CF915G Coprime Arrays （莫比乌斯反演）

CF915G Coprime Arrays

题解

（看了好半天终于看懂了）

我们先对于每一个i想，那么

我们设

我们用莫比乌斯反演

有了这个式子，可比可以求出△ans呢？我们注意到，由于那个(i/d)是下取整了的，所以i++后（下文的 i 是+1后的 i），只有当(d+1)|i 时答案有变化，于是

我们可以预处理a^n，以及用埃氏筛预处理△ans[i]

ＣＯＤＥ

#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#include<iostream>
#define MAXN 2000005
#define LL long long
#define rg register
#define lowbit(x) (-(x) & (x))
#define ENDL putchar('\n')
#pragma GCC optimize(2)
//#pragma G++ optimize(3)
//#define int LL
using namespace std;
inline int read() {
	int f = 1,x = 0;char s = getchar();
	while(s < '0' || s > '9') {if(s == '-')f = -1;s = getchar();}
	while(s >= '0' && s <= '9') {x = x * 10 - '0' + s;s = getchar();}
	return x * f;
}
LL zxy = 1e9 + 7;
LL n,m,i,j,k,s,o;
int p[MAXN],cnt;
int mu[MAXN];
LL po[MAXN],Delta[MAXN];
bool f[MAXN];
inline LL qkpow(LL a,LL b) {
	LL res = 1;
	while(b) {
		if(b & 1) res = res * a % zxy;
		b >>= 1;
		a = a * a % zxy;
	}
	return res % zxy;
}
inline void sieve(int n) {
	mu[1] = 1;
	for(int i = 2;i <= n;i ++) {
		if(!f[i]) {
			p[++ cnt] = i;
			mu[i] = -1;
		}
		for(int j = 1;j <= cnt && i * p[j] <= n;j ++) {
			int y = i * p[j];
			f[y] = 1;
			if(i % p[j] == 0) {
				mu[y] = 0;
				break;
			}
			mu[y] = -mu[i];
		}
	}
	return ;
}
signed main() {
	sieve(2000000);
	n = read();m = read();
	for(int i = 0;i <= 2000000;i ++) po[i] = qkpow(i,n);
	for(int i = 1;i <= m;i ++) {
		for(int j = i;j <= m;j += i) {
			Delta[j] = (Delta[j] + zxy +  mu[i] * 1ll * (po[j/i] - po[j/i - 1]) % zxy) % zxy;
		}
	}
	LL ans = 0,as = 0;
	for(int i = 1;i <= m;i ++) {
		as = (as + Delta[i]) % zxy;
		ans = (ans + (as ^ i)) % zxy;
	}
	printf("%lld\n",ans);
    return 0;
}