UVALive 4998 Simple Encryption --DFS

题意： 给出K1，求一个12位数（不含前导0）K2，使得K1^K2 mod (10^12) = K2.

解法： 求不动点问题。

有一个性质： 如果12位数K2满足如上式子的话，那么K2%1,K2%10,K2%100,...,K2%10^12都会满足如上式子。那么我们可以dfs从后往前一个一个找出这个数的每一位。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define SMod 1000000000000
#define ll long long
using namespace std;
#define N 10007

ll K1,K2;
long long mul(long long a,long long b,long long mod) {
    ll ite = (1LL<<20)-1;
    return (a*(b>>20)%mod*(1ll<<20)%mod+a*(b&(ite))%mod)%mod;
}

ll fastm(ll a,ll b,ll m) {
    ll res = 1LL;
    while(b) {
        if(b&1LL) res = mul(res,a,m);
        a = mul(a,a,m);
        b >>= 1;
    }
    return res;
}

ll wei[16],ans;

bool dfs(int c,ll now) {
    if(c == 13) {
        if(now >= wei[12]) { ans = now; return true; }
        return false;
    }
    ll W = wei[c];
    for(ll i=0;i<=9;i++) {
        ll tmp = W*i+now;
        if(fastm(K1,tmp,W) != tmp%W) continue;
        if(dfs(c+1,tmp)) return true;
    }
    return false;
}

int main()
{
    int t,cs = 1;
    wei[0] = 1LL;
    for(int i=1;i<=13;i++) wei[i] = wei[i-1]*10LL;
    while(scanf("%lld",&K1)!=EOF && K1) {
        dfs(0,0);
        printf("Case %d: Public Key = %lld Private Key = %lld
",cs++,K1,ans%SMod);
    }
    return 0;
}

还有一种循环迭代的方法，随机选取一个超过10^12的数，如1000000000007，将其代入计算，如果f(x)!=x，那么令x=f(x)，如此循环，能在短时间内找出合法解。不知道为啥。。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define SMod 1000000000000
#define ll long long
using namespace std;

ll K1,K2;
long long mul(long long a,long long b,long long mod) {
    ll ite = (1LL<<20)-1;
    return (a*(b>>20)%mod*(1ll<<20)%mod+a*(b&(ite))%mod)%mod;
}

ll fastm(ll a,ll b,ll m) {
    ll res = 1LL;
    while(b) {
        if(b&1LL) res = mul(res,a,m);
        a = mul(a,a,m);
        b >>= 1;
    }
    return res;
}
ll f(ll x) {
    return fastm(K1,x,SMod);
}
ll gao(ll x) {
    while(1) {
        ll fx = f(x);
        if(fx == x) return x;
        x = fx;
    }
}

int main()
{
    int t,cs = 1;
    while(scanf("%lld",&K1)!=EOF && K1) {
        printf("Case %d: Public Key = %lld Private Key = %lld
",cs++,K1,gao(1000000000007));
    }
}