)+MID(B3,16,1)*MOD(2^2,11)+MID(B3,17,1)*MOD(2^1,11),11),{0;1;2;3;4;5;6;7;8;9;10},{1;0;“X”;9;8;7;6;5;4;3;2})C语言代码#include<stdio.h>#include<stdlib.h>#include<math.h>int reIdNum(char id[]){ char check = ‘1’; int i = 16, sum = 0, result; for(i = 16; i >= 0; i–){ sum += ( (id[16-i] - ‘0’) * (int(pow(2,(i+1))) % 11) ); } switch (sum % 11){ case 0 : check = ‘1’; break; case 1 : check = ‘2’; break; case 2 : check = ‘X’; break; case 3 : check = ‘9’; break; case 4 : check = ‘8’; break; case 5 : check = ‘7’; break; case 6 : check = ‘6’; break; case 7 : check = ‘5’; break; case 8 : check = ‘4’; break; case 9 : check = ‘3’; break; case 10 : check = ‘2’; break; } if(id[17] == check){ result = 1; }else{ result = 0; } return (result);}int main(){ char id[19]; puts(“Please enter IdNumber:”); scanf("%s", id); if(reIdNum(id)){ puts(“The IdNumber is ture.”); }else{ puts(“The IdNumber is false.”); } system(“pause”); return (0);}