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 }
