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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;

#define LL long long
#define eps 1e-8
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mnx 1000100

bool check[mnx];
int mu[mnx], tot, prime[mnx];
void init(){
    mu[1] = 1;
    for( int i = 2; i < mnx; ++i ){
        if( !check[i] )
            prime[tot++] = i, mu[i] = -1;
        for( int j = 0; i * prime[j] < mnx &&j < tot; ++j ){
            check[i*prime[j]] = 1;
            if( i % prime[j] == 0 ){
                mu[i*prime[j]] = 0;
                break;
            }
            else mu[i*prime[j]] = -mu[i];
        }
    }
}
int main(){
    int cas;
    scanf( "%d", &cas );
    init();
    while( cas-- ){
        int n;
        scanf( "%d", &n );
        LL ans = 0;
        for( int i = 1; i <= n; ++i )
            ans += (LL)mu[i] * (n/i) * (n/i) * (n/i+3); //(+3是指x-y, x-z, y-z三个平面)
        printf( "%lld
", ans + 3 );
    }
    return 0;
}

HDU 1695 GCD

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。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;

#define LL long long
#define eps 1e-8
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mnx 100100

bool check[mnx];
int mu[mnx], tot, prime[mnx];
void init(){
    mu[1] = 1;
    for( int i = 2; i < mnx; ++i ){
        if( !check[i] )
            prime[tot++] = i, mu[i] = -1;
        for( int j = 0; i * prime[j] < mnx &&j < tot; ++j ){
            check[i*prime[j]] = 1;
            if( i % prime[j] == 0 ){
                mu[i*prime[j]] = 0;
                break;
            }
            else mu[i*prime[j]] = -mu[i];
        }
    }
}
int main(){
    int cas, kk = 1;
    scanf( "%d", &cas );
    init();
    while( cas-- ){
        LL n, m, nn, mm, k;
        scanf( "%I64d%I64d%I64d%I64d%I64d", &nn, &n, &mm, &m, &k );
        if( k == 0 ){
            printf( "Case %d: 0
", kk++ ); continue ;
        }
        n /= k, m /= k;
        if( n > m ) swap( n, m );
        LL ans = 0, ans1 = 0;
        for( int i = 1; i <= n; ++i )
            ans += (LL)mu[i] * (n/i) * (m/i);
        for( int i = 1; i <= n; ++i )
            ans1 += (LL)mu[i] * (n/i) * (n/i);
        printf( "Case %d: %I64d
", kk++, ans - ans1/2 );
    }
    return 0;
}

 HYSBZ 2301 Problem b

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

#define LL long long
#define eps 1e-8
#define pb push_back
#define pf push_front
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define mnx 51000

bool check[mnx];
int mu[mnx], tot, prime[mnx], s[mnx];
void init(){
    mu[1] = 1;
    for( int i = 2; i < mnx; ++i ){
        if( !check[i] )
            prime[tot++] = i, mu[i] = -1;
        for( int j = 0; j < tot && i * prime[j] < mnx; ++j ){
            check[i*prime[j]] = 1;
            if( i % prime[j] == 0 ){
                mu[i*prime[j]] = 0;
                break;
            }
            else mu[i*prime[j]] = -mu[i];
        }
    }
    for( int i = 1; i < mnx; ++i )
        s[i] = s[i-1] + mu[i];
}
int a, b, c, d, k;
LL calc( int L, int R ){
    L /= k, R /= k;
    LL ret = 0;
    if( L == 0 || R == 0 ) return 0;
    if( L > R ) swap( L, R );
    for( int i = 1, j; i <= L; i = j+1 ){
        j = min( L/(L/i), R/(R/i) );
        ret += (LL)( s[j] - s[i-1] ) * ( L/i ) * ( R/i );
    }
    return ret;
}
int main(){
    init();
    int cas;
    scanf( "%d", &cas );
    while( cas-- ){
        scanf( "%d%d%d%d%d", &a, &b, &c, &d, &k );
        LL ans = calc( b, d ) - calc( a-1, d ) - calc( b, c-1 ) + calc( a-1, c-1 );
        printf( "%lld
", ans );
    }
    return 0;
}