UVA - 12253 Simple Encryption （同余）
Some day in the future a personnamed Reuben wants to submit some interesting problems to the judging directorof ACM ICPC (Association for Copotron Mechanics International CollegiateProgramming Contest) World Finals so that they can be used in World Finals2031. This contest requires problem submitters to submit their problems inencrypted format. But in year 2030, encryption software is notwidely available on the Internet just to prevent terrorists from sendingencrypted messages. So he wants to use a simplified encryption technique tosubmit his problem. He uses several 12-digit encryption keys to encrypt hismessages. We would not disclose the encryption/decryption technique for safetybecause your next generations will be competing in World Finals by then. ButReuben also does not want to send his encryption key in a straightforwardemail. He wants to send a key (public) K1 that will implicitlydenote the actual key (private) K2. K2 issuch a 12 digit number so that (modulo 1012).Given the value of K1 your job is to help Reuben find thevalue of K2.
Input
The input file contains around 1600line of input. Each line contains an integer, which denotes the value of K1(0<K1<50001). A line containing a single zero terminatesinput.
Output
For eachline of input produce one line of output. This line contains the serial ofoutput, followed by the given public key K1andthen a possible actual private key K2. Look at the output forsample input for exact formatting. Inputs will be such that for given value of K1there will always be at least one value of K2.Note that K2 should always have 12 digits and will nothave any leading zeroes.
SampleInput Output for Sample Input
| 78 99 0 |
Case 1: Public Key = 78 Private Key = 308646916096 Case 2: Public Key = 99 Private Key = 817245479899
|
Problemsetter: ShahriarManzoor, Special Thanks: Derek Kisman
题意：输入正整数K1，找一个12位正整数K2，是的k1^k2%10^12 = K2%10^12。
思路：枚举每一位，通过同余判断DFS的过程
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const ll bit = (1ll<<20) - 1;
const ll inf = 1000000000000ll;
ll ans;
int n, k;
ll mul(ll a, ll b, ll m) {
return (a*(b>>20)%m*(1ll<<20)%m+a*(b&(bit))%m)%m;
}
ll pow_mod(ll a, ll b, ll m) {
ll ans = 1;
while (b) {
if (b & 1)
ans = mul(ans, a, m);
b >>= 1;
a = mul(a, a, m);
}
return ans;
}
int dfs(int deep, ll now, ll m) {
if (deep == 12) {
if (pow_mod(k, now, inf) == now && now >= inf/10) {
ans = now;
return 1;
}
return 0;
}
for (int i = 0; i <= 9; i++)
if (pow_mod(k, i*m + now, m) == now)
if (dfs(deep+1, i*m+now, m*10))
return 1;
return 0;
}
int main() {
int cas = 1;
while (scanf("%d", &n) != EOF && n) {
k = n;
printf("Case %d: ", cas++);
for (int i = 0; i <= 9; i++)
if (dfs(1, i, 10)) {
printf("Public Key = %d Private Key = %lld\n", k, ans);
break;
}
}
return 0;
}