LA 4998 Simple Encryption

题意：输入正整数$K_1(K_1 \leq 50000)$, 找一个$12$位正整数$K_2$（不能含有前导零）使得${K_1}^{K_2}\equiv K_2(mod10^{12})$。

例如，$K_1=78$和$99$时，$K_2$分别为$308646916096$和$817245479899$。

分析：这题还是需要搜索来寻找答案的，直接构造很困难。事实上，${K_1}^{K_2}\equiv K_2(mod10^{12})$，同时意味着：

${K_1}^{K_2}\equiv K_2(mod10^{i}), i \leq 12$。

并且有$\varphi(10^i)={10^i}(1-\frac{1}{2})(1-\frac{1}{5})=4\cdot10^{i-1}$

于是${K_1}^{K_2}\mod {10^i}={K_1}^{K_2\mod{\varphi(10^i)}+\varphi(10^i)}\mod {10^i}$

于是${K_1}^{K_2}\mod {10^i}={K_1}^{K_3}\mod {10^i}\iff\varphi(10^i) | (K_3 - K_2)$

满足上式的一个充分条件是$10^{i+1} | (K_3 - K_2)$

也就是说，我们从低到高枚举$K_2$第$i$位的值，同时枚举其第$i+1$位的值，那么只要从最低位到第$i+1$位关于分解后对应模方程是合法的，

在以后的构造过程中始终不会破坏当前的性质，因为高位的那些值对当前模取模值都是相同的。

那么我们就可以边枚举边验证进行dfs了。

交了之后发现TLE了。为了保证快速幂时乘法不溢出，又用了一个快速乘法操作，这个操作实际上是可以从$O(log(n))$优化到$O(1)$的。

首先考虑模数最大为$10^{12}$，不超过$2$进制中的$40$位，那么我们可以将待乘值拆分成高$20$位与低$20$位，确保在乘法时始终不超过$60$位，

用long long就不会溢出，具体如下：

$a\cdot b \mod n = a \cdot ((b>>20)\cdot(1<<20) + b \& ((1 <<20) - 1)\mod n$

$= ((a\cdot(b>>20)\mod n)\cdot(1<<20)\mod n) + (a\cdot(b \& ((1 <<20) - 1))))\mod n$

