【BZOJ2301】【HAOI2011】Problem b 莫比乌斯反演

Mission

对于给出的n个询问,每次求有多少个数对(x,y),满足a≤x≤b,c≤y≤d,且gcd(x,y) = k,gcd(x,y)函数为x和y的最大公约数。
1n500001ab500001cd500001k50000

Solution

裸的莫比乌斯反演
把询问拆分成四个子询问,然后莫比乌斯反演要用分块求解。

Code

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<string.h>
#define ll long long
using namespace std;
const char* fin="ex2301.in";
const char* fout="ex2301.out";
const ll inf=0x7fffffff;
const ll maxn=50007;
ll t,a,b,c,d,ind,i,j,k;
ll miu[maxn],p[maxn];
bool bz[maxn];
ll ans;
ll f(ll n,ll m){
    ll i,j,k,ans=0;
    if (n<=0 || m<=0) return 0;
    if (n>m) swap(n,m);
    for (i=1;ind*i<=n;){
        j=min(n/(ind*(n/(ind*i))),m/(ind*m/(ind*i)));
        ans+=(n/(ind*i))*(m/(ind*i))*(miu[j]-miu[i-1]);
        i=j+1;
    }
    return ans;
}
int main(){
    freopen(fin,"r",stdin);
    freopen(fout,"w",stdout);
    scanf("%lld",&t);
    miu[1]=1;
    for (i=2;i<maxn;i++){
        if (!bz[i]){
            p[++p[0]]=i;
            miu[i]=-1;
        }
        for (j=1;j<=p[0];j++){
            k=i*p[j];
            if (k>=maxn) break;
            bz[k]=true;
            if (i%p[j]==0){
                miu[k]=0;
                break;
            }else miu[k]=-miu[i];
        }
    }
    for (i=1;i<maxn;i++) miu[i]+=miu[i-1];
    while (t--){
        scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&ind);
        printf("%lld\n",f(b,d)-f(b,c-1)-f(a-1,d)+f(a-1,c-1));
    }
    return 0;
}
posted @ 2017-03-09 16:49  hiweibolu  阅读(...)  评论(...编辑  收藏