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：

 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <map>
 #include <set>
 #include <stack>
 #include <ctime>
 #include <cmath>
 #include <iostream>
 #include <assert.h>
 #define PI acos(-1.)
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define max(a, b) ((a) > (b) ? (a) : (b))
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define mp std :: make_pair
 #define st first
 #define nd second
 #define keyn (root->ch[1]->ch[0])
 #define lson (u << 1)
 #define rson (u << 1 | 1)
 #define pii std :: pair<int, int>
 #define pll pair<ll, ll>
 #define pb push_back
 #define type(x) __typeof(x.begin())
 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++)
 #define FOR(i, s, t) for(int i = (s); i <= (t); i++)
 #define ROF(i, t, s) for(int i = (t); i >= (s); i--)
 #define dbg(x) std::cout << x << std::endl
 #define dbg2(x, y) std::cout << x << " " << y << std::endl
 #define clr(x, i) memset(x, (i), sizeof(x))
 #define maximize(x, y) x = max((x), (y))
 #define minimize(x, y) x = min((x), (y))
 using namespace std;
 typedef long long ll;
 const int int_inf = 0x3f3f3f3f;
 const ll ll_inf = 0x3f3f3f3f3f3f3f3f;
 const int INT_INF = (int)((1ll << ) - );
 const double double_inf = 1e30;
 const double eps = 1e-;
 typedef unsigned long long ul;
 typedef unsigned int ui;
 inline int readint(){
     int x;
     scanf("%d", &x);
     return x;
 }
 inline int readstr(char *s){
     scanf("%s", s);
     return strlen(s);
 }
 //Here goes 2d geometry templates
 struct Point{
     double x, y;
     Point(double x = , double y = ) : x(x), y(y) {}
 };
 typedef Point Vector;
 Vector operator + (Vector A, Vector B){
     return Vector(A.x + B.x, A.y + B.y);
 }
 Vector operator - (Point A, Point B){
     return Vector(A.x - B.x, A.y - B.y);
 }
 Vector operator * (Vector A, double p){
     return Vector(A.x * p, A.y * p);
 }
 Vector operator / (Vector A, double p){
     return Vector(A.x / p, A.y / p);
 }
 bool operator < (const Point& a, const Point& b){
     return a.x < b.x || (a.x == b.x && a.y < b.y);
 }
 int dcmp(double x){
     if(abs(x) < eps) return ;
     return x <  ? - : ;
 }
 bool operator == (const Point& a, const Point& b){
     return dcmp(a.x - b.x) ==  && dcmp(a.y - b.y) == ;
 }
 double Dot(Vector A, Vector B){
     return A.x * B.x + A.y * B.y;
 }
 double Len(Vector A){
     return sqrt(Dot(A, A));
 }
 double Angle(Vector A, Vector B){
     return acos(Dot(A, B) / Len(A) / Len(B));
 }
 double Cross(Vector A, Vector B){
     return A.x * B.y - A.y * B.x;
 }
 double Area2(Point A, Point B, Point C){
     return Cross(B - A, C - A);
 }
 Vector Rotate(Vector A, double rad){
     //rotate counterclockwise
     return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
 }
 Vector Normal(Vector A){
     double L = Len(A);
     return Vector(-A.y / L, A.x / L);
 }
 void Normallize(Vector &A){
     double L = Len(A);
     A.x /= L, A.y /= L;
 }
 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w){
     Vector u = P - Q;
     double t = Cross(w, u) / Cross(v, w);
     return P + v * t;
 }
 double DistanceToLine(Point P, Point A, Point B){
     Vector v1 = B - A, v2 = P - A;
     return abs(Cross(v1, v2)) / Len(v1);
 }
 double DistanceToSegment(Point P, Point A, Point B){
     if(A == B) return Len(P - A);
     Vector v1 = B - A, v2 = P - A, v3 = P - B;
     if(dcmp(Dot(v1, v2)) < ) return Len(v2);
     else if(dcmp(Dot(v1, v3)) > ) return Len(v3);
     else return abs(Cross(v1, v2)) / Len(v1);
 }
 Point GetLineProjection(Point P, Point A, Point B){
     Vector v = B - A;
     return A + v * (Dot(v, P - A) / Dot(v, v));
 }
 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2){
     //Line1:(a1, a2) Line2:(b1,b2)
     double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
            c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
     return dcmp(c1) * dcmp(c2) <  && dcmp(c3) * dcmp(c4) < ;
 }
 bool OnSegment(Point p, Point a1, Point a2){
     return dcmp(Cross(a1 - p, a2 - p)) ==  && dcmp(Dot(a1 - p, a2 -p)) < ;
 }
 Vector GetBisector(Vector v, Vector w){
     Normallize(v), Normallize(w);
     return Vector((v.x + w.x) / , (v.y + w.y) / );
 }

