HDU 6134 Battlestation Operational

把原式里的艾弗森约定用莫比乌斯函数展开:

f(n)=∑i=1n∑j=1i⌈ij⌉[(i,j)=1]=∑i=1n∑j=1i⌈ij⌉∑d|(i,j)μ(d)

令:

i=k1∗d,j=k2∗d,g(n)=∑i=1n⌈ij⌉,h(n)=∑i=1ng(n)

则:

f(n)=∑d=1nμ(d)∑k1=1n/d∑k2k1⌈k1k2⌉=∑d=1nμ(d)∑k1=1n/dg(k1)=∑d=1nμ(d)h(n/d)

g(n)用分块加速求出，h(n)用前缀和求出，f(n)也用分块加速求出。

代码：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N = 1e6+5;const LL MOD = 1e9+7;bool prime[N];int p[N];LL g[N], mu[N];void init() {    for(int i = 2; i < N; i++) prime[i] = 1;    int k = 0;    mu[1] = 1;    for(int i = 2; i < N; i++) {        if(prime[i]) { p[k++] = i; mu[i] = -1;}        for(int j = 0; j<k && i*p[j]<N; j++) {            prime[i*p[j]] = 0;            if(i%p[j] == 0) { mu[i*p[j]] = 0; break; }            mu[i*p[j]] = -mu[i];        }    }    for(int i = 1; i < N-1; i++) {        g[i]++, g[i+1]--;        for(int j = 2; ; j++) {            int R = i*j, L = i*(j-1)+1;            if(R > N) R = N-2;            g[L] = (g[L]+j)%MOD; g[R+1] = (g[R+1]-j+MOD)%MOD;            if(R == N-2) break;        }    }    for(int i = 1; i < N; i++) (g[i] += g[i-1]+MOD)%=MOD;    for(int i = 1; i < N; i++) (g[i] += g[i-1]+MOD)%=MOD, (mu[i] += mu[i-1]+MOD)%=MOD;}int main() {    init();    int n;    while(~scanf("%d", &n)) {        LL ans = 0;        for(int i = 1; i <= n; i++) {            int j = n/(n/i);            (ans += (mu[j]-mu[i-1]+MOD)%MOD*g[n/i]%MOD)%=MOD;            i = j;        }        printf("%lld\n", ans);    }    return 0;}