标签:void img images freopen 分享 com main 前缀和 alt
看这个题解吧:http://blog.csdn.net/wubaizhe/article/details/77338332
代码里顺便把几个常用的线性筛附上了。
#include<cstdio> #include<algorithm> using namespace std; #define MOD 1000000007 #define N 1000000 bool notpri[N+5]; int pri[N+5],n,mu[N+5],sum[N+5]; typedef long long ll; void shai_mu()//线性筛莫比乌斯函数,并处理出前缀和 { notpri[1]=1; mu[1]=1; for(int i=2;i<=N;i++){ if(!notpri[i]){ pri[++pri[0]]=i; mu[i]=-1; } for(int j=1;j<=pri[0] && (ll)i*(ll)pri[j]<=(ll)N;j++){ notpri[i*pri[j]]=1; mu[i*pri[j]]=-mu[i]; if(i%pri[j]==0){ mu[i*pri[j]]=0; break; } } } sum[1]=mu[1]; for(int i=2;i<=N;i++){ sum[i]=sum[i-1]+mu[i]; } } int ysgs[N+5],facnum[N+5],d[N+5]/*d(i)是辅助数组,记录每个数的最小质因子的幂次*/; void shai_facnum()//线性筛每个数的约数个数 { facnum[1]=1; for(int i=2;i<=N;++i){ if(!notpri[i]){ facnum[i]=2; d[i]=1; } for(int j=1;j<=pri[0] && (ll)i*(ll)pri[j]<=(ll)N;++j){ if(i%pri[j]==0){ facnum[i*pri[j]]=facnum[i]/(d[i]+1)*(d[i]+2); d[i*pri[j]]=d[i]+1; break; } facnum[i*pri[j]]=facnum[i]*2; d[i*pri[j]]=1; } } } int g[N+5]; int main(){ //freopen("hdu6134.in","r",stdin); shai_mu(); shai_facnum(); g[1]=1; for(int i=2;i<=N;++i){ g[i]=(g[i-1]+(facnum[i-1]+1))%MOD; } for(int i=2;i<=N;++i){ g[i]=(g[i]+g[i-1])%MOD; } while(scanf("%d",&n)!=EOF){ int ans=0; for(int i=1;i<=n;){ ans=(ans+(int)((((ll)(sum[n/(n/i)]-sum[i-1]+(ll)MOD)%(ll)MOD)*(ll)g[n/i])%(ll)MOD))%MOD; i=n/(n/i)+1; } printf("%d\n",ans); } return 0; } /* 线性筛欧拉函数 void get_eular() { pnum = 0; for(int i = 2; i < MAX; i++) { if(!noprime[i]) { p[pnum ++] = i; phi[i] = i - 1; } for(int j = 0; j < pnum && i * p[j] < MAX; j++) { noprime[i * p[j]] = true; if(i % p[j] == 0) { phi[i * p[j]] = phi[i] * p[j]; break; } phi[i * p[j]] = phi[i] * (p[j] - 1); } } } */
【莫比乌斯反演】【线性筛】hdu6134 Battlestation Operational
标签:void img images freopen 分享 com main 前缀和 alt
原文地址:http://www.cnblogs.com/autsky-jadek/p/7491730.html