BZOJ 4407: 于神之怒加强版 [莫比乌斯反演 线性筛]

题意:提前给出\(k\),求\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m gcd(i,j)^k\)


套路推♂倒

\[\sum_{D=1}^n \sum_{d|D} d^k\mu(\frac{D}{d}) \frac{n}{D} \frac{m}{D} \]

是一个\(g = idk * \mu\)啊,单位幂函数和莫比乌斯函数的卷积!

\(g(1) = 1\)
\(g(p) = -1 + p^k\)
因为带着\(\mu\),只有sf才有贡献
所以\(p \mid i\)只能把\(p\)放到\(d^k\)里了,就是\(g(i)\cdot p\)
或者考虑\(g(p^k)\)只有1和p有贡献也可以,直接得到计算公式再递推

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N=5e6+5, INF=1e9, P=1e9+7;
#define pii pair<int, int>
#define MP make_pair 
#define fir first
#define sec second
typedef long long ll;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
 
int n, m, k;
int notp[N], p[N], mu[N];
ll pk[N], g[N];
inline ll Pow(ll a, int b) {
    ll ans = 1;
    for(; b; b>>=1, a=a*a%P)
        if(b&1) ans = ans*a%P;
    return ans;
}
inline ll powk(int a) {return pk[a] ? pk[a] : pk[a]=Pow(a, k);}
void sieve(int n) {
    mu[1] = 1; g[1] = 1;
    for(int i=2; i<=n; i++) {
        if(!notp[i]) p[++p[0]] = i, mu[i] = -1, g[i] = -1 + powk(i);
        for(int j=1; j<=p[0] && i*p[j]<=n; j++) {
            int t = i*p[j];
            notp[t] = 1;
            if(i%p[j] == 0) {
                g[t] = g[i]*powk(p[j])%P;
                mu[t] = 0;
                break;
            }
            g[t] = g[i]*g[p[j]]%P;
            mu[t] = -mu[i];
        }
    }
    for(int i=1; i<=n; i++) g[i] = (g[i] + g[i-1])%P;
}
ll cal(int n, int m) {
    ll ans=0; int r;
    for(int i=1; i<=n; i=r+1) {
        r = min(n/(n/i), m/(m/i));
        ans = (ans+ (g[r] - g[i-1]) * (n/i)%P * (m/i)%P )%P;
    }
    return (ans + P) %P;
}
int main() {
    //freopen("in","r",stdin);
    int T=read(); k=read(); 
    sieve(N-1);
    while(T--) {
        n=read(); m=read();
        if(n>m) swap(n, m);
        printf("%lld\n", cal(n, m));
    }
}
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posted @ 2017-03-24 15:20  Candy?  阅读(371)  评论(0编辑  收藏  举报