Ultraviolet A, 12253

Ultraviolet A, 12253

Link to the question; ultraviolet A 12253-Simple Encryption

Given K1, find a 12-bit K2 so that KK21 = K2 % 1012

Solution: enumeration by bit, without using the quick power modulo to judge the result.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const ll ite=(1<<20)-1;ll N;/*ll mul_mod (ll x, ll n, ll mod) {    ll ans = 0;    x %= mod;    n %= mod;    while (n) {        if (n&1)            ans = (ans + x) % mod;        x = 2 * x % mod;        n >>= 1;    }    return ans;}*/ll mul_mod(ll a, ll b, ll mod) {    return (a * (b>>20) % mod * (1ll<<20) %mod + a*(b&(ite)) % mod) % mod;}ll pow_mod (ll x, ll n, ll mod) {    ll ret = 1;    while (n) {        if (n&1)            ret = mul_mod(ret, x, mod);        x = mul_mod(x, x, mod);        n >>= 1;    }    return ret;}bool dfs (int d, ll u, ll mod) {    if (d == 12) {        if (u >= 1e11 && pow_mod(N, u, mod) == u) {            printf("%lld\n", u);            return true;        }        return false;    }    for (int i = 0; i < 10; i++) {        if (pow_mod(N, i * mod + u, mod) == u) {            if (dfs(d+1, i * mod + u, mod * 10))                return true;        }    }    return false;}int main () {    int cas = 1;    while (scanf("%lld", &N) == 1 && N) {        printf("Case %d: Public Key = %lld Private Key = ", cas++, N);        for (int i = 0; i < 10; i++) {            if (dfs(1, i, 10))                break;        }    }    return 0;}