您的位置:首页 > 其它

SRM 583 Div Level Two:IDNumberVerification

2013-06-23 16:40 375 查看
题目来源:http://community.topcoder.com/stat?c=problem_statement&pm=12610

这道题比较有意思,估计是中国人出的吧,以前都不知道身份证还这么麻烦,不过程序不难写。

#include <algorithm>
#include <iostream>
#include <queue>
#include <vector>
#include <list>
#include <string>
#include <cmath>
#include <limits>
#include <cstdlib>

using namespace std;
class IDNumberVerification
{
public:
string verify(string id, vector <string> regionCodes);
};

string IDNumberVerification::verify(string id, vector<string> regionCodes)
{
string region;
string year;
string monday;
string month, day;
string seq;
string checksum;
string gender;
int nyear, nmonth, nday, nchecksum, nseq;
int sum;
int days_notleap[] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int days_leap[] = {-1, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int *days;

region = id.substr(0, 6);
year = id.substr(6, 4);
monday = id.substr(10, 4);
month = id.substr(10, 2);
day = id.substr(12, 2);
seq = id.substr(14, 3);
checksum = id.substr(17, 1);
gender = id.substr(14, 3);
bool flag = false;
/* seq */
if (seq == "000") {
return "Invalid";
}
/* region */
for (int i = 0; i < regionCodes.size(); i++) {
if (region == regionCodes[i]) {
flag = true;
}
}
if (!flag) {
return "Invalid";
}
/* data */
bool leap = false;
nyear = atoi(year.c_str());

if (nyear < 1900 || nyear > 2011) {
return "Invalid";
}

if ( (nyear % 4 == 0 && nyear % 100 != 0) ||
(nyear % 400 == 0) ) {
leap = true;
}
if ("0229" == monday && !leap) {
return "Invalid";
}
days = days_notleap;
if (leap) {
days = days_leap;
}

nmonth = atoi(month.c_str());
nday = atoi(day.c_str());
if (nmonth > 12 || nmonth < 1) {
return "Invalid";
}

if (nday > days[nmonth] || nday < 1) {
return "Invalid";
}

/* checksum */
sum = 0;
for (int i = 0; i < 17; i++) {
sum = (sum * 2) + id[i] - '0';
}
sum = 2 * sum;

nchecksum = checksum[0] - '0';
if (checksum[0] == 'X') {
nchecksum = 10;
}
int rchecksum = 12 - sum % 11;
if (rchecksum == 11) {
rchecksum = 0;
}
if ( nchecksum != rchecksum ) {
return "Invalid";
}

/* gender */
nseq = atoi(seq.c_str());
if (nseq % 2 != 0) {
return "Male";
} else {
return "Female";
}
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: