BZOJ 2005 能量采集（莫比乌斯反演+分块）

思路：

#include<stdio.h>
#include <iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
#define eps 1e-8
typedef long long int lli;
using namespace std;

const int maxn = 1e5+10;
bool isprime[maxn];
int prime[maxn];
char miu[maxn];
void moblus(){
int cnt = 0;miu[1] = 1;
for(lli i = 2;i < maxn;i++){
if(!isprime[i]){
prime[cnt++] = i,miu[i] = -1;//phi[i] = i-1;
}
for(lli j = 0;j < cnt && i*prime[j] < maxn;j++){
lli x = prime[j];
isprime[i*x] = 1;
if(i%x){
miu[i*x] = -miu[i];
//phi[i*x] = phi[i] * phi[x];
}
else{
miu[i*x] = 0;
//phi[i*x] = phi[i] * x;
break;
}
}
}
}
int ff[maxn],sum[maxn];
int main(){
lli p,q;
moblus();
for(int i = 1;i <= maxn;i++){
sum[i] = sum[i-1] + miu[i];
}
lli a,b;
scanf("%lld%lld",&a,&b);
if(a>b) swap(a,b);
lli ans = 0,l,tempans = 0;
for(int i = 1;i <= a;i++){
lli n = a/i,m = b/i;
for(int j=1;j<=n;j=l+1){//分块加速
l= min(n/(n/j),m/(m/j));
tempans += (lli)(sum[l]-sum[j-1])*(n/j)*(m/j);
}
ans += (lli)((i-1)*2+1)*tempans;tempans = 0;
}
printf("%lld\n",ans);
return 0;
}