Hangzhou Electric Acm1211--rsa

This problem, similar to the decryption of ciphertext. The meaning of the topic is very clear.

Give you the number of P Q e l,l for the second row.

Calculate the N and FN first, and then calculate the D.

n = p * q;fn = (p-1) * (q-1).

Then according to the D * e% fn = 1, calculate d.

The key is to seek C. The number num to you is equal to c ^ d% n. c The corresponding ASCLL code is the decrypted character.

The following is the code for the AC:

#include <iostream>using namespace Std;int main () {int n, FN, p, Q, E, L, NUM, D;while (CIN >> p >> q > > E >> l) {n = q * p;fn = (p-1) * (q-1);d = 1;while (d * e% fn = 1) d++;for (int i = 0; i < L; i++) {cin > > Num;int c = 1;for (int j = 1; J <= D; j + +) {c = c * num;c = c% n;} cout << (char) c;} cout << Endl;} return 0;}

Hangzhou Electric Acm1211--rsa