import java.util.Base64;
import java.io.BufferedInputStream;
import java.nio.charset.StandardCharsets;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
//加密后的文本
String key = "d2hhdCBpcyB0aGUgcmVtYWluZGVyIHdoZW4gdGhlIG51bWJlciBpcyBkaXZpZGVkIGJ5IDIwMTc/";
String value = new String(Base64.getDecoder().decode(key),StandardCharsets.UTF_8);
//System.out.println(value);//本行取消注释，就可以看到题意，java解题开挂
Scanner sc = new Scanner (new BufferedInputStream(System.in));
int t = sc.nextInt();
while(t-- != 0) {
System.out.println(sc.nextInt() % 2017);
}
sc.close();
}
}
\
C++模拟base64加密算法，本来我是想写解密算法呢？额，一不小心写反了，写成了加密算法。。。。尴尬
const char str[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
string value = "what is the remainder when the number is divided by 2017?";
//string key = "d2hhdCBpcyB0aGUgcmVtYWluZGVyIHdoZW4gdGhlIG51bWJlciBpcyBkaXZpZGVkIGJ5IDIwMTc/";
string base64Encode(string source) { //参数为明文
string code01, key;
for (int i = 0; i < source.length(); ++i) {
int ch = source[i], num;
for (int j = 0; j < 8; ++j) {
num = (ch & 128) == 0 ? 0: 1;
ch <<= 1;
code01.append(1, num + '0');
}
}
int remainder = code01.length() % 6;
if (remainder != 0)
code01.insert(code01.end() - remainder, 6 - remainder, '0');
int cnt = code01.length() / 6;
for (int i = 0; i < code01.length(); i += 6) {
int num = 0;
for (int j = 0; j < 6; ++j)
num += (code01[i + j] - '0') * (1 << (5 - j));
char ch = str[num];
key.append(1, ch);
}
return key; //返回加密后的暗文
}
\
base64解码（其实解码，和加密算法类似）
const char str[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";;
string key = "d2hhdCBpcyB0aGUgcmVtYWluZGVyIHdoZW4gdGhlIG51bWJlciBpcyBkaXZpZGVkIGJ5IDIwMTc/";
string base64Decode(string key) { //参数为暗文
string code01, value;
int mod = 8;
for (int i = 0; i < key.length(); ++i) {
int ch = (strchr(str, key[i]) - str), num;
for (int j = 0; j < 6; ++j) {
num = (ch & 32) == 0 ? 0 : 1; // 128 - > 32
ch <<= 1;
code01.append(1, num + '0');
}
}
int remainder = code01.length() % mod;
if (remainder != 0)
code01.insert(code01.end() - remainder, mod - remainder, '0');
int cnt = code01.length() / mod;
for (int i = 0; i < code01.length(); i += mod) {
int num = 0;
for (int j = 0; j < mod; ++j)
num += (code01[i + j] - '0') * (1 << (mod - 1 - j));
char ch = num;
value.append(1, ch);
}
return value; //返回解密后的明文
}
\
