bzoj 1101 莫比乌斯反演

最裸的莫比乌斯

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e5 + ;
const int M = 1e6 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

int p[N], is[N], mbs[N], sum[N], n, m, d, ans, tot;

void init(){
    memset(is,  ,sizeof(is));
    mbs[]=;
    for(int i = ; i < N; i++) {
        if(is[i]){
            p[++tot] = i;
            mbs[i] = -;
        }
        for(int j = ; j <= tot && p[j] * i < N; j++){
            is[i * p[j]] = ;
            if(i % p[j] == ){
                mbs[i * p[j]] = ;
                break;
            } else {
                mbs[i * p[j]] = -mbs[i];
            }
        }
    }
    for(int i = ; i < N; i++) {
        sum[i] = mbs[i] + sum[i - ]; //维护前缀和
    }
}

int cal(int n, int m){  //求[1,n][1,m]区间内互质的(x, y)的对数
    int ans=;
    if(n > m) swap(n, m);
    for(int L = , R=; L <= n; L = R + ) {
        R = min(n / (n / L), m /(m / L));     // 分段加速
        ans += (sum[R] - sum[L - ]) * (n / L) * (m / L);
    }
    return ans;
}

int main(){
    init();
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d%d",&n,&m,&d);
        ans = cal(n / d, m / d);
        printf("%d\n", ans);
    }
    return ;
}
/*
*/