RSA hdu 1211 数论的水题（数据较弱）

RSA

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1138    Accepted Submission(s): 836


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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. 

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

101 103 7 11
7716 7746 7497 126 8486 4708 7746 623 7298 7357 3239

 

[align=left]Sample Output[/align]

I-LOVE-ACM.

 

[align=left]Author[/align]
JGShining（极光炫影）
 

[align=left]Source[/align]
杭电ACM省赛集训队选拔赛之热身赛
 

[align=left]Recommend[/align]
Eddy
 
 
 
 
这个题目的真的是相当水，我本来一开始正在发愁又要解线性同余方程了，可是当我看到别人的博客此题数据极
 
弱，让我颇有点不太适应的，我顿时觉得还是poj的数论题目的比较刺激啊，唉，Romance
however, is rocky
this
 
month!!!!感情不顺，只好做数论题目让我开心点！！！
 
n=p*q,      f(n)=(p-1)*(q-1),       d*e%f(n)=1, m=c^d%n,我们竟然按照这个式子模拟就能过了！！！！
 
让我等屌丝惊呆了！！！！！！！！！！！！！！！
 
 

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int i,j,p,q,f,e,l,c,m,n,d;
while(scanf("%d%d%d%d",&p,&q,&e,&l)!=EOF)
{

n=p*q;
f=(p-1)*(q-1);
d=1;
while(d*e%f!=1)
d++;
for(i=0;i<l;i++)
{
scanf("%d",&c);
int t=1;
for(j=0;j<d;j++)
{
t=(t*c)%n;
}
printf("%c",t);
}
cout<<endl;
}
return 0;
}