SPOJ

题目：x属于区间[1,a]，y属于区间[1,b]，问gcd(x,y)为质数的有多少个。(2,4) (4,2)算2个

思路：直接莫比乌斯+分块

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst int maxn=1e7+50;bool check[maxn];int prime[maxn];int mu[maxn],sum[maxn];int tot;void Moblus(int n){    mm(check,false);    mu[1]=1;    tot=0;    for(int i=2;i<=n;i++){        if(!check[i]){            prime[tot++]=i;            mu[i]=-1;        }        for(int j=0;j<tot;j++){            if(i*prime[j]>n)                break;            check[i*prime[j]]=true;            if(i%prime[j]==0){                mu[i*prime[j]]=0;                break;            }            else{                mu[i*prime[j]]=-mu[i];            }        }    }    sum[0]=0;    for(int i=1;i<=n;i++)        sum[i]=sum[i-1]+mu[i];}int main(){    int T,a,b;    Moblus(maxn-50);    scanf("%d",&T);    while(T--){        scanf("%d%d",&a,&b);        LL ans=0;        if(a>b) swap(a,b);        for(int i=0;i<tot&&prime[i]<=a;i++){            int na=a/prime[i],nb=b/prime[i];            for(int j=1,la;j<=na;j=la+1){                la=min(na/(na/j),nb/(nb/j));                ans+=(LL)(sum[la]-sum[j-1])*(na/j)*(nb/j);            }        }        printf("%lld\n",ans);    }    return 0;}