LintCode练习-2824 · 加密的情书-Java-大整数进制转换及异或运算

虽然用Java语言中的BigInteger可以直接对大数进行操作，但如果不是Java，是C/C++之流呢？

所以用Java语言写了一个进制转换及运算的代码。

 1 import org.jetbrains.annotations.NotNull;
 2 
 3 import java.math.BigInteger;
 4 import java.util.Arrays;
 5 import java.util.Scanner;
 6 
 7 import static java.lang.System.*;
 8 
 9 public class Main {
10     public static void main(String[] args) {
11         Scanner scanner = new Scanner(System.in);
12         String xxx = scanner.next();
13         int[] inV = new int[xxx.length()];
14         for (int i = 0;i<xxx.length();i++)
15             inV[i] = xxx.charAt(i) - '0';
16         int[] invs = {1,9,9,9,0,6,0,6};
17         int[] plaintext = decimalToBinary(inV,10,2);  // 十进制转二进制
18         int[] seckey = decimalToBinary(invs, 10,2);
19         int[] encrytext = myXOR(plaintext,seckey);  // 两个二进制异或运算
20         int[] binaryencry = decimalToBinary(encrytext,2,10); // 二进制转十进制
21         for (int i = 0; i < binaryencry.length; i++)
22             System.out.print(binaryencry[i]);
23         /**
24          * Java调包简单实现大整数运算
25          * BigInteger inputn = new BigInteger(xxx);
26          * BigInteger seckey = BigInteger.valueOf(19990606);
27          * System.out.print(inputn.xor(seckey));
28          */  
29     }
30     public static int[] myXOR(int[] inA, int[] inB) {
31         int lenMax, lenMin;
32         int[] inMax, inMin;
33         if (inA.length > inB.length){
34             lenMax = inA.length;
35             lenMin = inB.length;
36             inMax = inA;
37             inMin = inB;
38         }else {
39             lenMax = inB.length;
40             lenMin = inA.length;
41             inMax = inB;
42             inMin = inA;
43         }
44         int[] outC = new int[lenMax];
45         int index1 = -1;
46         for (int i = 0; i < lenMax-lenMin; i++) {
47             outC[i] = inMax[i] == 0 ? 0 : 1;
48             if (index1 == -1 && outC[i] == 1)
49                 index1 = i;
50         }
51         for (int i = lenMax - lenMin, j = 0; i < lenMax; i++, j++){
52             outC[i] = inMax[i] == inMin[j] ? 0 : 1;
53             if (index1 == -1 && outC[i] == 1)
54                 index1 = i;
55         }
56         int k = 0;
57         if (index1 == -1)
58             return outC;
59         int[] outV = new int[lenMax-index1];
60         for (int i = index1; i < lenMax; i++)
61             outV[k++] = outC[i];
62         return outV;
63     }
64     public static int[] decimalToBinary(@NotNull int[] inValue, int oldBase, int newBase){
65         // 十进制转二进制-对商求余 存储余数及更新原被除数
66         int[] out = new int[1000000];
67         int len = inValue.length;
68         int j, k= 0;
69         int shangs = 1;
70         while (shangs != 0){
71             shangs = 0;
72             for(int i = 0;i < len; i++){
73                 j = i + 1;
74                 int shang = inValue[i] / newBase;
75                 int yu = inValue[i] % newBase;
76                 if (i==(len-1))
77                     out[k++] = yu;
78                 else
79                     inValue[j] += yu*oldBase;
80                 inValue[i] = shang;
81                 shangs += shang;
82             }
83         }
84         int outRe[] = new int[k];
85         for(int i = 0;i < k;i++)
86             outRe[i] = out[k-i-1];
87         return outRe;
88     }
89    
90 }

 

posted @ 2021-12-28 16:49  achived  阅读(157)  评论(0)  编辑  收藏  举报