bzoj2154 Crash的數字表格

題目大意

給你n 和m ,求∑ni=1∑mj=1lcm(i,j) .

範圍 1≤n,m≤107 .

做法

易得ans=∑ni=1∑mj=1lcm(i,j)=∑ni=1∑mj=1i∗jgcd(i,j) .

令fx,y=∑xi=1∑yj=1,gcd(i,j)=1i∗j

枚舉gcd=d .

可得ans=∑min(n,m)d=1d2∗f(⌊nd⌋,⌊md⌋)d =∑min(n,m)d=1d∗f(⌊nd⌋,⌊md⌋) .

令gx,y=∑xi=1i∗∑yj=1j=x∗(x+1)2∗y∗(y+1)2 .

莫比烏斯反演得：
fx,y=∑min(x,y)i=1i2∗μi∗g⌊xi⌋,⌊yi⌋ .

這樣總時間就是O(n) 的。

參考代碼

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define maxn 10000005
#define ll long long
#define mo 20101009
using namespace std;

int n,m;

ll ans;

int miu[maxn];

int sum[maxn];

int pri[maxn];

bool bz[maxn];

ll query(ll x,ll y){
    return ((x*(x+1)) / 2 % mo) * ((y*(y+1)) / 2 % mo) % mo;
}

ll f(ll x,ll y){
    ll ret=0;
    ll last=0,now=1;
    while (last<x) {
        last=min(x/(x/now),y/(y/now));
        ret=(ret+(sum[last]-sum[now-1]+mo) % mo * query(x/now,y/now)) % mo;
        now=last+1;
    }
    return ret;
}

int main(){
    scanf("%d%d",&n,&m);
    if (n>m) swap(n,m);
    miu[1]=1;
    fo(i,2,n) {
        if (!bz[i]) {
            pri[++pri[0]]=i;
            miu[i]=-1;
        }
        fo(j,1,pri[0]) {
            if (i*pri[j]>n) break;
            bz[i*pri[j]]=1;
            miu[i*pri[j]]=miu[i]*miu[pri[j]];
            if (i % pri[j]==0) {
                miu[i*pri[j]]=0;
                break;
            }
        }
    }
    fo(i,1,n) sum[i]=(sum[i-1]+i*1ll*i*miu[i]) % mo;
    ll last=0,now=1;
    while (last<n) {
        last=min(n/(n/now),m/(m/now));
        ans=(ans+(last+now)*(last-now+1) / 2 % mo * f(n/now,m/now) % mo) % mo;
        ans=(ans+mo) % mo;
        now=last+1;
    }
    printf("%lld",ans);
    return 0;
}