1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 #define LL long long 6 using namespace std; 7 const LL Mod=1000000007; 8 const LL Maxn=60010; 9 LL Factor[35],cnt,n,m,tot,Rev,Kase,Prime[Maxn]; 10 bool vis[Maxn]; 11 12 inline LL Quick_Pow(LL x,LL y) 13 { 14 LL Ret=1; 15 while (true) 16 { 17 if (y&1) Ret=(Ret*x)%Mod; 18 x=(x*x)%Mod; y>>=1; 19 if (y==0) break; 20 } 21 return Ret; 22 } 23 24 inline void Make_Prime() 25 { 26 memset(vis,false,sizeof(vis)); 27 for (LL i=2;i<Maxn;i++) 28 { 29 if (!vis[i]) Prime[++tot]=i; 30 for (LL j=1;j<=tot && Prime[j]*i<Maxn;j++) 31 { 32 vis[Prime[j]*i]=true; 33 if (i%Prime[j]==0) break; 34 } 35 } 36 } 37 38 inline LL Calc(LL N) 39 { 40 LL Ret=N; 41 Ret=(Ret*(N+1))%Mod; 42 Ret=(Ret*(2*N+1))%Mod; 43 Ret=(Ret*((3*N*N+3*N-1)%Mod))%Mod; 44 Ret=(Ret*Rev)%Mod; 45 return Ret; 46 } 47 inline void Get_Factor(LL P) 48 { 49 cnt=0; 50 for (LL i=1;i<=tot && Prime[i]<=P;i++) 51 if (P%Prime[i]==0) 52 { 53 Factor[++cnt]=Prime[i]; 54 while (P%Prime[i]==0) P/=Prime[i]; 55 } 56 if (P!=1) Factor[++cnt]=P; 57 } 58 inline LL Pow2(LL x) {return (x*x)%Mod;} 59 inline LL Pow4(LL x) {return (Pow2(x)*Pow2(x))%Mod;} 60 LL Dfs(LL d,LL start) 61 { 62 LL Ret=0; 63 for (LL i=start;i<=cnt;i++) 64 { 65 LL tmp=Pow4(Factor[i]); 66 Ret=(Ret+(tmp*Calc(d/Factor[i]))%Mod)%Mod; 67 Ret=(Ret-(tmp*Dfs(d/Factor[i],i+1))%Mod+Mod)%Mod; 68 } 69 return Ret; 70 } 71 inline LL Solve() 72 { 73 Get_Factor(n); 74 return ((Calc(n)%Mod)-(Dfs(n,1))%Mod+Mod)%Mod; 75 } 76 int main() 77 { 78 scanf("%lld",&Kase); 79 Rev=Quick_Pow(30,Mod-2); 80 Make_Prime(); 81 82 for (LL kase=1;kase<=Kase;kase++) 83 { 84 scanf("%lld",&n); 85 printf("%lld ",Solve()); 86 } 87 return 0; 88 }
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)..)-..