HDU-1695-GCD(Mobius反演/容斥)


Sample Input
 
   
2 1 3 1 5 1 1 11014 1 14409 9
 

Sample Output
 
   
Case 1: 9 Case 2: 736427
Hint
For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
 


题意:a,b,c,d,m,  0<=x<=b,0<=y<=d,求满足gcd(x,y)=m这样的数对(x,y)的数量 

题解:

HDU-1695-GCD(Mobius反演/容斥)_第1张图片

 

CODE:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define LL long long

const int N = 100010;
int mu[N];
void getMu()
{
    for (int i = 1; i < N; ++i)
    {
        int target = i == 1 ? 1 : 0;
        int delta = target - mu[i];
        mu[i] = delta;
        for (int j = 2 * i; j < N; j += i)
            mu[j] += delta;
    }
}
bool check[N];
int prime[N];
void Moblus()
{
    memset(check,false, sizeof(check));
    mu[1]=1;
    int tot = 0;
    for(int i=2; i<=N; ++i)
    {
        if(!check[i])
        {
            prime[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0; jN)break;
            check[prime[j]*i] =true;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else
                mu[i*prime[j]]=-mu[i];
        }
    }
}
int a,b,c,m;
int main()
{
    //getMu();
    Moblus();
    int T;
    scanf("%d", &T);
    int cas=1;
    while(T--)
    {
        scanf("%d%d%d%d%d",&c,&a,&c,&b,&m);
        if(m==0){printf("Case %d: 0\n",cas++);continue;}
        if(a>b)swap(a,b);
        LL ans=0;
        a/=m,b/=m;
        for(int i=1; i<=a; ++i)
            ans-=(LL)mu[i]*(a/i)*(a/i);
        ans/=2;
        for(int i=1; i<=a; ++i)
            ans+=(LL)mu[i]*(a/i)*(b/i);
        printf("Case %d: %I64d\n",cas++,ans);
    }
    return 0;
}



另外容斥解法:


#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
using namespace std;
#define N 100005

int a,b,c,k,mod;
vector x[N];
bool is[N];
void get_prime()
{
    memset(is,0,sizeof(is));
    for(int i=0; ib)swap(a,b);
        for(int i=1; i<=b; ++i)
            //for(int j=1; j<=i&&j<=a; ++j)
            {
                ans+=solve(min(i,a),i);
            }
        printf("Case %d: %I64d\n",cas++,ans);
        //cout<


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