「bzoj 3944: Sum」

题目

杜教筛板子了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<tr1/unordered_map>
#define re register
#define maxn 5000005
#define LL long long
using namespace std::tr1;
unordered_map<int,int> ma;
unordered_map<int,LL> Ma;
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
char c=getchar();
int x=0;
while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();
return x;
}
int f[maxn],p[maxn>>1],mu[maxn];
LL phi[maxn];
int N[101];
int M,T;
int solve(int x)
{
if(x<=M) return mu[x];
if(ma.find(x)!=ma.end()) return ma[x];
int now=1;
for(re int l=2,r;l<=x;l=r+1)
{
r=x/(x/l);
now-=(r-l+1)*solve(x/l);
}
return ma[x]=now;
}
LL work(int x)
{
if(x<=M) return phi[x];
if(Ma.find(x)!=Ma.end()) return Ma[x];
LL ans=(LL)x*(x+1)/2;
for(re int l=2,r;l<=x;l=r+1)
{
r=x/(x/l);
ans-=(LL)(r-l+1)*work(x/l);
}
return Ma[x]=ans;
}
int main()
{
T=read();
for(re int i=1;i<=T;i++) N[i]=read(),M=max(M,N[i]);
M=5000000;
f[1]=mu[1]=phi[1]=1;
for(re int i=2;i<=M;i++)
{
if(!f[i]) p[++p[0]]=i,mu[i]=-1,phi[i]=i-1;
for(re int j=1;j<=p[0]&&p[j]*i<=M;j++)
{
f[p[j]*i]=1;
if(i%p[j]==0)
{
phi[p[j]*i]=p[j]*phi[i];
break;
}
mu[i*p[j]]=-1*mu[i];
phi[i*p[j]]=phi[i]*phi[p[j]];
}
}
for(re int i=1;i<=M;i++) mu[i]+=mu[i-1],phi[i]+=phi[i-1];
for(re int t=1;t<=T;t++) printf("%lld %d\n",work(N[t]),solve(N[t]));
return 0;
}

发现自己杜教筛常数太大了,b站卡不过去

于是怒写Min_25,发现还是卡着过的

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define maxn 100005
#define re register
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read() {
int x=0;char c=getchar();while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-48,c=getchar();return x;
}
int T,D,a[12];
LL n,m,w[maxn],id1[maxn],id2[maxn],g[maxn],d[maxn],pre[maxn];
int f[maxn],p[maxn>>1],tot;
inline LL Sphi(LL x,int y) {
if(x<=1||p[y]>x) return 0;
int t=(x<=D)?id1[x]:id2[n/x];
LL ans=d[t]-g[t]-(pre[y-1]-y+1);
for(re int k=y;k<=tot&&p[k]*p[k]<=x;k++) {
LL p1=p[k],now=p[k]-1;
for(re int e=1;p1<=x;e++,p1*=p[k],now*=p[k])
ans+=now*(Sphi(x/p1,k+1)+(e>1));
}
return ans;
}
inline int Smu(LL x,int y) {
if(x<=1||p[y]>x) return 0;
int t=(x<=D)?id1[x]:id2[n/x];
int ans=-1*g[t]+y-1;
for(re int k=y;k<=tot&&p[k]*p[k]<=x;k++) {
ans-=Smu(x/p[k],k+1);
}
return ans;
}
int main() {
T=read();
for(re int i=1;i<=T;i++) a[i]=read(),D=max(D,a[i]);
f[1]=1;D=std::sqrt(D)+1;
for(re int i=2;i<=D;i++) {
if(!f[i]) p[++tot]=i,pre[tot]=pre[tot-1]+i;
for(re int j=1;j<=tot&&p[j]*i<=D;j++) {
f[p[j]*i]=1;if(i%p[j]==0) break;
}
}
for(re int tt=1;tt<=T;tt++) {
n=a[tt];m=0;
if(!n) {puts("0 0");continue;}
for(re LL l=1,r;l<=n;l=r+1) {
r=n/(n/l);w[++m]=n/l;
if(w[m]<=D) id1[w[m]]=m;
else id2[n/w[m]]=m;
g[m]=w[m]-1ll;
d[m]=(w[m]+2ll)*(w[m]-1ll);d[m]/=2ll;
}
for(re int j=1;j<=tot&&p[j]*p[j]<=n;j++)
for(re int i=1;i<=m&&p[j]*p[j]<=w[i];i++) {
int k=(w[i]/p[j]<=D)?id1[w[i]/p[j]]:id2[n/(w[i]/p[j])];
g[i]=g[i]-g[k]+j-1;
d[i]=d[i]-(LL)p[j]*(d[k]-pre[j-1]);
}
printf("%lld %d\n",Sphi(n,1)+1ll,Smu(n,1)+1ll);
}
return 0;
}
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