洛谷 P3312 [SDOI2014]数表

式子化出来是$sum_{T=1}^m{lfloor}frac{n}{T}{ floor}{lfloor}frac{m}{T}{ floor}sum_{k|T}mu(frac{T}{k})[sigma(k)<=a]sigma(k)$

如果没有a的限制的话，显然只要把后面那个sigma预处理出来

有了a的限制呢？考虑把询问按a从小到大排序，然后每次把所有满足$lasta<sigma(k)<=nowa$（lasta指上个询问的a，nowa指当前询问的a）的加入后面那部分的贡献（开始时对于每个T预处理有哪些k满足$sigma(k)=T$，这里就暴力枚举k使得$sigma(k)$=当前要加入贡献的值，再暴力枚举k的倍数更新）；用树状数组维护前缀和；复杂度是n*log^2+T*sqrt*log

错误记录：

1.121行<=n写成<=m；导致除以0；本地测试的时候只用了n=m的数据，所以没测出来

2.108行没规定这个<=500000

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
const ll N=100000;
const ll M=100000;
struct Q
{
    ll n,m,a,num;
}q[M+100];
ll qq;
ll ans[M+100];
ll mu[N+100],len,prime[N+100];
ll f1[N+100],f2[N+100],h[N+100];
//分别为最小质因子的p^k，1+p^0+p^1+..+p^k，函数值
vector<int> dd[501000];
bool nprime[N+100];
bool c1(const Q &a,const Q &b)    {return a.a<b.a;}
const ll md=1LL<<31;
#define lowbit(x) ((x)&(-x))
ll d[N+100];
void add(ll p,ll x)
{
    for(;p<=N;p+=lowbit(p))    d[p]+=x;
}
ll sum(ll p)
{
    ll ans=0;
    for(;p>0;p-=lowbit(p))    ans+=d[p];
    return ans;
}
//ll Mod(ll a,ll b)
//{
//    if(a>=0)    return a%b;
//    else if(a%b==0)    return 0;
//    else    return b+a%b;
//}
int main()
{
    ll i,j,k,n,m,an;
    mu[1]=1;h[1]=1;
    for(i=2;i<=N;i++)
    {
        if(!nprime[i])
        {
            prime[++len]=i;mu[i]=-1;
            f1[i]=i;f2[i]=i+1;h[i]=i+1;
        }
        for(j=1;j<=len&&i*prime[j]<=N;j++)
        {
            nprime[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                f1[i*prime[j]]=f1[i]*prime[j];
                f2[i*prime[j]]=f2[i]+f1[i*prime[j]];
                h[i*prime[j]]=h[i]/f2[i]*f2[i*prime[j]];
                break;
            }
            else
            {
                mu[i*prime[j]]=-mu[i];
                f1[i*prime[j]]=prime[j];
                f2[i*prime[j]]=1+prime[j];
                h[i*prime[j]]=h[i]*f2[i*prime[j]];
            }
        }
    }
    for(i=1;i<=N;i++)
        dd[h[i]].pb(i);
    /*
    ll ttt=0;
    for(i=1;i<=20000;i++)    ttt=max(ttt,h[i]);
    printf("%lld",ttt);
    {
        ll a,n,m,k;
        while(1)
        {
            scanf("%lld%lld%lld",&n,&m,&a);
            if(n>m)    swap(n,m);
            ll ans=0;
            for(ll T=1;T<=m;T++)
            {
                ll a1=0;
                for(k=1;k<=T;k++)
                    if(T%k==0)
                    {
                        a1+=mu[T/k]*(h[k]<=a)*h[k];
                    }
                ans+=(n/T)*(m/T)*a1;
            }
            printf("%lld
",ans);
        }
    }*/
    scanf("%lld",&qq);
    for(i=1;i<=qq;i++)    scanf("%lld%lld%lld",&q[i].n,&q[i].m,&q[i].a),q[i].num=i;
    sort(q+1,q+qq+1,c1);
    for(i=1,j=0;i<=qq;i++)
    {
        while(j+1<=q[i].a&&j+1<=500000)
        {
            j++;
            for(auto d:dd[j])
            {
                for(k=d;k<=N;k+=d)
                {
                    add(k,mu[k/d]*j);
                }
            }
        }
        if(q[i].n>q[i].m)    swap(q[i].n,q[i].m);
        n=q[i].n;m=q[i].m;an=0;
        for(ll i=1,j;i<=n;i=j+1)
        {
            j=min(n/(n/i),m/(m/i));
            an+=(n/i)*(m/i)*(sum(j)-sum(i-1));
        }
        ans[q[i].num]=an;
        //printf("a%lld
",an);
    }
    for(i=1;i<=qq;i++)    printf("%lld
",ans[i]%md);
    return 0;
}