HDU 6706 huntian oy
题意 求以下式子的值，T组数据各个字母满足1 ≤ n , a , b ≤109 ，a,b互质
思路：
卡常毒瘤题，出题人时限卡的非常紧，考场上推出来又T又WA
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn=1e6+10,mod=1e9+7; 5 const int inv2=500000004,inv3=333333336; 6 int phi[maxn],n,ans; 7 8 vector<int> p;bool np[maxn]; 9 map<int,int> Phi; 10 11 void init(){ 12 phi[1]=1; 13 for(int i=2;i<maxn;++i){ 14 if(!np[i]){ 15 p.push_back(i); 16 phi[i]=i-1; 17 } 18 for(int j=0;j<p.size()&&p[j]*i<maxn;++j){ 19 np[i*p[j]]=true; 20 if(i%p[j]==0){ 21 phi[i*p[j]]=phi[i]*p[j]; 22 break; 23 } 24 phi[i*p[j]]=phi[i]*(p[j]-1); 25 } 26 } 27 for(int i=1;i<maxn;++i) 28 phi[i]=(1ll*phi[i]*i%mod+phi[i-1])%mod; 29 } 30 int read(){ 31 int x=0;char c=getchar(); 32 for(;!isdigit(c);c=getchar()); 33 for(;isdigit(c);c=getchar())x=(x<<3)+(x<<1)+(c^48); 34 return x; 35 } 36 void Write(int x){ 37 if(x<=0)return; 38 Write(x/10);putchar(x%10+'0'); 39 } 40 void write(int x){ 41 if(x==0)putchar('0'); 42 else Write(x); 43 } 44 int Sum2(int n){ 45 int ret=n; 46 ret=(ll)ret*inv2%mod; 47 ret=(ll)ret*(n+1)%mod; 48 return ret; 49 } 50 int Sum3(int n){ 51 int ret=Sum2(n),tmp=(2ll*n+1)%mod; 52 ret=(ll)ret*inv3%mod; 53 ret=(ll)ret*tmp%mod; 54 return ret; 55 } 56 int phii(int n){ 57 if(n<maxn)return phi[n]; 58 if(Phi.count(n))return Phi[n]; 59 int ret=Sum3(n); 60 for(int i=2,r,tmp;i<=n;i=r+1){ 61 r=min(n,n/(n/i)); 62 tmp=(ll)phii(n/i)*(Sum2(r)-Sum2(i-1))%mod; 63 ret-=tmp; 64 if(ret<0)ret+=mod; 65 if(ret>=mod)ret-=mod; 66 } 67 return Phi[n]=ret; 68 } 69 void solve(){ 70 n=read();read();read(); 71 Phi.clear(); 72 ans=(phii(n)-1+mod)%mod; 73 ans=(ll)inv2*ans%mod; 74 write(ans);puts(""); 75 } 76 int main(){ 77 init(); 78 int T;scanf("%d",&T); 79 for(;T--;)solve(); 80 return 0; 81 }
