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 }
code：
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。
时间： 2024-10-21 16:31:42