这样优化之后就可以通过了。

  1 #include <algorithm>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <string>
  5 #include <queue>
  6 #include <map>
  7 #include <set>
  8 #include <ctime>
  9 #include <cmath>
 10 #include <iostream>
 11 #include <assert.h>
 12 #define PI acos(-1.)
 13 #pragma comment(linker, "/STACK:102400000,102400000")
 14 #define max(a, b) ((a) > (b) ? (a) : (b))
 15 #define min(a, b) ((a) < (b) ? (a) : (b))
 16 #define mp make_pair
 17 #define st first
 18 #define nd second
 19 #define keyn (root->ch[1]->ch[0])
 20 #define lson (u << 1)
 21 #define rson (u << 1 | 1)
 22 #define pii pair<int, int>
 23 #define pll pair<ll, ll>
 24 #define pb push_back
 25 #define type(x) __typeof(x.begin())
 26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
 27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
 28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
 29 #define dbg(x) cout << x << endl
 30 #define dbg2(x, y) cout << x << " " << y << endl
 31 #define clr(x, i) memset(x, (i), sizeof(x))
 32 #define maximize(x, y) x = max((x), (y))
 33 #define minimize(x, y) x = min((x), (y))
 34 using namespace std;
 35 typedef long long ll;
 36 const int int_inf = 0x3f3f3f3f;
 37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
 38 const int INT_INF = (int)((1ll << 31) - 1);
 39 const double double_inf = 1e30;
 40 const double eps = 1e-14;
 41 typedef unsigned long long ul;
 42 inline int readint(){
 43     int x;
 44     scanf("%d", &x);
 45     return x;
 46 }
 47 inline int readstr(char *s){
 48     scanf("%s", s);
 49     return strlen(s);
 50 }
 51 //Here goes 2d geometry templates
 52 struct Point{
 53     double x, y;
 54     Point(double x = 0, double y = 0) : x(x), y(y) {}
 55 };
 56 typedef Point Vector;
 57 Vector operator + (Vector A, Vector B){
 58     return Vector(A.x + B.x, A.y + B.y);
 59 }
 60 Vector operator - (Point A, Point B){
 61     return Vector(A.x - B.x, A.y - B.y);
 62 }
 63 Vector operator * (Vector A, double p){
 64     return Vector(A.x * p, A.y * p);
 65 }
 66 Vector operator / (Vector A, double p){
 67     return Vector(A.x / p, A.y / p);
 68 }
 69 bool operator < (const Point& a, const Point& b){
 70     return a.x < b.x || (a.x == b.x && a.y < b.y);
 71 }
 72 int dcmp(double x){
 73     if(abs(x) < eps) return 0;
 74     return x < 0 ? -1 : 1;
 75 }
 76 bool operator == (const Point& a, const Point& b){
 77     return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
 78 }
 79 double Dot(Vector A, Vector B){
 80     return A.x * B.x + A.y * B.y;
 81 }
 82 double Len(Vector A){
 83     return sqrt(Dot(A, A));
 84 }
 85 double Angle(Vector A, Vector B){
 86     return acos(Dot(A, B) / Len(A) / Len(B));
 87 }
 88 double Cross(Vector A, Vector B){
 89     return A.x * B.y - A.y * B.x;
 90 }
 91 double Area2(Point A, Point B, Point C){
 92     return Cross(B - A, C - A);
 93 }
 94 Vector Rotate(Vector A, double rad){
 95     //rotate counterclockwise
 96     return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
 97 }
 98 Vector Normal(Vector A){
 99     double L = Len(A);
100     return Vector(-A.y / L, A.x / L);
101 }
102 void Normallize(Vector &A){
103     double L = Len(A);
104     A.x /= L, A.y /= L;
105 }
106 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){
107     Vector u = P - Q;
108     double t = Cross(w, u) / Cross(v, w);
109     return P + v * t;
110 }
111 double DistanceToLine(Point P, Point A, Point B){
112     Vector v1 = B - A, v2 = P - A;
113     return abs(Cross(v1, v2)) / Len(v1);
114 }
115 double DistanceToSegment(Point P, Point A, Point B){
116     if(A == B) return Len(P - A);
117     Vector v1 = B - A, v2 = P - A, v3 = P - B;
118     if(dcmp(Dot(v1, v2)) < 0) return Len(v2);
119     else if(dcmp(Dot(v1, v3)) > 0) return Len(v3);
120     else return abs(Cross(v1, v2)) / Len(v1);
121 }
122 Point GetLineProjection(Point P, Point A, Point B){
123     Vector v = B - A;
124     return A + v * (Dot(v, P - A) / Dot(v, v));
125 }
126 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
127     //Line1:(a1, a2) Line2:(b1,b2)
128     double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
129            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
130     return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
131 }
132 bool OnSegment(Point p, Point a1, Point a2){
133     return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 -p)) < 0;
134 }
135 Vector GetBisector(Vector v, Vector w){
136     Normallize(v), Normallize(w);
137     return Vector((v.x + w.x) / 2, (v.y + w.y) / 2);
138 }
139
