HDU 4059 容斥初步练习

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <algorithm>
 #define LL long long
 using namespace std;
 const LL Mod=;
 const LL Maxn=;
 LL Factor[],cnt,n,m,tot,Rev,Kase,Prime[Maxn];
 bool vis[Maxn];

 inline LL Quick_Pow(LL x,LL y)
 {
     LL Ret=;
     while (true)
     {
         if (y&) Ret=(Ret*x)%Mod;
         x=(x*x)%Mod; y>>=;
         if (y==) break;
     }
     return Ret;
 }

 inline void Make_Prime()
 {
     memset(vis,false,sizeof(vis));
     for (LL i=;i<Maxn;i++)
     {
         if (!vis[i]) Prime[++tot]=i;
         for (LL j=;j<=tot && Prime[j]*i<Maxn;j++)
         {
             vis[Prime[j]*i]=true;
             if (i%Prime[j]==) break;
         }
     }
 }

 inline LL Calc(LL N)
 {
     LL Ret=N;
     Ret=(Ret*(N+))%Mod;
     Ret=(Ret*(*N+))%Mod;
     Ret=(Ret*((*N*N+*N-)%Mod))%Mod;
     Ret=(Ret*Rev)%Mod;
     return Ret;
 }
 inline void Get_Factor(LL P)
 {
     cnt=;
     for (LL i=;i<=tot && Prime[i]<=P;i++)
         if (P%Prime[i]==)
         {
             Factor[++cnt]=Prime[i];
             while (P%Prime[i]==) P/=Prime[i];
         }
     if (P!=) Factor[++cnt]=P;
 }
 inline LL Pow2(LL x) {return (x*x)%Mod;}
 inline LL Pow4(LL x) {return (Pow2(x)*Pow2(x))%Mod;}
 LL Dfs(LL d,LL start)
 {
     LL Ret=;
     for (LL i=start;i<=cnt;i++)
     {
         LL tmp=Pow4(Factor[i]);
         Ret=(Ret+(tmp*Calc(d/Factor[i]))%Mod)%Mod;
         Ret=(Ret-(tmp*Dfs(d/Factor[i],i+))%Mod+Mod)%Mod;
     }
     return Ret;
 }
 inline LL Solve()
 {
     Get_Factor(n);
     return ((Calc(n)%Mod)-(Dfs(n,))%Mod+Mod)%Mod;
 }
 int main()
 {
     scanf("%lld",&Kase);
     Rev=Quick_Pow(,Mod-);
     Make_Prime();

     for (LL kase=;kase<=Kase;kase++)
     {
         scanf("%lld",&n);
         printf("%lld\n",Solve());
     }
     return ;
 }

HDU 4059

求的是与n互质的数的四次方的和。 首先四次方有个公式。把1~n的然后减去n的约数的四次方即可，这就需要用到容斥了。

Sum(2)-(Sum(2*3)+Sum(2*5)+Sum(2*7)..)+(Sum(2*3*5)+Sum(2*3*7)+Sum(3*5*7)..)-..