hdu 5780 gcd
题意：给定$x, n$满足$1 \leq x, n \leq 1000000$，求$\sum{(x^a-1,x^b-1)}$对$1e9+7$取模后的值，其中$1 \leq a, b \leq n$。
分析：首先不难有$(x^a - 1, x ^ b - 1) = x^{(a,b)}-1$（证明方法可沿欧几里得定理思路），那么我们只需要考虑$(a,b) = d$即可，设$f(d)$为使得$(a, b) = d$的对数，那么不难有$ans = \sum_{d = 1}^{n}{f(d)(x^d-1)}$。下面考虑计算$f(d)$，由于$(a, b) = d$满足$d \mid a, d \mid b, (\frac{a}{d}, \frac{b}{d}) = 1$，于是对于满足$1\leq a \leq b \leq n$的序对$(a, b)$，其总数为$\sum_{i=1}^{\frac{n}{d}}{\varphi(i)}$，因为我们可以在互质的对上乘以系数$d$使其映射到对$(a, b)$。那么对于题目中的要求，总数为$2\sum_{i=1}^{\frac{n}{d}}{\varphi(i)}-1$，那么我们就只用到了欧拉函数的前缀和，注意到$f(d)$只与$\frac{n}{d}$有关，而右边可以使用等比求和，于是可以分段计算，每段对应的$\frac{n}{d}$固定。那么这样做预处理是$O(n)$的，每次询问的复杂度是$O(\sqrt{n})$的。代码如下：
1 #include <algorithm> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <queue> 6 #include <map> 7 #include <set> 8 #include <stack> 9 #include <ctime> 10 #include <cmath> 11 #include <iostream> 12 #include <assert.h> 13 #define PI acos(-1.) 14 #pragma comment(linker, "/STACK:102400000,102400000") 15 #define max(a, b) ((a) > (b) ? (a) : (b)) 16 #define min(a, b) ((a) < (b) ? (a) : (b)) 17 #define mp std :: make_pair 18 #define st first 19 #define nd second 20 #define keyn (root->ch[1]->ch[0]) 21 #define lson (u << 1) 22 #define rson (u << 1 | 1) 23 #define pii std :: pair<int, int> 24 #define pll pair<ll, ll> 25 #define pb push_back 26 #define type(x) __typeof(x.begin()) 27 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++) 28 #define FOR(i, s, t) for(int i = (s); i <= (t); i++) 29 #define ROF(i, t, s) for(int i = (t); i >= (s); i--) 30 #define dbg(x) std::cout << x << std::endl 31 #define dbg2(x, y) std::cout << x << " " << y << std::endl 32 #define clr(x, i) memset(x, (i), sizeof(x)) 33 #define maximize(x, y) x = max((x), (y)) 34 #define minimize(x, y) x = min((x), (y)) 35 using namespace std; 36 typedef long long ll; 37 const int int_inf = 0x3f3f3f3f; 38 const ll ll_inf = 0x3f3f3f3f3f3f3f3f; 39 const int INT_INF = (int)((1ll << 31) - 1); 40 const double double_inf = 1e30; 41 const double eps = 1e-14; 42 typedef unsigned long long ul; 43 typedef unsigned int ui; 44 inline int readint(){ 45 int x; 46 scanf("%d", &x); 47 return x; 48 } 49 inline int readstr(char *s){ 50 scanf("%s", s); 51 return strlen(s); 52 } 53 //Here goes 2d geometry templates 54 struct Point{ 55 double x, y; 56 Point(double x = 0, double y = 0) : x(x), y(y) {} 57 }; 58 typedef Point Vector; 59 Vector operator + (Vector A, Vector B){ 60 return Vector(A.x + B.x, A.y + B.y); 61 } 62 Vector operator - (Point A, Point B){ 63 return Vector(A.x - B.x, A.y - B.y); 64 } 65 Vector operator * (Vector A, double p){ 66 return Vector(A.x * p, A.y * p); 67 } 68 Vector operator / (Vector A, double p){ 69 return Vector(A.x / p, A.y / p); 70 } 71 bool operator < (const Point& a, const Point& b){ 72 return a.x < b.x || (a.x == b.x && a.y < b.y); 73 } 74 int dcmp(double x){ 75 if(abs(x) < eps) return 0; 76 return x < 0 ? -1 : 1; 77 } 78 bool operator == (const Point& a, const Point& b){ 79 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; 80 } 81 double Dot(Vector A, Vector B){ 82 return A.x * B.x + A.y * B.y; 83 } 84 double Len(Vector A){ 85 return sqrt(Dot(A, A)); 86 } 87 double Angle(Vector