bzoj 2154 Crash的数字表格 反演+数学

版权声明：本文为博主原创文章，转载请附上原博客链接。 https://blog.csdn.net/Dale_zero/article/details/82390370

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=2154

在这位大神的blog里才看懂，有时间再自己写一下

https://www.cnblogs.com/cjyyb/p/8253033.html

#include<bits/stdc++.h>
#include<math.h>
#define MOD  20101009
#define For(i,m,n) for(int i=m;i<=n;i++)
#define LL long long
#define lan(a,b) memset(a,b,sizeof(a))
#define sqr(x) (x*x)
using namespace std;

const int maxn=1e7+5;
int mu[maxn],pri[maxn],tot;
int smu[maxn];
bool zs[maxn];
int a[maxn];
int n,m;
void Getmu()
{
    zs[1]=true;mu[1]=1;
    for(int i=2;i<=n;++i)
    {
        if(!zs[i]){pri[++tot]=i;mu[i]=-1;}
        for(int j=1;j<=tot&&i*pri[j]<=n;++j)
        {
            zs[i*pri[j]]=true;
            if(i%pri[j])mu[i*pri[j]]=-mu[i];
            else{mu[i*pri[j]]=0;break;}
        }
    }
    for(int i=1;i<=n;++i)smu[i]=(smu[i-1]+mu[i]+MOD)%MOD;
}

LL pin(LL x,LL y)
{
    LL d=1;
    LL ans=0;
    while(d<=x)
    {
        LL r=min(x/(x/d),y/(y/d));
        LL len=r-d+1;
        LL tem=1LL*(1LL*(1+x/d)*(x/d)/2%MOD)*(1LL*(1+y/d)*(y/d)/2%MOD)%MOD;
        ans+=1LL*(a[r]-a[d-1])%MOD*tem%MOD;
        ans%=MOD;
        d=r+1;
    }
    return (ans+MOD)%MOD;
}

int main()
{
    while(~scanf("%lld%lld",&n,&m))
    {
        tot=0;
        Getmu();
        if(n>m)
            swap(n,m);
        For(i,1,n)
            a[i]=(a[i-1]+1LL*i*i%MOD*mu[i]%MOD+MOD)%MOD;
        LL d=1;
        LL ans=0;
        while(d<=n)
        {
//            printf("d=%lld,ans=%lld\n",d,ans);
            LL r=min(n/(n/d),m/(m/d));
            LL tem=1LL*(d+r)*(r-d+1)/2%MOD;
            ans=(ans+1LL*pin(n/d,m/d)*tem%MOD)%MOD;
            d=r+1;
        }
        printf("%lld\n",ans);
    }
    return 0;
}