• 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$1leq a leq b leq n$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$2sum_{i=1}^{frac{n}{d}}{varphi(i)}-1$,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
    ", kases);
    318     while(kases--){
    319         int sz = 2e4;
    320         int m = 1e5;
    321         printf("%d %d
    ", sz, m);
    322         FOR(i, 1, sz) printf("%d ", Rand(100, 1));
    323         printf("
    ");
    324         FOR(i, 1, sz) printf("%d ", Rand(1e9, 1));
    325         printf("
    ");
    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
    ", 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
    ", 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。

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  • 原文地址:https://www.cnblogs.com/astoninfer/p/5772431.html
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