 bool OnLine(Point p, Point a1, Point a2){
     Vector v1 = p - a1, v2 = a2 - a1;
     double tem = Cross(v1, v2);
     return dcmp(tem) == ;
 }
 struct Line{
     Point p;
     Vector v;
     Point point(double t){
         return Point(p.x + t * v.x, p.y + t * v.y);
     }
     Line(Point p, Vector v) : p(p), v(v) {}
 };
 struct Circle{
     Point c;
     double r;
     Circle(Point c, double r) : c(c), r(r) {}
     Circle(int x, int y, int _r){
         c = Point(x, y);
         r = _r;
     }
     Point point(double a){
         return Point(c.x + cos(a) * r, c.y + sin(a) * r);
     }
 };
 int GetLineCircleIntersection(Line L, Circle C, double &t1, double& t2, std :: vector<Point>& sol){
     double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y - C.c.y;
     double e = a * a + c * c, f =  * (a * b + c * d), g = b * b + d * d - C.r * C.r;
     double delta = f * f -  * e * g;
     if(dcmp(delta) < ) return ;
     if(dcmp(delta) == ){
         t1 = t2 = -f / ( * e); sol.pb(L.point(t1));
         return ;
     }
     t1 = (-f - sqrt(delta)) / ( * e); sol.pb(L.point(t1));
     t2 = (-f + sqrt(delta)) / ( * e); sol.pb(L.point(t2));
     return ;
 }
 double angle(Vector v){
     return atan2(v.y, v.x);
     //(-pi, pi]
 }
 int GetCircleCircleIntersection(Circle C1, Circle C2, std :: vector<Point>& sol){
     double d = Len(C1.c - C2.c);
     if(dcmp(d) == ){
         if(dcmp(C1.r - C2.r) == ) return -; //two circle duplicates
         return ; //two circles share identical center
     }
     if(dcmp(C1.r + C2.r - d) < ) return ; //too close
     if(dcmp(abs(C1.r - C2.r) - d) > ) return ; //too far away
     double a = angle(C2.c - C1.c); // angle of vector(C1, C2)
     double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / ( * C1.r * d));
     Point p1 = C1.point(a - da), p2 = C1.point(a + da);
     sol.pb(p1);
     if(p1 == p2) return ;
     sol.pb(p2);
     return ;
 }
 int GetPointCircleTangents(Point p, Circle C, Vector* v){
     Vector u = C.c - p;
     double dist = Len(u);
     if(dist < C.r) return ;//p is inside the circle, no tangents
     else if(dcmp(dist - C.r) == ){
         // p is on the circles, one tangent only
         v[] = Rotate(u, PI / );
         return ;
     }else{
         double ang = asin(C.r / dist);
         v[] = Rotate(u, -ang);
         v[] = Rotate(u, +ang);
         return ;
     }
 }
 int GetCircleCircleTangents(Circle A, Circle B, Point* a, Point* b){
     //a[i] store point of tangency on Circle A of tangent i
     //b[i] store point of tangency on Circle B of tangent i
     //six conditions is in consideration
     int cnt = ;
     if(A.r < B.r) { std :: swap(A, B); std :: swap(a, b); }
     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);
     int rdiff = A.r - B.r;
     int rsum = A.r + B.r;
     if(d2 < rdiff * rdiff) return ; // one circle is inside the other
     double base = atan2(B.c.y - A.c.y, B.c.x - A.c.x);
     if(d2 ==  && A.r == B.r) return -; // two circle duplicates
     if(d2 == rdiff * rdiff){ // internal tangency
         a[cnt] = A.point(base); b[cnt] = B.point(base); cnt++;
         return ;
     }
     double ang = acos((A.r - B.r) / sqrt(d2));
     a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang);
     a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang);
     if(d2 == rsum * rsum){
         //one internal tangent
         a[cnt] = A.point(base);
         b[cnt++] = B.point(base + PI);
     }else if(d2 > rsum * rsum){
         //two internal tangents
         double ang = acos((A.r + B.r) / sqrt(d2));
         a[cnt] = A.point(base + ang); b[cnt++] = B.point(base + ang + PI);
         a[cnt] = A.point(base - ang); b[cnt++] = B.point(base - ang + PI);
     }
     return cnt;
 }
 Point ReadPoint(){
     double x, y;
     scanf("%lf%lf", &x, &y);
     return Point(x, y);
 }
 Circle ReadCircle(){
     double x, y, r;
     scanf("%lf%lf%lf", &x, &y, &r);
     return Circle(x, y, r);
 }
 //Here goes 3d geometry templates
 struct Point3{
     double x, y, z;
     Point3(double x = , double y = , double z = ) : x(x), y(y), z(z) {}
 };
 typedef Point3 Vector3;
 Vector3 operator + (Vector3 A, Vector3 B){
     return Vector3(A.x + B.x, A.y + B.y, A.z + B.z);
 }
 Vector3 operator - (Vector3 A, Vector3 B){
     return Vector3(A.x - B.x, A.y - B.y, A.z - B.z);
 }
 Vector3 operator * (Vector3 A, double p){
     return Vector3(A.x * p, A.y * p, A.z * p);
 }
 Vector3 operator / (Vector3 A, double p){
     return Vector3(A.x / p, A.y / p, A.z / p);
 }
 double Dot3(Vector3 A, Vector3 B){
     return A.x * B.x + A.y * B.y + A.z * B.z;
 }
 double Len3(Vector3 A){
     return sqrt(Dot3(A, A));
 }
 double Angle3(Vector3 A, Vector3 B){
     return acos(Dot3(A, B) / Len3(A) / Len3(B));
 }
 double DistanceToPlane(const Point3& p, const Point3 &p0, const Vector3& n){
     return abs(Dot3(p - p0, n));
 }
 Point3 GetPlaneProjection(const Point3 &p, const Point3 &p0, const Vector3 &n){
     return p - n * Dot3(p - p0, n);
 }
 Point3 GetLinePlaneIntersection(Point3 p1, Point3 p2, Point3 p0, Vector3 n){
     Vector3 v = p2 - p1;
     double t = (Dot3(n, p0 - p1) / Dot3(n, p2 - p1));
     return p1 + v * t;//if t in range [0, 1], intersection on segment
 }
 Vector3 Cross(Vector3 A, Vector3 B){
     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);
 }
 double Area3(Point3 A, Point3 B, Point3 C){
     return Len3(Cross(B - A, C - A));
 }
 class cmpt{
 public:
     bool operator () (const int &x, const int &y) const{
         return x > y;
     }
 };

