# bzoj 2693 jzptab 莫比乌斯反演

code:

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iomanip>
using namespace std;
#define LL long long
#define up(i,j,n) for(LL i=j;i<=n;i++)
#define pii pair<LL,LL>
#define db double
#define eps 1e-4
#define FILE "dealing"
LL x=0,f=1,ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0',ch=getchar();}
return x*f;
}
const LL maxn=(LL)(1e7+3),inf=1000000000000000LL,limit=(LL)(1e7+2),mod=100000009;
bool cmin(LL& a,LL b){return a>b?a=b,true:false;}
bool cmax(LL& a,LL b){return a<b?a=b,true:false;}
LL n,m;
LL mu[maxn],b[maxn],H[maxn],prime[2000000],tail=0;
LL sum(LL x,LL y){
x%=mod,y%=mod;
return ((((x+1)*x)>>1)%mod)*((((y+1)*y)>>1)%mod)%mod;
}
void getmu(){
mu[1]=1;H[1]=1;
up(i,2,limit){
if(!b[i])prime[++tail]=i,H[i]=(i-i*i)%mod;
for(LL j=1;prime[j]*i<=limit;j++){
b[i*prime[j]]=1;
if(i%prime[j]){
H[i*prime[j]]=(H[i]*H[prime[j]])%mod;
}
else {
H[i*prime[j]]=H[i]*prime[j];
break;
}
}
}
up(i,1,limit)H[i]=(H[i]+H[i-1])%mod;
}
int main(){
freopen(FILE".in","r",stdin);
freopen(FILE".out","w",stdout);
getmu();
while(q--){
if(n>m)swap(n,m);
LL ans=0;
up(i,1,n){
LL pos=min(n/(n/i),m/(m/i));
ans+=(H[pos]-H[i-1])*sum(n/i,m/i)%mod;
ans%=mod;
i=pos;
}
printf("%lld\n",(ans+mod)%mod);
}
return 0;
}


posted @ 2017-03-06 13:38  CHADLZX  阅读(94)  评论(0编辑  收藏  举报