140 bool OnLine(Point p, Point a1, Point a2){
141     Vector v1 = p - a1, v2 = a2 - a1;
142     double tem = Cross(v1, v2);
143     return dcmp(tem) == 0;
144 }
145 struct Line{
146     Point p;
147     Vector v;
148     Point point(double t){
149         return Point(p.x + t * v.x, p.y + t * v.y);
150     }
151     Line(Point p, Vector v) : p(p), v(v) {}
152 };
153 struct Circle{
154     Point c;
155     double r;
156     Circle(Point c, double r) : c(c), r(r) {}
157     Circle(int x, int y, int _r){
158         c = Point(x, y);
159         r = _r;
160     }
161     Point point(double a){
162         return Point(c.x + cos(a) * r, c.y + sin(a) * r);
163     }
164 };
165 int GetLineCircleIntersection(Line L, Circle C, double &t1, double& t2, vector<Point>& sol){
166     double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
167     double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r;
168     double delta = f * f - 4 * e * g;
169     if(dcmp(delta) < 0) return 0;
170     if(dcmp(delta) == 0){
171         t1 = t2 = -f / (2 * e); sol.pb(L.point(t1));
172         return 1;
173     }
174     t1 = (-f - sqrt(delta)) / (2 * e); sol.pb(L.point(t1));
175     t2 = (-f + sqrt(delta)) / (2 * e); sol.pb(L.point(t2));
176     return 2;
177 }
178 double angle(Vector v){
179     return atan2(v.y, v.x);
180     //(-pi, pi]
181 }
182 int GetCircleCircleIntersection(Circle C1, Circle C2, vector<Point>& sol){
183     double d = Len(C1.c - C2.c);
184     if(dcmp(d) == 0){
185         if(dcmp(C1.r - C2.r) == 0) return -1; //two circle duplicates
186         return 0; //two circles share identical center
187     }
188     if(dcmp(C1.r + C2.r - d) < 0) return 0; //too close
189     if(dcmp(abs(C1.r - C2.r) - d) > 0) return 0; //too far away
190     double a = angle(C2.c - C1.c); // angle of vector(C1, C2)
191     double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
192     Point p1 = C1.point(a - da), p2 = C1.point(a + da);
193     sol.pb(p1);
194     if(p1 == p2) return 1;
195     sol.pb(p2);
196     return 2;
197 }
198 int GetPointCircleTangents(Point p, Circle C, Vector* v){
199     Vector u = C.c - p;
200     double dist = Len(u);
201     if(dist < C.r) return 0;//p is inside the circle, no tangents
202     else if(dcmp(dist - C.r) == 0){
203         // p is on the circles, one tangent only
204         v[0] = Rotate(u, PI / 2);
205         return 1;
206     }else{
207         double ang = asin(C.r / dist);
208         v[0] = Rotate(u, -ang);
209         v[1] = Rotate(u, +ang);
210         return 2;
211     }
212 }
213 int GetCircleCircleTangents(Circle A, Circle B, Point* a, Point* b){
214     //a[i] store point of tangency on Circle A of tangent i
215     //b[i] store point of tangency on Circle B of tangent i
216     //six conditions is in consideration
217     int cnt = 0;
218     if(A.r < B.r) { swap(A, B); swap(a, b); }
219     int d2 = (A.c.x - B.c.x) * (A.c.x - B.c.x) + (A.c.y - B.c.y) * (A.c.y - B.c.y);
220     int rdiff = A.r - B.r;
221     int rsum = A.r + B.r;
222     if(d2 < rdiff * rdiff) return 0; // one circle is inside the other
223     double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x);
224     if(d2 == 0 && A.r == B.r) return -1; // two circle duplicates
225     if(d2 == rdiff * rdiff){ // internal tangency
226         a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++;
227         return 1;
228     }
229     double ang = acos((A.r - B.r) / sqrt(d2));
230     a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang);
231     a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang);
232     if(d2 == rsum * rsum){
233         //one internal tangent
234         a[cnt] = A.point(base);
235         b[cnt++] = B.point(base + PI);
236     }else if(d2 > rsum * rsum){
237         //two internal tangents
238         double ang = acos((A.r + B.r) / sqrt(d2));
239         a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang + PI);
240         a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang + PI);
241     }
242     return cnt;
243 }
244 Point ReadPoint(){
245     double x, y;
246     scanf("%lf%lf", &x, &y);
247     return Point(x, y);
248 }
249 Circle ReadCircle(){
250     double x, y, r;
251     scanf("%lf%lf%lf", &x, &y, &r);
252     return Circle(x, y, r);
253 }
254 //Here goes 3d geometry templates
255 struct Point3{
256     double x, y, z;
257     Point3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) {}
258 };
259 typedef Point3 Vector3;
260 Vector3 operator + (Vector3 A, Vector3 B){
261     return Vector3(A.x + B.x, A.y + B.y, A.z + B.z);
262 }
