Problem Description RSA is one of the most powerful methods to encrypt data. The RSA algorithm is described as follow: 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 |
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
long long ju(long long a,long long d,long long n){
long long s=a,sum=1;
while(d){
if(d&1){
sum=sum*s%n;
}
s=s*s%n;
d>>=1;
}
return sum;
}
void gcd_(long long a,long long b,int &x,int &y){
if(b==0){
x=1;
y=0;
return ;
}
gcd_(b,a%b,x,y);
long long temp=x;
x=y;
y=temp-a/b*y;
return ;
}
long long inv(long long s,long long d){
int x,y;
gcd_(s,d,x,y);
return (x%d+d)%d;
}
int main(){
long long p,q,e,l;
while(scanf("%lld %lld %lld %lld",&p,&q,&e,&l)!=EOF){
long long fn=(p-1)*(q-1);
long long n=p*q;
long long d=inv(e,fn);
long long temp;
while(l--){
scanf("%lld",&temp);
printf("%c",(int)ju(temp,d,n)%n);
}
printf("\n");
}
}