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  • HYSBZ

    链接

    对于gcd(i,j)的位置来说,对答案的贡献是2*(gcd(i,j)-1)+1,所以答案ans

    ans=Σ(1<=i<=n)(1<=j<=m)2*(gcd(i,j)-1)+1

    ans=2*Σ(1<=i<=n)(1<=j<=m)gcd(i,j)-n*m

    前者可以通过莫比乌斯反演来计算,便很容易得出答案

    //#pragma comment(linker, "/stack:200000000")
    //#pragma GCC optimize("Ofast,no-stack-protector")
    //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
    //#pragma GCC optimize("unroll-loops")
    #include<bits/stdc++.h>
    #define fi first
    #define se second
    #define mp make_pair
    #define pb push_back
    #define pi acos(-1.0)
    #define ll long long
    #define mod 1000000007
    #define C 0.5772156649
    #define ls l,m,rt<<1
    #define rs m+1,r,rt<<1|1
    #define pil pair<int,ll>
    #define pii pair<int,int>
    #define ull unsigned long long
    #define base 1000000000000000000
    #define fio ios::sync_with_stdio(false);cin.tie(0)
    
    using namespace std;
    
    const double g=10.0,eps=1e-12;
    const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;
    
    int mu[N],prime[N],sum[N];
    bool mark[N];
    int num[N];
    void init()
    {
        mu[1]=1;
        int cnt=0;
        for(int i=2;i<N;i++)
        {
            if(!mark[i])prime[++cnt]=i,mu[i]=-1,num[i]=1;
            for(int j=1;j<=cnt;j++)
            {
                int t=i*prime[j];
                if(t>N)break;
                mark[t]=1;
                num[t]=num[i]+1;
                if(i%prime[j]==0){mu[t]=0;break;}
                else mu[t]=-mu[i];
            }
        }
        for(int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
    }
    int main()
    {
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        ll ans=0;
        for(int j=1;j<=max(n,m);j++)
        {
            ll te=0;
            int ten=n/j,tem=m/j;
            for(int i=1,last;i<=min(ten,tem);i=last+1)
            {
                last=min(ten/(ten/i),tem/(tem/i));
                te+=(ll)(sum[last]-sum[i-1])*(ten/i)*(tem/i);
            }
    //        printf("%lld
    ",te);
            ans+=te*j;
        }
        printf("%lld
    ",2*ans-(ll)n*m);
        return 0;
    }
    /********************
    
    ********************/
    View Code
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  • 原文地址:https://www.cnblogs.com/acjiumeng/p/8444273.html
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