263 Vector3 operator - (Vector3 A, Vector3 B){
264     return Vector3(A.x - B.x, A.y - B.y, A.z - B.z);
265 }
266 Vector3 operator * (Vector3 A, double p){
267     return Vector3(A.x * p, A.y * p, A.z * p);
268 }
269 Vector3 operator / (Vector3 A, double p){
270     return Vector3(A.x / p, A.y / p, A.z / p);
271 }
272 double Dot3(Vector3 A, Vector3 B){
273     return A.x * B.x + A.y * B.y + A.z * B.z;
274 }
275 double Len3(Vector3 A){
276     return sqrt(Dot3(A, A));
277 }
278 double Angle3(Vector3 A, Vector3 B){
279     return acos(Dot3(A, B) / Len3(A) / Len3(B));
280 }
281 double DistanceToPlane(const Point3& p, const Point3 &p0, const Vector3& n){
282     return abs(Dot3(p - p0, n));
283 }
284 Point3 GetPlaneProjection(const Point3 &p, const Point3 &p0, const Vector3 &n){
285     return p - n * Dot3(p - p0, n);
286 }
287 Point3 GetLinePlaneIntersection(Point3 p1, Point3 p2, Point3 p0, Vector3 n){
288     Vector3 v = p2 - p1;
289     double t = (Dot3(n, p0 - p1) / Dot3(n, p2 - p1));
290     return p1 + v * t;//if t in range [0, 1], intersection on segment
291 }
292 Vector3 Cross(Vector3 A, Vector3 B){
293     return Vector3(A.y * B.z - A.z * B.y, A.z * B.x - A.x * B.z, A.x * B.y - A.y * B.x);
294 }
295 double Area3(Point3 A, Point3 B, Point3 C){
296     return Len3(Cross(B - A, C - A));
297 }
298 class cmpt{
299 public:
300     bool operator () (const int &x, const int &y) const{
301         return x > y;
302     }
303 };
304
305 int Rand(int x, int o){
306     //if o set, return [1, x], else return [0, x - 1]
307     if(!x) return 0;
308     int tem = (int)((double)rand() / RAND_MAX * x) % x;
309     return o ? tem + 1 : tem;
310 }
311 ////////////////////////////////////////////////////////////////////////////////////
312 ////////////////////////////////////////////////////////////////////////////////////
313 void data_gen(){
314     srand(time(0));
315     freopen("in.txt", "w", stdout);
316     int times = 100;
317     printf("%d\n", times);
318     while(times--){
319         int r = Rand(1000, 1), a = Rand(1000, 1), c = Rand(1000, 1);
320         int b = Rand(r, 1), d = Rand(r, 1);
321         int m = Rand(100, 1), n = Rand(m, 1);
322         printf("%d %d %d %d %d %d %d\n", n, m, a, b, c, d, r);
323     }
324 }
325
326 struct cmpx{
327     bool operator () (int x, int y) { return x > y; }
328 };
329 int debug = 1;
330 int dx[] = {0, 0, 1, 1};
331 int dy[] = {1, 0, 0, 1};
332 //-------------------------------------------------------------------------
333 ll multi(ll x, ll y, ll mod){
334     ll lhs = x * (y >> 20) % mod * (1ll << 20) % mod;
335     ll rhs = x * (y & ((1ll << 20) - 1)) % mod;
336     return (lhs + rhs) % mod;
337 }
338
339 ll __power(ll a, ll p, ll mod){
340     ll ans = 1;
341     a %= mod;
342     while(p){
343         if(p & 1) ans = (ans * a) % mod;
344         p >>= 1;
345         a = (a * a) % mod;
346     }
347     return ans;
348 }
349
350 ll power(ll a, ll p, ll mod){
351     ll ans = 1;
352     while(p){
353         if(p & 1) ans = multi(ans, a, mod);
354         p >>= 1;
355         a = multi(a, a, mod);
356     }
357     return ans % mod;
358 }
359
360 ll n;
361 ll ans[50000 + 10];
362 bool ok;
363 void dfs(int pos, ll pre, ll mod){
364     if(pos == 12){
365         if(pre < 1e11 || power(n, pre, mod) != pre) return;
366         ans[n] = pre;
367         ok = 1;
368         return;
369     }
370     FOR(i, 0, 9){
371         ll nex = mod * i + pre;
372         //try to set next position as i
373         ll lhs = power(n, nex, mod);
374         ll rhs = nex % mod;
375         if(lhs != rhs) continue;
376         dfs(pos + 1, nex, mod * 10);
377         if(ok) return;
378     }
379 }
380 void solve(){
381     if(ans[n] != -1) return;
382     ok = 0;
383     dfs(0, 0, 1);
384 }
385
386 //-------------------------------------------------------------------------
387 int main(){
388     //data_gen(); return 0;
389     //C(); return 0;
390     debug = 0;
391     ///////////////////////////////////////////////////////////////////////////////////////////////////////////////
392     if(debug) freopen("in.txt", "r", stdin);
393     //freopen("in.txt", "w", stdout);
394     int kase = 0;
395     clr(ans, -1);
396     while(~scanf("%lld", &n) && n){
397         solve();
398         printf("Case %d: Public Key = %lld Private Key = %lld\n", ++kase, n, ans[n]);
399     }
400     //////////////////////////////////////////////////////////////////////////////////////////////////////////////
401     return 0;
402 }

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