A, Vector B){ 88 return acos(Dot(A, B) / Len(A) / Len(B)); 89 } 90 double Cross(Vector A, Vector B){ 91 return A.x * B.y - A.y * B.x; 92 } 93 double Area2(Point A, Point B, Point C){ 94 return Cross(B - A, C - A); 95 } 96 Vector Rotate(Vector A, double rad){ 97 //rotate counterclockwise 98 return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); 99 } 100 Vector Normal(Vector A){ 101 double L = Len(A); 102 return Vector(-A.y / L, A.x / L); 103 } 104 void Normallize(Vector &A){ 105 double L = Len(A); 106 A.x /= L, A.y /= L; 107 } 108 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){ 109 Vector u = P - Q; 110 double t = Cross(w, u) / Cross(v, w); 111 return P + v * t; 112 } 113 double DistanceToLine(Point P, Point A, Point B){ 114 Vector v1 = B - A, v2 = P - A; 115 return abs(Cross(v1, v2)) / Len(v1); 116 } 117 double DistanceToSegment(Point P, Point A, Point B){ 118 if(A == B) return Len(P - A); 119 Vector v1 = B - A, v2 = P - A, v3 = P - B; 120 if(dcmp(Dot(v1, v2)) < 0) return Len(v2); 121 else if(dcmp(Dot(v1, v3)) > 0) return Len(v3); 122 else return abs(Cross(v1, v2)) / Len(v1); 123 } 124 Point GetLineProjection(Point P, Point A, Point B){ 125 Vector v = B - A; 126 return A + v * (Dot(v, P - A) / Dot(v, v)); 127 } 128 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){ 129 //Line1:(a1, a2) Line2:(b1,b2) 130 double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1), 131 c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1); 132 return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0; 133 } 134 bool OnSegment(Point p, Point a1, Point a2){ 135 return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 -p)) < 0; 136 } 137 Vector GetBisector(Vector v, Vector w){ 138 Normallize(v), Normallize(w); 139 return Vector((v.x + w.x) / 2, (v.y + w.y) / 2); 140 } 141 142 bool OnLine(Point p, Point a1, Point a2){ 143 Vector v1 = p - a1, v2 = a2 - a1; 144 double tem = Cross(v1, v2); 145 return dcmp(tem) == 0; 146 } 147 struct Line{ 148 Point p; 149 Vector v; 150 Point point(double t){ 151 return Point(p.x + t * v.x, p.y + t * v.y); 152 } 153 Line(Point p, Vector v) : p(p), v(v) {} 154 }; 155 struct Circle{ 156 Point c; 157 double r; 158 Circle(Point c, double r) : c(c), r(r) {} 159 Circle(int x, int y, int _r){ 160 c = Point(x, y); 161 r = _r; 162 } 163 Point point(double a){ 164 return Point(c.x + cos(a) * r, c.y + sin(a) * r); 165 } 166 }; 167 int GetLineCircleIntersection(Line L, Circle C, double &t1, double& t2, std :: vector<Point>& sol){ 168 double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y; 169 double e = a * a + c * c, f = 2 * (a * b + c * d), g = b * b + d * d - C.r * C.r; 170 double delta = f * f - 4 * e * g; 171 if(dcmp(delta) < 0) return 0; 172 if(dcmp(delta) == 0){ 173 t1 = t2 = -f / (2 * e); sol.pb(L.point(t1)); 174 return 1; 175 } 176 t1 = (-f - sqrt(delta)) / (2 * e); sol.pb(L.point(t1)); 177 t2 = (-f + sqrt(delta)) / (2 * e); sol.pb(L.point(t2)); 178 return 2; 179 } 180 double angle(Vector v){ 181 return atan2(v.y, v.x); 182 //(-pi, pi] 183 } 184 int GetCircleCircleIntersection(Circle C1, Circle C2, std :: vector<Point>& sol){ 185 double d = Len(C1.c - C2.c); 186 