「POI2007」ZAP-Queries「莫比乌斯反演」

题目传送门

题意

T组询问求，

∑i=1a∑j=1b[gcd(i,j)=d] ∑ i = 1 a ∑ j = 1 b [ g c d ( i , j ) = d ]

， T,a,b,d≤5×104 T , a , b , d ≤ 5 × 10 4

题解

不妨设 a≤b a ≤ b .

∑i=1a∑j=1b[gcd(i,j)=d] ∑ i = 1 a ∑ j = 1 b [ g c d ( i , j ) = d ]

=∑i=1⌊ad⌋∑j=1⌊bd⌋[gcd(i′,j′)=1] = ∑ i = 1 ⌊ a d ⌋ ∑ j = 1 ⌊ b d ⌋ [ g c d ( i ′ , j ′ ) = 1 ]

=∑i=1⌊ad⌋∑j=1⌊bd⌋e(gcd(i,j)) = ∑ i = 1 ⌊ a d ⌋ ∑ j = 1 ⌊ b d ⌋ e ( g c d ( i , j ) )

=∑i=1⌊ad⌋∑j=1⌊bd⌋∑k|gcd(i,j)e(gcd(i,j)) = ∑ i = 1 ⌊ a d ⌋ ∑ j = 1 ⌊ b d ⌋ ∑ k | g c d ( i , j ) e ( g c d ( i , j ) )

=∑i=1⌊ad⌋∑j=1⌊bd⌋∑k|i,k|jμ(k) = ∑ i = 1 ⌊ a d ⌋ ∑ j = 1 ⌊ b d ⌋ ∑ k | i , k | j μ ( k )

=∑k=1⌊ad⌋μ(k)⌊adk⌋⌊bdk⌋ = ∑ k = 1 ⌊ a d ⌋ μ ( k ) ⌊ a d k ⌋ ⌊ b d k ⌋

然后整除分块做一下就行.

#include 
#include 
using namespace std;

typedef long long LL;

const int MAXN = 50010;

bool tag[MAXN];
int pr[MAXN], tot, mu[MAXN];
LL mus[MAXN];

void init(int n) {
    tag[1] = true;
    mus[1] = mu[1] = 1;
    for(int i = 2; i <= n; i ++) {
        if(!tag[i]) {
            pr[++ tot] = i;
            mu[i] = -1;
        }
        for(int j = 1; j <= tot && i * pr[j] <= n; j ++) {
            tag[i * pr[j]] = true;
            if(i % pr[j] == 0) {
                mu[i * pr[j]] = 0;
                break ;
            }
            mu[i * pr[j]] = - mu[i];
        }
        mus[i] = mus[i - 1] + mu[i];
    }
}

LL calc(int a, int b, int d) {
    if(a > b) a ^= b ^= a ^= b;
    a /= d; b /= d;
    LL ans = 0;
    for(int i = 1, j; i <= a; i = j + 1) {
        j = min(a / (a / i), b / (b / i));
        ans += (mus[j] - mus[i - 1]) * (a / i) * (b / i);
    }
    return ans;
}

int main() {
    init(50000);
    int n, a, b, d;
    for(scanf("%d", &n); n --; ) {
        scanf("%d%d%d", &a, &b, &d);
        printf("%lld\n", calc(a, b, d));
    }
    return 0;
}