莫比乌斯反演
资料 http://wenku.baidu.com/view/542961fdba0d4a7302763ad5.html
还有一些就看百度百科或者其他人的blog。并不是很懂
SPOJ VLATTICE Visible Lattice Points
POJ 3090是二维的，这个是三维版本。二维的可以用欧拉函数做。三维用莫比乌斯反演做，具体解释看 这里。我并没有很懂，感觉莫比乌斯反演好牛逼啊
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define lson l, m, rt<<1 13 #define rson m+1, r, rt<<1|1 14 #define mnx 1000100 15 16 bool check[mnx]; 17 int mu[mnx], tot, prime[mnx]; 18 void init(){ 19 mu[1] = 1; 20 for( int i = 2; i < mnx; ++i ){ 21 if( !check[i] ) 22 prime[tot++] = i, mu[i] = -1; 23 for( int j = 0; i * prime[j] < mnx &&j < tot; ++j ){ 24 check[i*prime[j]] = 1; 25 if( i % prime[j] == 0 ){ 26 mu[i*prime[j]] = 0; 27 break; 28 } 29 else mu[i*prime[j]] = -mu[i]; 30 } 31 } 32 } 33 int main(){ 34 int cas; 35 scanf( "%d", &cas ); 36 init(); 37 while( cas-- ){ 38 int n; 39 scanf( "%d", &n ); 40 LL ans = 0; 41 for( int i = 1; i <= n; ++i ) 42 ans += (LL)mu[i] * (n/i) * (n/i) * (n/i+3); //(+3是指x-y, x-z, y-z三个平面) 43 printf( "%lld\n", ans + 3 ); 44 } 45 return 0; 46 }
做了上题，这题就没什么难度了。已经用容斥搞过了，不过容斥比莫比乌斯反演慢很多。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 7 8 using namespace std; 9 10 #define LL long long 11 #define eps 1e-8 12 #define lson l, m, rt<<1 13 #define rson m+1, r, rt<<1|1 14 #define mnx 100100 15 16 bool check[mnx]; 17 int mu[mnx], tot, prime[mnx]; 18 void init(){ 19 mu[1] = 1; 20 for( int i = 2; i < mnx; ++i ){ 21 if( !check[i] ) 22 prime[tot++] = i, mu[i] = -1; 23 for( int j = 0; i * prime[j] < mnx &&j < tot; ++j ){ 24 check[i*prime[j]] = 1; 25 if( i % prime[j] == 0 ){ 26 mu[i*prime[j]] = 0; 27 break; 28 } 29 else mu[i*prime[j]] = -mu[i]; 30 } 31 } 32 } 33 int main(){ 34 int cas, kk = 1; 35 scanf( "%d", &cas ); 36 init(); 37 while( cas-- ){ 38 LL n, m, nn, mm, k; 39 scanf( "%I64d%I64d%I64d%I64d%I64d", &nn, &n, &mm, &m, &k ); 40 if( k == 0 ){ 41 printf( "Case %d: 0\n", kk++ ); continue ; 42 } 43 n /= k, m /= k; 44 if( n > m ) swap( n, m ); 45 LL ans = 0, ans1 = 0; 46 for( int i = 1; i <= n; ++i ) 47 ans += (LL)mu[i] * (n/i) * (m/i); 48 for( int i = 1; i <= n; ++i ) 49 ans1 += (LL)mu[i] * (n/i) * (n/i); 50 printf( "Case %d: %I64d\n", kk++, ans - ans1/2 ); 51 } 52 return 0; 53 }
中文题。对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，且gcd(x,y) = k。
上题已经做了1 <= x <= b，1 <= y <= d，且gcd( x, y ) = k的。因此，对a <= x <= b, c <= y <= d，可以用容斥搞。ans = calc( b, d ) - calc( a-1, d ) - calc( b, c-1 ) + calc( a-1, c-1 );但是题目有5w组case，每组case都是O(n)的复杂度，会tle，因此还需要再优化。用分块的方法，因为对于一段区间，( L / i ), ( R / i )的值可能是一样的。所以我们先预处理莫比乌斯函数的前缀和，然后对于i，令j = min( L/(L/i), R/(R/i) )，区间 [ i，j ]的商都是一样的，所以ret += (s[j] - s[i-1] ) * (L/i) * (R/i);这样处理后复杂度的O(sqrt(n))，5w组case就可以过了
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <vector> 6 #include <algorithm> 7 #include <queue> 8 9 using namespace std; 10 11 #define LL long long 12 #define eps 1e-8 13 #define pb push_back 14 #define pf push_front 15 #define lson l, m, rt<<1 16 #define rson m+1, r, rt<<1|1 17 #define mnx 51000 18 19 bool check[mnx]; 20 int mu[mnx], tot, prime[mnx], s[mnx]; 21 void init(){ 22 mu[1] = 1; 23 for( int i = 2; i < mnx; ++i ){ 24 if( !check[i] ) 25 prime[tot++] = i, mu[i] = -1; 26 for( int j = 0; j < tot && i * prime[j] < mnx; ++j ){ 27 check[i*prime[j]] = 1; 28 if( i % prime[j] == 0 ){ 29 mu[i*prime[j]] = 0; 30 break; 31 } 32 else mu[i*prime[j]] = -mu[i]; 33 } 34 } 35 for( int i = 1; i < mnx; ++i ) 36 s[i] = s[i-1] + mu[i]; 37 } 38 int a, b, c, d, k; 39 LL calc( int L, int R ){ 40 L /= k, R /= k; 41 LL ret = 0; 42 if( L == 0 || R == 0 ) return 0; 43 if( L > R ) swap( L, R ); 44 for( int i = 1, j; i <= L; i = j+1 ){ 45 j = min( L/(L/i), R/(R/i) ); 46 ret += (LL)( s[j] - s[i-1] ) * ( L/i ) * ( R/i ); 47 } 48 return ret; 49 } 50 int main(){ 51 init(); 52 int cas; 53 scanf( "%d", &cas ); 54 while( cas-- ){ 55 scanf( "%d%d%d%d%d", &a, &b, &c, &d, &k ); 56 LL ans = calc( b, d ) - calc( a-1, d ) - calc( b, c-1 ) + calc( a-1, c-1 ); 57 printf( "%lld\n", ans ); 58 } 59 return 0; 60 }