if(dcmp(d) == 0){ 187 if(dcmp(C1.r - C2.r) == 0) return -1; //two circle duplicates 188 return 0; //two circles share identical center 189 } 190 if(dcmp(C1.r + C2.r - d) < 0) return 0; //too close 191 if(dcmp(abs(C1.r - C2.r) - d) > 0) return 0; //too far away 192 double a = angle(C2.c - C1.c); // angle of vector(C1, C2) 193 double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); 194 Point p1 = C1.point(a - da), p2 = C1.point(a + da); 195 sol.pb(p1); 196 if(p1 == p2) return 1; 197 sol.pb(p2); 198 return 2; 199 } 200 int GetPointCircleTangents(Point p, Circle C, Vector* v){ 201 Vector u = C.c - p; 202 double dist = Len(u); 203 if(dist < C.r) return 0;//p is inside the circle, no tangents 204 else if(dcmp(dist - C.r) == 0){ 205 // p is on the circles, one tangent only 206 v[0] = Rotate(u, PI / 2); 207 return 1; 208 }else{ 209 double ang = asin(C.r / dist); 210 v[0] = Rotate(u, -ang); 211 v[1] = Rotate(u, +ang); 212 return 2; 213 } 214 } 215 int GetCircleCircleTangents(Circle A, Circle B, Point* a, Point* b){ 216 //a[i] store point of tangency on Circle A of tangent i 217 //b[i] store point of tangency on Circle B of tangent i 218 //six conditions is in consideration 219 int cnt = 0; 220 if(A.r < B.r) { std :: swap(A, B); std :: swap(a, b); } 221 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); 222 int rdiff = A.r - B.r; 223 int rsum = A.r + B.r; 224 if(d2 < rdiff * rdiff) return 0; // one circle is inside the other 225 double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x); 226 if(d2 == 0 && A.r == B.r) return -1; // two circle duplicates 227 if(d2 == rdiff * rdiff){ // internal tangency 228 a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++; 229 return 1; 230 } 231 double ang = acos((A.r - B.r) / sqrt(d2)); 232 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang); 233 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang); 234 if(d2 == rsum * rsum){ 235 //one internal tangent 236 a[cnt] = A.point(base); 237 b[cnt++] = B.point(base + PI); 238 }else if(d2 > rsum * rsum){ 239 //two internal tangents 240 double ang = acos((A.r + B.r) / sqrt(d2)); 241 a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang + PI); 242 a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang + PI); 243 } 244 return cnt; 245 } 246 Point ReadPoint(){ 247 double x, y; 248 scanf("%lf%lf", &x, &y); 249 return Point(x, y); 250 } 251 Circle ReadCircle(){ 252 double x, y, r; 253 scanf("%lf%lf%lf", &x, &y, &r); 254 return Circle(x, y, r); 255 } 256 //Here goes 3d geometry templates 257 struct Point3{ 258 double x, y, z; 259 Point3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) {} 260 }; 261 typedef Point3 Vector3; 262 Vector3 operator + (Vector3 A, Vector3 B){ 263 return Vector3(A.x + B.x, A.y + B.y, A.z + B.z); 264 } 265 Vector3 operator - (Vector3 A, Vector3 B){ 266 return Vector3(A.x - B.x, A.y - B.y, A.z - B.z); 267 } 268 Vector3 operator * (Vector3 A, double p){ 269 return Vector3(A.x * p, A.y * p, A.z * p); 270 } 271 Vector3 operator / (Vector3 A, double p){ 272 return Vector3(A.x / p, A.y / p, A.z / p); 273 } 274 double Dot3(Vector3 A, Vector3 B){ 275 return A.x * B.x + A.y * B.y + A.z * B.z; 