洛谷 P3312 [SDOI2014]数表

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

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=;
 const ll M=;
 struct Q
 {
     ll n,m,a,num;
 }q[M+];
 ll qq;
 ll ans[M+];
 ll mu[N+],len,prime[N+];
 ll f1[N+],f2[N+],h[N+];
 //分别为最小质因子的p^k，1+p^0+p^1+..+p^k，函数值
 vector<int> dd[];
 bool nprime[N+];
 bool c1(const Q &a,const Q &b)    {return a.a<b.a;}
 const ll md=1LL<<;
 #define lowbit(x) ((x)&(-x))
 ll d[N+];
 void add(ll p,ll x)
 {
     for(;p<=N;p+=lowbit(p))    d[p]+=x;
 }
 ll sum(ll p)
 {
     ll ans=;
     for(;p>;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[]=;h[]=;
     for(i=;i<=N;i++)
     {
         if(!nprime[i])
         {
             prime[++len]=i;mu[i]=-;
             f1[i]=i;f2[i]=i+;h[i]=i+;
         }
         for(j=;j<=len&&i*prime[j]<=N;j++)
         {
             nprime[i*prime[j]]=;
             if(i%prime[j]==)
             {
                 mu[i*prime[j]]=;
                 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]]=+prime[j];
                 h[i*prime[j]]=h[i]*f2[i*prime[j]];
             }
         }
     }
     for(i=;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\n",ans);
         }
     }*/
     scanf("%lld",&qq);
     for(i=;i<=qq;i++)    scanf("%lld%lld%lld",&q[i].n,&q[i].m,&q[i].a),q[i].num=i;
     sort(q+,q+qq+,c1);
     for(i=,j=;i<=qq;i++)
     {
         while(j+<=q[i].a&&j+<=)
         {
             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=;
         for(ll i=,j;i<=n;i=j+)
         {
             j=min(n/(n/i),m/(m/i));
             an+=(n/i)*(m/i)*(sum(j)-sum(i-));
         }
         ans[q[i].num]=an;
         //printf("a%lld\n",an);
     }
     for(i=;i<=qq;i++)    printf("%lld\n",ans[i]%md);
     return ;
 }