HDUOJ-------(1211)RSA
RSA
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1135 Accepted Submission(s): 833
Problem Description
RSA is one of the most powerful methods to encrypt data. The RSA algorithm is described as follow: > choose two large prime integer p, q > calculate n = p × q, calculate F(n) = (p - 1) × (q - 1) > choose an integer e(1 < e < F(n)), making gcd(e, F(n)) = 1, e will be the public key > calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key You can encrypt data with this method : C = E(m) = me mod n When you want to decrypt data, use this method : M = D(c) = cd mod n Here, c is an integer ASCII value of a letter of cryptograph and m is an integer ASCII value of a letter of plain text. Now given p, q, e and some cryptograph, your task is to "translate" the cryptograph into plain text.
Input
Each case will begin with four integers p, q, e, l followed by a line of cryptograph. The integers p, q, e, l will be in the range of 32-bit integer. The cryptograph consists of l integers separated by blanks.
Output
For each case, output the plain text in a single line. You may assume that the correct result of plain text are visual ASCII letters, you should output them as visualable letters with no blank between them.
Sample Input
101 103 7 11 7716 7746 7497 126 8486 4708 7746 623 7298 7357 3239
Sample Output
I-LOVE-ACM.
Author
JGShining(极光炫影)
Source
Recommend
Eddy
何为快速幂.....自己百度去... 自己根据以往经验敲的...所以自己验证去吧...
说下简单的快速幂吧....
//求 ans=a的b次幂mod n
quick_pow(int a,int b,int ans,int n)
{
while(a)
{
if(a&1)
{
ans*=b;
ans%=n;
a--;
}
else
{
b*=b;
b%=n;
b>>=1;
}
}
}
扩展欧几里得代码:
1 // ax+by=c; c/q=k;(k=1,2,3,.....n)
2 int x,y,q;
3 void exgcd(int a ,int b)
4 {
5 if(b==0)
6 {
7 y=0,x=1,q=a;
8 }
9 else
10 {
11 int temp=x;
12 x=y;
13 y=temp-a/b*y;
14 }
15 }
简单题。。。。
代码:
1 #include<stdio.h>
2 #include<string.h>
3 #include<stdlib.h>
4 #define LL _int64
5 LL d,a,b;
6 /*ax+by=c*/
7 /*欧几里得扩展*/
8 void exgcd(LL x ,LL y)
9 {
10 if(y==0)
11 {
12 a=1,b=0,d=x;
13 }
14 else
15 {
16 exgcd(y,x%y);
17 LL temp=a;
18 a=b,b=temp-x/y*b;
19 }
20 }
21 int main()
22 {
23 LL p,q,e,len,key;
24 while(scanf("%I64d %I64d %I64d %I64d",&p,&q,&e,&len)!=EOF)
25 {
26 exgcd(e,(p-1)*(q-1));
27 while(a<0)
28 {
29 a+=(p-1)*(q-1);
30 }
31 LL n=p*q;
32 LL cnt=a;
33 while(len--)
34 {
35 scanf("%I64d",&key);
36 key%=n;
37 LL ans=1;
38 /*可以用快速幂*/
39 a=cnt;
40 while(a)
41 {
42 if(a&1)
43 {
44 ans*=key;
45 ans%=n;
46 a--;
47 }
48 else
49 {
50 key*=key;
51 key%=n;
52 a>>=1L;
53 }
54 }
55 putchar(ans);
56 }
57 putchar(10);
58 }
59 return 0;
60 }
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法