276 } 277 double Len3(Vector3 A){ 278 return sqrt(Dot3(A, A)); 279 } 280 double Angle3(Vector3 A, Vector3 B){ 281 return acos(Dot3(A, B) / Len3(A) / Len3(B)); 282 } 283 double DistanceToPlane(const Point3& p, const Point3 &p0, const Vector3& n){ 284 return abs(Dot3(p - p0, n)); 285 } 286 Point3 GetPlaneProjection(const Point3 &p, const Point3 &p0, const Vector3 &n){ 287 return p - n * Dot3(p - p0, n); 288 } 289 Point3 GetLinePlaneIntersection(Point3 p1, Point3 p2, Point3 p0, Vector3 n){ 290 Vector3 v = p2 - p1; 291 double t = (Dot3(n, p0 - p1) / Dot3(n, p2 - p1)); 292 return p1 + v * t;//if t in range [0, 1], intersection on segment 293 } 294 Vector3 Cross(Vector3 A, Vector3 B){ 295 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); 296 } 297 double Area3(Point3 A, Point3 B, Point3 C){ 298 return Len3(Cross(B - A, C - A)); 299 } 300 class cmpt{ 301 public: 302 bool operator () (const int &x, const int &y) const{ 303 return x > y; 304 } 305 }; 306 307 int Rand(int x, int o){ 308 //if o set, return [1, x], else return [0, x - 1] 309 if(!x) return 0; 310 int tem = (int)((double)rand() / RAND_MAX * x) % x; 311 return o ? tem + 1 : tem; 312 } 313 void data_gen(){ 314 srand(time(0)); 315 freopen("in.txt", "w", stdout); 316 int kases = 10; 317 printf("%d\n", kases); 318 while(kases--){ 319 int sz = 2e4; 320 int m = 1e5; 321 printf("%d %d\n", sz, m); 322 FOR(i, 1, sz) printf("%d ", Rand(100, 1)); 323 printf("\n"); 324 FOR(i, 1, sz) printf("%d ", Rand(1e9, 1)); 325 printf("\n"); 326 FOR(i, 1, m){ 327 int l = Rand(sz, 1); 328 int r = Rand(sz, 1); 329 int c = Rand(1e9, 1); 330 printf("%d %d %d %d\n", l, r, c, Rand(100, 1)); 331 } 332 } 333 } 334 335 struct cmpx{ 336 bool operator () (int x, int y) { return x > y; } 337 }; 338 const int mod = 1e9 + 7; 339 const int maxn = 1e6 + 10; 340 ll phi[maxn]; 341 void init_miu(){ 342 clr(phi, 0); 343 phi[1] = 1; 344 FOR(i, 2, maxn - 1) if(!phi[i]){ 345 for(int j = i; j < maxn; j += i){ 346 if(!phi[j]) phi[j] = j; 347 phi[j] = phi[j] / i * (i - 1); 348 } 349 } 350 FOR(i, 1, maxn - 1) phi[i] = (phi[i] + phi[i - 1]) % mod; 351 } 352 ll power(ll a, ll p, ll mod){ 353 ll res = 1; 354 a %= mod; 355 while(p){ 356 if(p & 1) res = res * a % mod; 357 p >>= 1; 358 a = a * a % mod; 359 } 360 return res; 361 } 362 363 ll cal2(int x, int n){ 364 ll ans = 0; 365 int p = 1; 366 while(p <= n){ 367 int np = n / (n / p); 368 int delta = np - p + 1; 369 ll cnt = (2 * phi[n / p] + mod - 1) % mod; 370 ll lhs = 0; 371 if(x > 1){ 372 lhs = power(x, p, mod) * (power(x, delta, mod) + mod - 1) % mod; 373 lhs = lhs * power(x - 1, mod - 2, mod) % mod; 374 lhs = (lhs - delta + mod) % mod; 375 } 376 ans = (ans + lhs * cnt % mod) % mod; 377 p = np + 1; 378 } 379 return ans; 380 } 381 int main(){ 382 //data_gen(); return 0; 383 //C(); return 0; 384 int debug = 0; 385 if(debug) freopen("in.txt", "r", stdin); 386 //freopen("out.txt", "w", stdout); 387 init_miu(); 388 int T = readint(); 389 while(T--){ 390 int x = readint(), n = readint(); 391 printf("%lld\n", cal2(x, n)); 392 } 393 return 0; 394 }
当然本题我用莫比乌斯反演也勉强通过了（用了900多ms，仅仅通过了1次），使用了两次分段求和，每次询问的复杂度介于$O(\sqrt{n})$与$O(n)$之间。虽然不是正解，仍然可以看出欧拉函数与莫比乌斯函数之间关系相通。