 int Rand(int x, int o){
     //if o set, return [1, x], else return [0, x - 1]
     if(!x) return ;
     int tem = (int)((double)rand() / RAND_MAX * x) % x;
     return o ? tem +  : tem;
 }
 void data_gen(){
     srand(time());
     freopen("in.txt", "w", stdout);
     int kases = ;
     printf("%d\n", kases);
     while(kases--){
         int sz = 2e4;
         int m = 1e5;
         printf("%d %d\n", sz, m);
         FOR(i, , sz) printf("%d ", Rand(, ));
         printf("\n");
         FOR(i, , sz) printf("%d ", Rand(1e9, ));
         printf("\n");
         FOR(i, , m){
             int l = Rand(sz, );
             int r = Rand(sz, );
             int c = Rand(1e9, );
             printf("%d %d %d %d\n", l, r, c, Rand(, ));
         }
     }
 }

 struct cmpx{
     bool operator () (int x, int y) { return x > y; }
 };
 const int mod = 1e9 + ;
 const int maxn = 1e6 + ;
 ll phi[maxn];
 void init_miu(){
     clr(phi, );
     phi[] = ;
     FOR(i, , maxn - ) if(!phi[i]){
         for(int j = i; j < maxn; j += i){
             if(!phi[j]) phi[j] = j;
             phi[j] = phi[j] / i * (i - );
         }
     }
     FOR(i, , maxn - ) phi[i] = (phi[i] + phi[i - ]) % mod;
 }
 ll power(ll a, ll p, ll mod){
     ll res = ;
     a %= mod;
     while(p){
         if(p & ) res = res * a % mod;
         p >>= ;
         a = a * a % mod;
     }
     return res;
 }

 ll cal2(int x, int n){
     ll ans = ;
     int p = ;
     while(p <= n){
         int np = n / (n / p);
         int delta = np - p + ;
         ll cnt = ( * phi[n / p] + mod - ) % mod;
         ll lhs = ;
         if(x > ){
             lhs = power(x, p, mod) * (power(x, delta, mod) + mod - ) % mod;
             lhs = lhs * power(x - , mod - , mod) % mod;
             lhs = (lhs - delta + mod) % mod;
         }
         ans = (ans + lhs * cnt % mod) % mod;
         p = np + ;
     }
     return ans;
 }
 int main(){
     //data_gen(); return 0;
     //C(); return 0;
     int debug = ;
     if(debug) freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
     init_miu();
     int T = readint();
     while(T--){
         int x = readint(), n = readint();
         printf("%lld\n", cal2(x, n));
     }
     return ;
 }

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。